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PROFESSOR: Ladies and gentlemen,
welcome to this
00:00:23.500 --> 00:00:26.070
lecture on nonlinear finite
element analysis of solids and
00:00:26.070 --> 00:00:27.610
structures.
00:00:27.610 --> 00:00:29.800
In the previous lectures,
we discussed the general
00:00:29.800 --> 00:00:32.950
continuum mechanics formulations
that we use in
00:00:32.950 --> 00:00:35.200
nonlinear finite element
analysis.
00:00:35.200 --> 00:00:37.910
And we also derived some
element matrices.
00:00:37.910 --> 00:00:39.770
In this lecture, I'd like
to discuss this
00:00:39.770 --> 00:00:41.840
you the truss element.
00:00:41.840 --> 00:00:45.650
The truss element is a very
interesting element and a very
00:00:45.650 --> 00:00:46.415
important element.
00:00:46.415 --> 00:00:49.890
It is important because it's
used in the analysis of many
00:00:49.890 --> 00:00:52.890
truss structures and also
in the analysis of cable
00:00:52.890 --> 00:00:53.940
structures.
00:00:53.940 --> 00:00:57.080
It's an interesting element to
study from a theoretical point
00:00:57.080 --> 00:01:01.230
of view because we can use the
general continuum mechanics
00:01:01.230 --> 00:01:06.700
equations analytically and
derive directly the finite
00:01:06.700 --> 00:01:09.660
element equations and the finite
matrices corresponding
00:01:09.660 --> 00:01:11.650
to truss element.
00:01:11.650 --> 00:01:14.700
These are derived
in closed form.
00:01:14.700 --> 00:01:17.550
We can study these matrices
and get some insight, some
00:01:17.550 --> 00:01:20.850
physical insight, into what
really these individual terms
00:01:20.850 --> 00:01:24.010
in the continuum mechanics
equations mean.
00:01:24.010 --> 00:01:26.370
We want to study in this lecture
the updated Lagrangian
00:01:26.370 --> 00:01:27.860
formulation.
00:01:27.860 --> 00:01:31.010
And in the next lecture, the
total Lagrangian formulation.
00:01:31.010 --> 00:01:34.480
I mentioned earlier in a
lecture that these two
00:01:34.480 --> 00:01:38.870
formulations really reduce to
exactly the same matrices
00:01:38.870 --> 00:01:39.620
provided certain
00:01:39.620 --> 00:01:41.710
transformations routes are followed.
00:01:41.710 --> 00:01:45.580
And for the truss element, we
can actually very nicely
00:01:45.580 --> 00:01:48.890
analytically demonstrate
that all is true.
00:01:48.890 --> 00:01:51.280
So let us study now the updated
Lagrangian formulation
00:01:51.280 --> 00:01:52.730
and later on the total
Lagrangian
00:01:52.730 --> 00:01:55.310
formulation of the element.
00:01:55.310 --> 00:01:58.750
The assumptions that we are
using for the truss element
00:01:58.750 --> 00:02:01.440
are that the stresses are
transmitted only in the
00:02:01.440 --> 00:02:04.620
direction normal to
the cross section.
00:02:04.620 --> 00:02:07.520
The stress is constant over
the cross section.
00:02:07.520 --> 00:02:10.190
And the cross sectional area
remains constant doing
00:02:10.190 --> 00:02:12.450
deformations.
00:02:12.450 --> 00:02:15.820
Of course, these first two
assumptions here are those
00:02:15.820 --> 00:02:18.660
that we are also using
in linear analysis.
00:02:18.660 --> 00:02:21.740
This we add now as an
assumption, meaning that we
00:02:21.740 --> 00:02:24.840
really only consider small
strain problems.
00:02:24.840 --> 00:02:27.620
We will consider large rotation,
large displacement,
00:02:27.620 --> 00:02:30.740
but only small strain
problems.
00:02:30.740 --> 00:02:33.780
Here we have a typical
element.
00:02:33.780 --> 00:02:40.710
We align that to another truss
element with the x1 in its
00:02:40.710 --> 00:02:42.470
original configuration.
00:02:42.470 --> 00:02:45.420
And the elemental goes through
large deformations as you can
00:02:45.420 --> 00:02:48.680
see and the large
rotation, theta.
00:02:48.680 --> 00:02:52.950
This is node 1, originally
here, moves there.
00:02:52.950 --> 00:02:55.810
This node 2, originally
here, moves there.
00:02:55.810 --> 00:02:59.050
Notice that the length of the
element is L in its original
00:02:59.050 --> 00:03:02.010
configuration and it
is also L in the
00:03:02.010 --> 00:03:04.350
configuration at time t.
00:03:04.350 --> 00:03:09.840
We use Young's modulus, an
elastic material, in other
00:03:09.840 --> 00:03:12.620
words, we assume for
the element.
00:03:12.620 --> 00:03:17.750
The cross sectional area is A.
And once again, we also use
00:03:17.750 --> 00:03:22.660
the fact that the element lies
in the x1, x2 space.
00:03:22.660 --> 00:03:26.815
The same kind of deviation that
I'm now following through
00:03:26.815 --> 00:03:28.940
could, of course, also be
generalized to the three
00:03:28.940 --> 00:03:31.460
dimensional case, in other
words were you have three
00:03:31.460 --> 00:03:33.730
axes, x3 being included.
00:03:33.730 --> 00:03:36.840
But all of the relevant
information, particularly the
00:03:36.840 --> 00:03:40.190
physical insight that we want to
get from this deviation, is
00:03:40.190 --> 00:03:43.520
very well obtained by just
considering element lying in
00:03:43.520 --> 00:03:48.380
the x1, x2 plane and moving
in that plane.
00:03:48.380 --> 00:03:51.960
If we now look at the
deformations of the element,
00:03:51.960 --> 00:03:55.720
we can see that at times 0
it was here once again.
00:03:55.720 --> 00:04:00.960
The displacement off node 2
is into the x1 direction
00:04:00.960 --> 00:04:06.490
signified by this symbol here,
time t, u, into the 1
00:04:06.490 --> 00:04:09.220
direction at node 2.
00:04:09.220 --> 00:04:13.320
Into the 2 direction,
we have u2.
00:04:13.320 --> 00:04:17.180
The lower 2 means 2 direction,
2 of course being upwards.
00:04:17.180 --> 00:04:19.610
The upper 2 means node 2.
00:04:19.610 --> 00:04:22.079
The t, once again,
means at time t.
00:04:24.850 --> 00:04:30.450
If you look at the left node
here, node 1, that one moves
00:04:30.450 --> 00:04:32.250
this way and that way.
00:04:32.250 --> 00:04:35.000
We use, of course, a
similar symbolism
00:04:35.000 --> 00:04:37.440
to denote this movement.
00:04:37.440 --> 00:04:41.990
So this is the deformation from
time 0 to times t given
00:04:41.990 --> 00:04:45.250
by these nodal point
displacements.
00:04:45.250 --> 00:04:51.530
Now from time t to time t plus
delta t, we obtain am addition
00:04:51.530 --> 00:04:52.410
deformation.
00:04:52.410 --> 00:04:55.990
And that additional deformation
is given by these
00:04:55.990 --> 00:05:02.870
displacements, at node 2 here
and at node 1 here.
00:05:02.870 --> 00:05:06.000
Notice that, in other words,
from time t to times t plus
00:05:06.000 --> 00:05:10.140
delta t, the truss has moved
from the red configuration to
00:05:10.140 --> 00:05:12.410
the blue configuration.
00:05:12.410 --> 00:05:15.740
Of course, note that these
displacements are measured in
00:05:15.740 --> 00:05:17.690
the stationary coordinate
frame.
00:05:17.690 --> 00:05:21.690
I made a big point in the
earlier lectures out of the
00:05:21.690 --> 00:05:24.980
fact that the coordinate frame
remains stationary and the
00:05:24.980 --> 00:05:28.760
elements move through the
coordinate frame.
00:05:28.760 --> 00:05:31.000
Of course, what we're interested
in doing is to
00:05:31.000 --> 00:05:33.720
calculate, in the finite
element solution, these
00:05:33.720 --> 00:05:36.720
incremental displacements.
00:05:36.720 --> 00:05:38.180
We assume that the
configuration
00:05:38.180 --> 00:05:39.780
at time t is known.
00:05:39.780 --> 00:05:42.760
We want to calculate the
new configuration.
00:05:42.760 --> 00:05:45.040
That means we want to calculate
these incremental
00:05:45.040 --> 00:05:46.810
displacements.
00:05:46.810 --> 00:05:50.680
And we achieve that by setting
up the appropriate matrices
00:05:50.680 --> 00:05:52.860
for the element.
00:05:52.860 --> 00:05:57.930
Well to develop the appropriate
matrices for the
00:05:57.930 --> 00:06:00.290
updated Lagrangian formulation
which I would like to discuss
00:06:00.290 --> 00:06:03.760
in this lecture, it is best
to introduce an auxiliary
00:06:03.760 --> 00:06:04.630
coordinate frame.
00:06:04.630 --> 00:06:08.540
And that auxiliary coordinate
frame is one that is aligned
00:06:08.540 --> 00:06:10.950
with one axis along
the element.
00:06:10.950 --> 00:06:15.780
We introduce x1 curl
and x2 curl.
00:06:15.780 --> 00:06:20.870
The curl always meaning body
aligned coordinate frame.
00:06:20.870 --> 00:06:23.990
Notice that this is now the
rotation here that the element
00:06:23.990 --> 00:06:25.250
has undergone.
00:06:25.250 --> 00:06:28.030
And of course, we really want to
get the stiffness matrix in
00:06:28.030 --> 00:06:31.640
the x1, x2 frame.
00:06:31.640 --> 00:06:34.700
x2, of course, being
perpendicular to x1, not shown
00:06:34.700 --> 00:06:37.500
on this viewgraph, but shown
on the earlier viewgraphs.
00:06:37.500 --> 00:06:41.310
We know that if we have
calculated the matrices in
00:06:41.310 --> 00:06:45.100
this curl coordinate frame, the
body attached coordinate
00:06:45.100 --> 00:06:48.270
frame, we very easily can
transform to the stationary
00:06:48.270 --> 00:06:49.810
coordinate frame.
00:06:49.810 --> 00:06:53.830
So let us now concentrate on
finding the K-matrix, the
00:06:53.830 --> 00:06:56.540
F-vector, corresponding
to this body
00:06:56.540 --> 00:06:57.790
attached coordinate frame.
00:07:00.560 --> 00:07:04.370
To do so, we look back to our
continuum mechanics equations
00:07:04.370 --> 00:07:06.910
that we developed in
earlier lectures.
00:07:06.910 --> 00:07:10.400
Here we have the basic continuum
mechanics equation,
00:07:10.400 --> 00:07:13.160
the principal of virtual work
written for the update
00:07:13.160 --> 00:07:16.910
Lagrangian formulation in the
curled coordinate frame, see
00:07:16.910 --> 00:07:19.370
that curl there.
00:07:19.370 --> 00:07:23.000
And this was the starting point
for the development off
00:07:23.000 --> 00:07:26.400
all of the equations that we're
using in the update
00:07:26.400 --> 00:07:27.490
Lagrangian formulation.
00:07:27.490 --> 00:07:32.150
The linearization resided
in this equation.
00:07:32.150 --> 00:07:35.930
And we talked about this
equation quite a bit.
00:07:35.930 --> 00:07:39.330
Except, of course, we did not
have to curls there because we
00:07:39.330 --> 00:07:45.110
were talking about the x1, x2,
x3 uncurled coordinate frame.
00:07:45.110 --> 00:07:47.500
Now we're talking about this
curled coordinate frame, so I
00:07:47.500 --> 00:07:49.620
simply put a curl on there.
00:07:49.620 --> 00:07:54.510
But otherwise, the terms
are identical.
00:07:54.510 --> 00:07:58.040
Because for the truss as I
state earlier the only
00:07:58.040 --> 00:08:02.690
non-zero stress is the stress
along the length of the truss
00:08:02.690 --> 00:08:05.670
which acts normally to the cross
section of the truss, we
00:08:05.670 --> 00:08:09.890
can simplify this general
equation to this equation.
00:08:09.890 --> 00:08:14.670
The only non-zero stress
is tau curl 1 1.
00:08:14.670 --> 00:08:19.050
The only strain that we call
stress, only small strain that
00:08:19.050 --> 00:08:22.710
we call stress, is e curl 1 1.
00:08:22.710 --> 00:08:25.070
Of course a t on the left-hand
side because it's referred to
00:08:25.070 --> 00:08:27.560
as the configuration
at time t.
00:08:27.560 --> 00:08:31.450
So this very general equation
reduces to a much simpler
00:08:31.450 --> 00:08:34.870
equation already
as shown here.
00:08:34.870 --> 00:08:39.429
What we are out now to do is to
evaluate these quantities
00:08:39.429 --> 00:08:43.590
here and there using the finite
element interpolation.
00:08:46.110 --> 00:08:50.630
First let us now look at what
are some of the terms that we
00:08:50.630 --> 00:08:52.470
easily can identify.
00:08:52.470 --> 00:08:55.470
We identify this tensile
term here to be
00:08:55.470 --> 00:08:57.470
simply Young's modulus.
00:08:57.470 --> 00:09:01.300
This stress term to be simply
the force in the truss divided
00:09:01.300 --> 00:09:03.010
by the cross sectional area.
00:09:03.010 --> 00:09:05.010
The volume of the truss
is given here.
00:09:05.010 --> 00:09:07.850
Notice, once again, the length
of the truss is assumed to be
00:09:07.850 --> 00:09:10.500
constant, therefore we only
consider small strain
00:09:10.500 --> 00:09:13.400
conditions, I mentioned
that earlier.
00:09:13.400 --> 00:09:17.060
If we now use this information
in the general equation, we
00:09:17.060 --> 00:09:19.220
directly arrive at this
equation here.
00:09:21.800 --> 00:09:24.450
You simply substitute and you
will see immediately that
00:09:24.450 --> 00:09:25.740
these are the terms.
00:09:25.740 --> 00:09:29.060
Of course, what we now have to
evaluate are these curled
00:09:29.060 --> 00:09:32.240
terms here.
00:09:32.240 --> 00:09:37.820
To proceed, we identify that
the e curl 1 1, the linear
00:09:37.820 --> 00:09:40.900
strain term, is simply given
by the linear strain
00:09:40.900 --> 00:09:44.230
displacement matrix with a curl
on it times the nodal
00:09:44.230 --> 00:09:46.330
point displacement vector.
00:09:46.330 --> 00:09:49.130
These are the nodal point
displacements.
00:09:49.130 --> 00:09:50.910
There are four such
displacements because we have
00:09:50.910 --> 00:09:54.540
two nodes and two displacements
per node.
00:09:54.540 --> 00:09:59.020
Notice that I have a curl here
signifying that we're talking
00:09:59.020 --> 00:10:02.150
about the curled coordinate
frame.
00:10:02.150 --> 00:10:05.930
And there's a hat on top off
that curl as you can see here.
00:10:05.930 --> 00:10:09.900
That hat means these are
discrete nodal point
00:10:09.900 --> 00:10:12.180
displacements.
00:10:12.180 --> 00:10:16.480
This term here is evaluated
via that term.
00:10:16.480 --> 00:10:21.900
Notice that we construct the
BNL in such a way that this
00:10:21.900 --> 00:10:25.570
right-hand side is equal
to that left-hand side.
00:10:25.570 --> 00:10:27.990
I mentioned that also in the
earlier lectures for the two
00:10:27.990 --> 00:10:29.610
dimensional and the
three dimensional
00:10:29.610 --> 00:10:31.560
finite element cases.
00:10:31.560 --> 00:10:35.950
Well the vector of nodal point
displacements, this vector
00:10:35.950 --> 00:10:38.910
here, is listed out here.
00:10:38.910 --> 00:10:41.350
Notice it contains the four
displacements that I referred
00:10:41.350 --> 00:10:42.260
to earlier.
00:10:42.260 --> 00:10:45.890
And these are the displacements,
u curl
00:10:45.890 --> 00:10:49.470
1 1, u curl 2 1.
00:10:49.470 --> 00:10:54.610
This upper 1 always denoting
nodal point, the lower 1 and 2
00:10:54.610 --> 00:10:56.630
meaning coordinate directions.
00:10:56.630 --> 00:10:58.175
Similarly for these two terms.
00:11:00.880 --> 00:11:05.540
To evaluate now these terms,
we recognize that the total
00:11:05.540 --> 00:11:08.520
strain, the total incremental
Green-Lagrange strain I should
00:11:08.520 --> 00:11:12.080
say, is given by this
relationship here from our
00:11:12.080 --> 00:11:14.990
general continuum mechanics
equation.
00:11:14.990 --> 00:11:17.970
The linear strain term
is given here.
00:11:17.970 --> 00:11:20.840
And the nonlinear strain
term is the rest.
00:11:20.840 --> 00:11:24.750
The rest meaning taking this
total, subtracting the linear
00:11:24.750 --> 00:11:27.460
one, you're left
with that one.
00:11:27.460 --> 00:11:29.640
And once again, these
expressions are directly
00:11:29.640 --> 00:11:33.490
obtained by just taking the
general continuum mechanics
00:11:33.490 --> 00:11:37.670
equations and varying the
indices the way you want to
00:11:37.670 --> 00:11:38.670
see them varied.
00:11:38.670 --> 00:11:43.310
1 1, for i and j, if you
would refer back.
00:11:43.310 --> 00:11:48.060
And k, the k that used earlier
goes over 1 and 2 in this
00:11:48.060 --> 00:11:49.260
particular case.
00:11:49.260 --> 00:11:52.050
In a three dimensional case,
of course, you would have k
00:11:52.050 --> 00:11:54.500
also going to 3.
00:11:54.500 --> 00:11:57.590
The variation on this term
resides directly in this
00:11:57.590 --> 00:11:59.170
expression.
00:11:59.170 --> 00:12:02.720
And we notice that this
expression can also be written
00:12:02.720 --> 00:12:05.020
in matrix form.
00:12:05.020 --> 00:12:08.840
One row vector times
one column vector.
00:12:08.840 --> 00:12:13.060
It is convenient to work now
with this matrix form because
00:12:13.060 --> 00:12:16.360
we want to bring it into a
form BNL transposed BNL.
00:12:19.010 --> 00:12:25.670
Before doing so, let us identify
one important point,
00:12:25.670 --> 00:12:28.510
namely that the displacement
derivatives are
00:12:28.510 --> 00:12:30.170
constant along the truss.
00:12:30.170 --> 00:12:34.310
The reason for it is that we
have only two nodes and these
00:12:34.310 --> 00:12:37.710
two notes can only specify
a linear variation and
00:12:37.710 --> 00:12:40.250
displacement between the two
nodes, meaning a linear
00:12:40.250 --> 00:12:44.380
variation along the truss and
the displacement derivatives
00:12:44.380 --> 00:12:46.050
are therefore constant.
00:12:46.050 --> 00:12:49.700
For example, this displacement
derivative is simply obtained
00:12:49.700 --> 00:12:52.730
by taking the difference in the
nodal point displacements,
00:12:52.730 --> 00:12:55.630
of course, in direction 1
because we're talking about 1
00:12:55.630 --> 00:12:59.220
here, over the length L.
00:12:59.220 --> 00:13:06.930
So this is directly obtained
as u curl 1 comma 1.
00:13:06.930 --> 00:13:10.970
Similarly, we also evaluate u
curl 2, 1 and you see the
00:13:10.970 --> 00:13:15.010
lower part here gives
us that expression.
00:13:15.010 --> 00:13:17.920
Of course, the upper row here
is nothing else than the
00:13:17.920 --> 00:13:20.610
rewriting of that equation.
00:13:20.610 --> 00:13:25.580
If you rewrite this equation in
matrix form, you directly
00:13:25.580 --> 00:13:28.650
obtain this relationship.
00:13:28.650 --> 00:13:32.120
Or another check would be simply
takes this vector,
00:13:32.120 --> 00:13:37.240
multiply this matrix by that
vector, and the upper part of
00:13:37.240 --> 00:13:40.730
that multiplication will
directly be this result.
00:13:43.580 --> 00:13:46.640
If we now use this information,
we directly
00:13:46.640 --> 00:13:52.130
extract the BL matrix, which of
course links up the linear
00:13:52.130 --> 00:13:56.240
strain increment with the nodal
point displacements.
00:13:56.240 --> 00:13:58.450
Here there is a small error.
00:13:58.450 --> 00:13:59.490
Let me just point it out.
00:13:59.490 --> 00:14:06.200
This bracket, of course, should
go to there because the
00:14:06.200 --> 00:14:09.530
B matrix does not contain
the nodal point
00:14:09.530 --> 00:14:10.680
displacement vector.
00:14:10.680 --> 00:14:13.470
It only goes up to there.
00:14:13.470 --> 00:14:17.860
Because the e curl 1 1 is equal
to B matrix times the
00:14:17.860 --> 00:14:20.250
nodal point displacement
vector.
00:14:20.250 --> 00:14:22.210
Similarly we can write
now the variation in
00:14:22.210 --> 00:14:24.760
eta 1 1 in this form.
00:14:24.760 --> 00:14:29.780
Notice that this part here
captures this amount and this
00:14:29.780 --> 00:14:32.910
part here captures
that amount.
00:14:32.910 --> 00:14:35.500
This one, of course, we had
already derived earlier.
00:14:35.500 --> 00:14:38.710
We just plug it in now.
00:14:38.710 --> 00:14:40.810
With this information
we can not directly
00:14:40.810 --> 00:14:42.890
develop the k matrices.
00:14:42.890 --> 00:14:46.850
We notice that the linear strain
stiffness matrix is
00:14:46.850 --> 00:14:49.140
obtained from this expression.
00:14:49.140 --> 00:14:50.530
And here you have it.
00:14:50.530 --> 00:14:54.320
You simply substitute for these
terms and obtain this
00:14:54.320 --> 00:14:56.890
matrix, much in the
same way as we are
00:14:56.890 --> 00:14:58.140
dealing in linear analysis.
00:15:04.040 --> 00:15:06.360
The nonlinear strain stiffness
matrix, that's the one we want
00:15:06.360 --> 00:15:10.870
to look at next, is obtained
from this relationship.
00:15:10.870 --> 00:15:14.540
And we simply now substitute
what we have derived for this
00:15:14.540 --> 00:15:19.180
term and directly obtain
this expression here.
00:15:19.180 --> 00:15:23.020
And what's under this blue
bracket, of course, is our
00:15:23.020 --> 00:15:24.570
nonlinear strain stiffness
matrix.
00:15:28.000 --> 00:15:32.770
Finally, the force vector is
obtained from this expression.
00:15:32.770 --> 00:15:36.960
And you, once again, simply
plug in and obtain this
00:15:36.960 --> 00:15:40.210
expression here where what's
under the blue
00:15:40.210 --> 00:15:42.480
is the force vector.
00:15:42.480 --> 00:15:46.430
The beauty is that we have
started with a fairly
00:15:46.430 --> 00:15:50.170
complicated continuum
mechanics equation.
00:15:50.170 --> 00:15:54.430
We have specialized it to the
truss element, recognizing
00:15:54.430 --> 00:15:57.460
that only certain terms really
need to be included or are
00:15:57.460 --> 00:15:59.730
included in this particular
case.
00:15:59.730 --> 00:16:03.280
We have evaluated them in a
fairly simple, straightforward
00:16:03.280 --> 00:16:07.980
manner, plugged in, and obtained
now matrices that can
00:16:07.980 --> 00:16:11.310
of course be analytically
evaluated the way that they
00:16:11.310 --> 00:16:12.430
have it done.
00:16:12.430 --> 00:16:16.050
And now we can in fact even
think about these matrices and
00:16:16.050 --> 00:16:18.790
try to physically interpret
their meaning.
00:16:18.790 --> 00:16:21.130
We will do that just now.
00:16:21.130 --> 00:16:24.990
However, before getting into
that, we want to get back to
00:16:24.990 --> 00:16:28.150
one point, namely that we have
sued so far a curled
00:16:28.150 --> 00:16:29.950
coordinate system.
00:16:29.950 --> 00:16:32.790
And as we mentioned earlier, we
will have to transform from
00:16:32.790 --> 00:16:35.780
that curled coordinate system
to the stationary global
00:16:35.780 --> 00:16:37.090
coordinate system.
00:16:37.090 --> 00:16:39.660
That is achieved by this
transformation.
00:16:39.660 --> 00:16:43.140
Notice that the curled
displacements are nothing else
00:16:43.140 --> 00:16:46.860
than transformation matrix
times the uncurled
00:16:46.860 --> 00:16:47.790
displacement.
00:16:47.790 --> 00:16:49.700
This transformation matrix
is very well
00:16:49.700 --> 00:16:52.550
known from linear analysis.
00:16:52.550 --> 00:16:54.650
It simply transforms
displacement from one
00:16:54.650 --> 00:16:57.750
coordinate system
into another.
00:16:57.750 --> 00:17:02.400
This is the one that holds
anywhere along the element for
00:17:02.400 --> 00:17:04.230
the continuous displacement.
00:17:04.230 --> 00:17:06.170
But it also holds
at the nodes.
00:17:06.170 --> 00:17:09.750
So if we apply that relationship
at the nodes, be
00:17:09.750 --> 00:17:13.000
directly obtain this
transformation here.
00:17:13.000 --> 00:17:16.619
The curled displacements at the
nodes are given in terms
00:17:16.619 --> 00:17:20.329
of the transformation matrix
times the uncurled
00:17:20.329 --> 00:17:22.010
displacement at the nodes.
00:17:22.010 --> 00:17:27.730
Of course, we want to get the
matrix expressions related to
00:17:27.730 --> 00:17:32.070
this uncurled displacement
vector.
00:17:32.070 --> 00:17:35.880
Well having obtained this
relationship, we go back to
00:17:35.880 --> 00:17:39.740
the basic equations, recognizing
that this part
00:17:39.740 --> 00:17:45.310
here is coming from the linear
strain stiffness matrix.
00:17:45.310 --> 00:17:50.500
We substitute for u curl,
the uncurled vectors
00:17:50.500 --> 00:17:51.780
with the T's in front.
00:17:51.780 --> 00:17:54.800
There is a transpose
appearing now here.
00:17:54.800 --> 00:17:57.320
This capital transpose has to
be there because there's a
00:17:57.320 --> 00:17:59.090
transpose there.
00:17:59.090 --> 00:18:04.700
And what's in here underlined
in red is the linear strain
00:18:04.700 --> 00:18:07.980
stiffness matrix of the element
corresponding to the
00:18:07.980 --> 00:18:11.330
stationery global coordinate
system.
00:18:11.330 --> 00:18:14.270
We proceed much the same way
for the nonlinear strain
00:18:14.270 --> 00:18:15.660
stiffness matrix.
00:18:15.660 --> 00:18:19.810
And what's in here now is what
we want to get, the nonlinear
00:18:19.810 --> 00:18:22.530
strain stiffness matrix
corresponding to the global
00:18:22.530 --> 00:18:24.520
stationary coordinate frame.
00:18:24.520 --> 00:18:27.680
Same thing holds of course
for the F-vector.
00:18:27.680 --> 00:18:31.750
And here we have the F-vector
now in the uncurled global
00:18:31.750 --> 00:18:33.530
stationary coordinate frame.
00:18:33.530 --> 00:18:40.050
So it's really this part, that
part, and this part that we're
00:18:40.050 --> 00:18:43.570
using in our finite element
analysis when we assemble
00:18:43.570 --> 00:18:47.330
truss elements into a global
assemblage of truss elements.
00:18:50.500 --> 00:18:53.590
Performing the indicated matrix
multiplications gives
00:18:53.590 --> 00:18:55.350
this expression here.
00:18:55.350 --> 00:18:59.140
In fact, this is the same
stiffness matrix that we use
00:18:59.140 --> 00:19:04.670
in linear analysis when we
have an element that is
00:19:04.670 --> 00:19:09.280
oriented at an angle theta
to the global x-axis, the
00:19:09.280 --> 00:19:11.840
horizontal x-axis.
00:19:11.840 --> 00:19:16.030
So you might have very well
seen this matrix before.
00:19:16.030 --> 00:19:21.060
The nonlinear strain stiffness
matrix looks this way.
00:19:21.060 --> 00:19:26.120
Notice there is a tP over
L and then these terms.
00:19:26.120 --> 00:19:28.050
Of course, it's also
symmetric.
00:19:28.050 --> 00:19:30.710
And this is the matrix
corresponding to the global
00:19:30.710 --> 00:19:32.550
coordinate frame now.
00:19:32.550 --> 00:19:36.040
We notice immediately that this
matrix is in fact the
00:19:36.040 --> 00:19:39.240
same matrix that we already had
evaluated in the curved
00:19:39.240 --> 00:19:39.960
coordinate frame.
00:19:39.960 --> 00:19:42.550
We will get back to
that just now.
00:19:42.550 --> 00:19:46.610
The F-vector comes
out to be this.
00:19:46.610 --> 00:19:50.310
Let's now look at these
terms physically.
00:19:50.310 --> 00:19:53.730
Here we have a single truss
element pinned at the
00:19:53.730 --> 00:19:55.490
left-hand side.
00:19:55.490 --> 00:20:00.270
And a load R is applied to that
truss at the other end.
00:20:00.270 --> 00:20:04.490
Of course, this direction of
the load must be along the
00:20:04.490 --> 00:20:06.030
truss element.
00:20:06.030 --> 00:20:09.750
Or rather the truss element
aligns its direction so that
00:20:09.750 --> 00:20:13.160
it balances this load
that is applied.
00:20:13.160 --> 00:20:15.760
The internal force is tP.
00:20:15.760 --> 00:20:18.000
tP balances tR.
00:20:18.000 --> 00:20:21.080
And tP, of course, acts along
the truss element.
00:20:21.080 --> 00:20:24.500
Now if you look at this node
here, node 2, we notice
00:20:24.500 --> 00:20:28.600
immediately that this tR vector
can be decomposed.
00:20:28.600 --> 00:20:33.420
Or this tR can be decomposed
into tR cosine theta along
00:20:33.420 --> 00:20:37.240
this axis and tR sin theta
along that axis.
00:20:37.240 --> 00:20:42.240
P is acting here and P can also
be decomposed as shown.
00:20:42.240 --> 00:20:45.920
Now if we look here, we find
therefore that the tR vector
00:20:45.920 --> 00:20:49.200
corresponding to the global
coordinate system contains
00:20:49.200 --> 00:20:53.590
these two entries shown here.
00:20:53.590 --> 00:20:58.450
And the tF from our earlier
viewgraph at node 2 was indeed
00:20:58.450 --> 00:21:00.720
simply this part.
00:21:00.720 --> 00:21:06.750
We notice therefore that tR is
equal to tF as show here or tR
00:21:06.750 --> 00:21:08.640
minus ttF equal to 0.
00:21:08.640 --> 00:21:12.000
And that is, of course, what
has to hold when we satisfy
00:21:12.000 --> 00:21:15.440
equilibrium at that node.
00:21:15.440 --> 00:21:16.730
So we have a nice
00:21:16.730 --> 00:21:19.340
interpretation of this F-vector.
00:21:19.340 --> 00:21:22.610
And in fact, we can see that our
arithmetic of getting that
00:21:22.610 --> 00:21:24.580
F-vector has been correct.
00:21:27.470 --> 00:21:31.190
Let's look now at the KNL
matrix, the nonlinear strain
00:21:31.190 --> 00:21:34.450
stiffness matrix.
00:21:34.450 --> 00:21:40.430
We already pointed out earlier
that the KNL in the uncurled
00:21:40.430 --> 00:21:45.420
frame was equal to the KNL
in the curled frame.
00:21:45.420 --> 00:21:48.560
And we can ask ourself, of
course, why is that so?
00:21:48.560 --> 00:21:50.790
Well the reason is
the following.
00:21:50.790 --> 00:21:56.060
Let's look again at this truss
element pinned here and we
00:21:56.060 --> 00:22:02.330
look at node 2, then we see that
this vector here, which
00:22:02.330 --> 00:22:08.150
is by Pythagoras the component
or the resultant vector of
00:22:08.150 --> 00:22:14.240
these two displacements,
has this length.
00:22:14.240 --> 00:22:17.720
Now this length can be evaluated
as shown here.
00:22:20.530 --> 00:22:23.730
Because the derivatives are
constants, we discussed that
00:22:23.730 --> 00:22:29.020
earlier, we can write this
derivative squared plus that
00:22:29.020 --> 00:22:32.790
derivative squared, square root
out of it, times L, being
00:22:32.790 --> 00:22:34.830
equal to that.
00:22:34.830 --> 00:22:37.890
But then we recognize that
what we're seeing here is
00:22:37.890 --> 00:22:42.010
nothing else than that term
there, in other words, our
00:22:42.010 --> 00:22:45.090
nonlinear strain increment.
00:22:45.090 --> 00:22:50.740
Which in this particular case
we can see, for the given
00:22:50.740 --> 00:22:54.820
displacement is constant and
independent of the coordinate
00:22:54.820 --> 00:22:56.130
frame used.
00:22:56.130 --> 00:23:00.500
And that is really the reason
why the KNL matrix is constant
00:23:00.500 --> 00:23:05.180
for any coordinate frame
that we are using.
00:23:05.180 --> 00:23:08.610
Let us now try to understand
what the elements in the KNL
00:23:08.610 --> 00:23:11.900
matrix physically mean.
00:23:11.900 --> 00:23:17.140
They, in fact, give the change
in force, the required change
00:23:17.140 --> 00:23:20.360
in the externally applied
force, when
00:23:20.360 --> 00:23:22.380
the element is rotated.
00:23:22.380 --> 00:23:26.940
Here we have an element at time
t, once again fixed at
00:23:26.940 --> 00:23:32.110
the left end, and subjected
to a force tR.
00:23:32.110 --> 00:23:34.590
Of course, this force tR
is balanced by the
00:23:34.590 --> 00:23:37.440
internal force tP.
00:23:37.440 --> 00:23:40.640
Let us now impose a displacement
such that the
00:23:40.640 --> 00:23:45.230
element rotates about
this point.
00:23:45.230 --> 00:23:49.710
We reach the configuration
at time t plus delta t by
00:23:49.710 --> 00:23:52.340
imposing this displacement
here, which we
00:23:52.340 --> 00:23:55.340
denote u curl 2 2.
00:23:55.340 --> 00:23:58.690
Now in this configuration,
a new force has
00:23:58.690 --> 00:24:02.290
to act on this element.
00:24:02.290 --> 00:24:05.810
And this new force, denoted
as t plus delta
00:24:05.810 --> 00:24:11.450
tR, is shown here.
00:24:11.450 --> 00:24:14.740
Here we have the force
t plus delta tR.
00:24:14.740 --> 00:24:20.600
Notice it is aligned with the
red element at time t plus
00:24:20.600 --> 00:24:23.330
delta t now.
00:24:23.330 --> 00:24:27.440
The change in the force from
configuration t to t plus
00:24:27.440 --> 00:24:31.830
delta t is given
by this vector.
00:24:31.830 --> 00:24:36.170
And the element in that vector
can be calculated by taking
00:24:36.170 --> 00:24:40.430
moment equilibrium
about this point.
00:24:40.430 --> 00:24:43.900
We do just that right here.
00:24:43.900 --> 00:24:46.980
This is the equation of moment
equilibrium about the fixed
00:24:46.980 --> 00:24:48.540
point of the truss.
00:24:48.540 --> 00:24:53.870
Notice that, of course, this
displacement is small.
00:24:53.870 --> 00:24:58.090
And from this, we directly
obtain that delta R is given
00:24:58.090 --> 00:25:00.340
as shown here on the
right-hand side.
00:25:00.340 --> 00:25:04.200
And this is actually the entry
4, 4 of the KNL matrix.
00:25:06.960 --> 00:25:10.090
Now the same information is
also expressed here on the
00:25:10.090 --> 00:25:15.010
right-hand side where we have
written the delta R as a
00:25:15.010 --> 00:25:20.990
matrix product of KNL times u
hat being equal to t plus
00:25:20.990 --> 00:25:22.240
delta tR minus tR.
00:25:26.490 --> 00:25:29.770
So this completes our discussion
of the finite
00:25:29.770 --> 00:25:33.400
element matrices of the truss
element corresponding to the
00:25:33.400 --> 00:25:35.790
updated Lagrangian
formulation.
00:25:35.790 --> 00:25:38.330
In the next lecture, we will
actually discuss the total
00:25:38.330 --> 00:25:40.420
Lagrangian formulation
of the truss element.
00:25:40.420 --> 00:25:43.850
And we will find that
identically the same matrices
00:25:43.850 --> 00:25:47.350
are obtained as in the updated
Lagrangian formulation.
00:25:47.350 --> 00:25:49.610
That will be a very interesting
theoretical point,
00:25:49.610 --> 00:25:51.980
as well as practical point.
00:25:51.980 --> 00:25:55.930
Let us now look at an example
of using the truss element.
00:25:55.930 --> 00:25:59.410
And the example is a
simple one, and yet
00:25:59.410 --> 00:26:01.220
quite practical one.
00:26:01.220 --> 00:26:03.470
We will analyze a prestressed
cable.
00:26:03.470 --> 00:26:07.110
The cable has a length 2L.
00:26:07.110 --> 00:26:11.460
At the midpoint, a load
of 2 tR is supplied.
00:26:11.460 --> 00:26:14.010
The Young's modulus of the
cable is E, the cross
00:26:14.010 --> 00:26:17.800
sectional area of
the cable is A.
00:26:17.800 --> 00:26:20.920
We can model one half of the
cable because of symmetry
00:26:20.920 --> 00:26:23.290
conditions as shown down here.
00:26:23.290 --> 00:26:25.930
Notice that, of course, we
assume the transverse
00:26:25.930 --> 00:26:29.470
displacement to be small, the
angle therefore to be small,
00:26:29.470 --> 00:26:32.690
because we assume that the
length of the cable element
00:26:32.690 --> 00:26:35.230
remains the same.
00:26:35.230 --> 00:26:38.450
In other words, it does not
change from time zero to time
00:26:38.450 --> 00:26:41.520
t to time t plus delta
t and so on.
00:26:41.520 --> 00:26:46.140
Using the U.L. Formation, we
obtain directly from the
00:26:46.140 --> 00:26:49.080
general matrices that we
discussed us now, this
00:26:49.080 --> 00:26:51.610
equation here.
00:26:51.610 --> 00:26:55.280
Which gives the tangent
stiffness matrix times the
00:26:55.280 --> 00:26:59.090
displacement increment equal to
the externally applied load
00:26:59.090 --> 00:27:03.650
minus the nodal point force
corresponding to the internal
00:27:03.650 --> 00:27:07.300
element stress, the internal
element force tP, of course,
00:27:07.300 --> 00:27:09.420
in this case.
00:27:09.420 --> 00:27:13.250
Notice that this is the
equilibrium equation that
00:27:13.250 --> 00:27:17.590
carries us from time t to
time t plus delta t.
00:27:17.590 --> 00:27:23.190
We would iterate on this, of
course, introducing an
00:27:23.190 --> 00:27:26.490
iteration counter here on
the right-hand side.
00:27:26.490 --> 00:27:29.670
We have discussed all that in
earlier lectures to obtain an
00:27:29.670 --> 00:27:34.100
accurate solution for time t
plus delta t, but these are
00:27:34.100 --> 00:27:36.490
some details that we don't
need to look at now.
00:27:36.490 --> 00:27:40.400
For the moment, let's just look
at this equation here, in
00:27:40.400 --> 00:27:46.370
which we imply basically a
simple step by step forward
00:27:46.370 --> 00:27:50.090
incrementation of load
without iteration.
00:27:50.090 --> 00:27:53.090
In practice once again, we would
actually iterate to make
00:27:53.090 --> 00:27:56.010
sure that we are satisfying
equilibrium at the end
00:27:56.010 --> 00:27:58.150
of each time step.
00:27:58.150 --> 00:28:01.490
Of particular interest is the
configuration at time 0.
00:28:01.490 --> 00:28:04.770
And there we recognized that
the linear strain stiffness
00:28:04.770 --> 00:28:07.060
matrix up here is 0.
00:28:07.060 --> 00:28:10.520
And that's all we have in terms
of stiffness is the
00:28:10.520 --> 00:28:12.860
nonlinear strain stiffness.
00:28:12.860 --> 00:28:17.430
And that is expressed in
this equation here.
00:28:17.430 --> 00:28:23.230
Notice that therefore if P0 is
equal to 0, in other words,
00:28:23.230 --> 00:28:26.060
there's no prestress in the
cable, then initially it has
00:28:26.060 --> 00:28:27.310
no stiffness.
00:28:29.710 --> 00:28:34.880
Of course, this element here,
let's look at it here at time
00:28:34.880 --> 00:28:39.840
t, this element here will
increase ad the deformations
00:28:39.840 --> 00:28:43.400
increase because tP
will increase.
00:28:43.400 --> 00:28:46.290
And this element here will
also increase as the
00:28:46.290 --> 00:28:47.540
deformations increase.
00:28:51.010 --> 00:28:54.770
And in fact, this is expressed
once more here.
00:28:54.770 --> 00:28:57.930
This total matrix will in
fact be increasing.
00:28:57.930 --> 00:29:00.900
We say the cable stiffens
and the load is applied.
00:29:04.440 --> 00:29:09.030
If theta becomes very large, in
fact if as theta goes to 90
00:29:09.030 --> 00:29:13.540
degrees, the stiffness becomes
quite large and the stiffness
00:29:13.540 --> 00:29:19.000
approaches as theta goes to 90
degrees EA over L. Now this,
00:29:19.000 --> 00:29:22.800
of course, is rather theoretical
because we assume
00:29:22.800 --> 00:29:25.880
in our formulation that the
length remains constant and
00:29:25.880 --> 00:29:27.950
therefore that the deformations
are small, as I
00:29:27.950 --> 00:29:30.830
pointed out earlier.
00:29:30.830 --> 00:29:35.130
Let us now look at the actual
load displacement curve that
00:29:35.130 --> 00:29:40.100
is calculated for this
particular truss structure,
00:29:40.100 --> 00:29:42.520
which is really a cable
structure, we could look also
00:29:42.520 --> 00:29:45.620
at it as a trust structure
because it is made up of two
00:29:45.620 --> 00:29:47.750
truss elements.
00:29:47.750 --> 00:29:51.070
Here we have the deflection
plotted.
00:29:51.070 --> 00:29:54.060
And here we have the applied
force plotted for
00:29:54.060 --> 00:29:55.310
a particular case.
00:29:57.810 --> 00:30:03.140
Notice that L is 120 inches,
A is 1 square inch.
00:30:03.140 --> 00:30:05.550
Notice that the maximum
displacement we're looking at
00:30:05.550 --> 00:30:08.460
here is about 2 inches.
00:30:08.460 --> 00:30:12.720
So 2 inches over the length, 120
inches, means really small
00:30:12.720 --> 00:30:13.970
deformations.
00:30:15.420 --> 00:30:20.290
But notice the stiffening effect
shown by the increase
00:30:20.290 --> 00:30:24.350
in the slope of this curve.
00:30:24.350 --> 00:30:28.440
We can actually show the
stiffness matrix components as
00:30:28.440 --> 00:30:29.790
a function of the
applied load.
00:30:29.790 --> 00:30:32.420
And this is a very interesting
viewgraph.
00:30:32.420 --> 00:30:39.510
Here we see that at 0 load,
the only stiffness that is
00:30:39.510 --> 00:30:42.110
there is given by KNL.
00:30:42.110 --> 00:30:45.270
We pointed that out
already earlier.
00:30:45.270 --> 00:30:48.050
KL is equal to 0.
00:30:48.050 --> 00:30:53.290
As the load increases,
KNL increases as
00:30:53.290 --> 00:30:56.220
shown by the blue curve.
00:30:56.220 --> 00:31:02.180
As a load increases, also KL
increases as shown here.
00:31:02.180 --> 00:31:05.210
And the sum of these two curves,
of course, gives us
00:31:05.210 --> 00:31:07.430
the red curve.
00:31:07.430 --> 00:31:13.330
And the red curve is the total
tangent stiffness matrix.
00:31:13.330 --> 00:31:18.840
At any particular applied force,
we can directly read
00:31:18.840 --> 00:31:21.560
off the stiffness
corresponding to
00:31:21.560 --> 00:31:22.640
that applied force.
00:31:22.640 --> 00:31:26.330
Or, more naturally, we would
of course look for the
00:31:26.330 --> 00:31:27.620
corresponding displacement.
00:31:27.620 --> 00:31:30.570
In other words, at a particular
applied force, we
00:31:30.570 --> 00:31:32.560
have a corresponding
displacement.
00:31:32.560 --> 00:31:35.890
And for that corresponding
displacement, we
00:31:35.890 --> 00:31:38.570
would have a stiffness.
00:31:38.570 --> 00:31:41.910
Of course, these curves are
directly obtained from the
00:31:41.910 --> 00:31:45.090
expression that I just
showed you earlier.
00:31:45.090 --> 00:31:47.710
Well, this brings us to the
end of this lecture.
00:31:47.710 --> 00:31:50.470
In the next lecture then, as I
pointed out already, I'd like
00:31:50.470 --> 00:31:53.670
to look with you at the total
Lagrangian formulation of the
00:31:53.670 --> 00:31:55.140
truss element.
00:31:55.140 --> 00:31:59.170
And as I mentioned earlier, we
will find something very
00:31:59.170 --> 00:32:03.080
interesting, namely that
reducing all of these complex
00:32:03.080 --> 00:32:06.650
continuum mechanics equations
down to what we really need
00:32:06.650 --> 00:32:10.570
for the truss element, we will
find that the same matrices in
00:32:10.570 --> 00:32:13.790
a total Lagrangian formulation
are obtained as in the updated
00:32:13.790 --> 00:32:16.140
Lagrangian formulation.
00:32:16.140 --> 00:32:17.470
Thank you very much for
your attention.