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PROFESSOR: Ladies and gentlemen, welcome to this
lecture on nonlinear finite element analysis of solids and
structures.
In this lecture, I would like to continue with our
discussion of the use of constitutive relations used in
nonlinear analysis.
In the previous lecture, we already considered the
modeling of elastic and hyper elastic materials, subjected
to large displacements, large rotations, and large strains.
I'd like to not turn our attention to the use of the
updated Lagrangian formulation.
Of course, the kinematic relations for the total and
updated Lagrangian formulations, we discussed in
an earlier lecture already.
In particular, I would like now, in very general terms, to
ask the following question--
is it possible to obtain, using the updated Lagrangian
formulation, identically the same numerical results for
each load step and each iteration, as are obtained
when we use a total Lagrangian formulation?
Let's look at this question here, now, in
quite gentle terms.
And I have rephrased it here once more, so as to really
make very clear what I mean.
Let's say that we have a Program 1 that uses only the
total Lagrangian formulation.
And the constitutive relations that we are employing in that
total Lagrangian formulation are defined as shown here.
So total stress at time, t, of course, second Piola-Kirchhoff
stress, is given as a function of displacements.
And the tangent material relationship, relating the
increment in the second Piola-Kirchhoff stress and the
increment in the Green-Lagrange strain is as
shown here.
The program results for a particular physical problem
might look systematically as shown here.
Of course, these two relationships here have been
obtained from a set of physical laboratory
experiments.
This is how Program 1 works.
Now we have also, say, a Program 2.
And this Program 2 only uses the updated Lagrangian
formulation.
Now, remember our laboratory physical experimental results
have given us the considered affiliations for the towards
Lagrangian formulation.
However, we now have a program that only is operating on the
updated Lagrangian formulation.
The constitutive relations are, in this case, well, like
that and like that, where I have given here three dots,
because we want to now study what they should be looking
like in order to obtain the same results as we are
attaining with Program 1.
In other words, how can we obtain, with this Program 2,
identically the same results as are obtained
using Program 1.
This is a question that we really are asking.
To answer this question, we have to go back to some
derivations that we discussed in an earlier lecture.
And there, we discussed that this is here the governing
continuum mechanics equation, corresponding to the total
Lagrangian formulation.
Here we have the incremental stress strain law going into
the formulation.
Notice this is a tangent material tensor.
Here we have the second Piola-Kirchhoff stress, and
here, as well.
And of course, here on the right hand side, we have to
total external loads entering the total
external virtual work.
In the updated Lagrangian formulation, this is the
equation that we derived earlier.
Here, again, is a total external virtual work.
And here, a constitutive relation enters, corresponding
to this formulation.
And here we have the Cauchy stress that is being used in
the updated Lagrangian formulation.
The terms used in the formulations are summarized on
this view graph.
In the TL formulation, of course, we're integrating over
the original volume.
In the updated Lagrangian formulation, we're integrating
over the current volume, time t.
And there is this transformation, where the mass
density's at time t, and the original mass density, enters.
Notice here are listed the linear and the nonlinear
increment in the Green-Lagrange strain.
Here, there also is a linear and a nonlinear increment in
the Green-Lagrange, but for the updated Lagrangian
formulation, and here, for the total Lagrangian formulation.
There's a relationship between these
increments, as shown here.
And this relationship I will focus our attention upon
further on the next view graph, because I'd like you to
clearly understand how we obtain this relationship.
But we notice is that the deformation gradient enters in
this equation, as well as in this equation.
The variations on these linear strain increments and
nonlinear strain increments of the total Lagrangian
formulation and the updated Lagrangian formulation are
also related via the deformation gradient.
In fact, twice the deformation gradient enters here.
Let's look at how we derive this kinematic relationships.
Well, we use that there is a fundamental property of
Green-Lagrange strain as expressed in this equation.
If you know a Green-Lagrange strain at a point, then we can
pick any material fiber with these components.
And we know, if these are the origin of components of the
material fiber, at that point, that after motion, in other
words, in configuration t, corresponding to time t, that
material fiber has this length, where the original
length is given here.
This is a very basic equation that we want to study a bit
more on the next view graph.
But let me already point out that we can, directly, use
this basic relation, and, of course, apply it also
corresponding to time t, plus delta t.
Notice here, t, here, t plus delta t.
Here t, here, t plus delta t.
Otherwise, same relationship.
And we also can apply incrementally.
Notice this a Green-Lagrange strain increment, from time t,
to time t plus delta t.
But refer to time t.
That's why we have the little t down here.
Up here, we always referred the increments in strains, or
the total strains, if it is a total strain, to the
configuration at time zero.
Now we are referring the quantity to time t.
And, if we simply apply this same basic relationship here,
we directly obtain this equation here for the
increment in Green-Lagrange strain from time, t, to time,
t plus delta t.
Refer to time t configuration, this relationship holds.
Let's look at this relationship here.
What does it really mean physically?
Well, it means the following--
here we have our stationary coordinate frame, x1, x2, x3.
And at time zero, we have a material fiber here, any
material fiber.
But let's just pick one.
And that has a length of 0 ds.
The components of this vector, the vector that is shown here,
of length 0 ds, the components are shown here.
And that vector we call a d 0 x.
This material fiber moves from time 0 to time t, to this
configuration.
The new components are listed here--
dtx2, dtx1, dtx3.
Of course, there's a relationship between the new
components and the original components of
the material fiber.
And here, goes into the deformation gradient.
Now the equation that I just referred to really relates
this length and that length.
And the way we obtain it is to look at the definition of the
Green-Lagrange strain.
strain The Green-Lagrange strain is defined as follows--
I'm going to simply write it in here.
You see here, we have--
oops.
I should have not to use that delta t, because I'm
looking at time t.
This is the equation for the Green-Lagrange strain, from
here to there.
And what I'm now going to do is to premultiply this
relationship by d0x, transposed, and
postmultiplied by d0x.
Of course, I have to do the same thing here.
d0x, transposed, goes in front of the bracket.
And d0x goes into the back of the bracket.
Let me make clear what I'm doing here.
I'm putting this in here.
And I'm putting that on the back.
Notice I'm using a transposed here and a transposed there.
The basic relation of the Green-Lagrange strain, once
again, is this being equal to that part with out the d0x
transposed in front, and without the d0x on the back.
But now we notice, directly, that this part here and this
part here is nothing else then dtx.
If you recognize that this is dtx, that is dtx, you
immediately get on the right hand side, one half,
bracket open, d--
well, I should say t ds, squared, minus 0ds squared.
That's what you have on the right hand side.
And on the left hand side, you have what
I showed you earlier.
Let me go back one view graph, just to make sure that we
understand each other.
On the right hand side, you immediately obtain this term
minus that term.
And on the left hand side, you end up with this term.
So this is the relationship, a very basic relationship, that
we want to use in our derivation.
And it is obtained by simply looking at the Green-Lagrange
strain, the way it's defined, and the way we have defined it
in an earlier lecture, and premultiplying by d0x
transposed, and postmultiplying by d0x.
Of course, what you're seeing there, in that relationship I
showed you, you see the components of this term here,
written out in terms of i and j components.
Well, so what we are doing, then, is to follow a material
fiber, original length 0 ds, through it's motion to t ds
and to t plus delta t ds.
And that's expressed by the three formulas that are on the
previous view graph.
Hence, if we subtract from the second formula, the first
formula, we directly obtain this relationship here.
And we now use this term here, or this equation, linking up
dtx to d0x and expand these two terms
to obtain this equation.
Now, notice that this equation has to hold for
any material fiber.
And if it does hold for any material fiber, it immediately
follows that this equation here has to hold it, because
we can apply it to a material fiber that is oriented along
the x1 axis, then another one that is oriented along the x2
axis, then another one that is oriented along the x3
axis, and so on.
And surely, this relationship here must then hold.
If we look at this relationship more closely, we
see that we can write it out as a linear strain increment
and as a nonlinear strain increment on the left hand
side and on the right hand side of the equation.
And this shows, then, directly that the 0 eij must be equal
to twice a deformation gradient times ters.
The reason being that this term here is a function of the
incremental displacement only.
Linear functional incremental displacement, so is this.
And if you simply equate those terms that are linear, we get
that this part must be equal to that part.
And if you create all the nonlinear parts, we find that
0ei, e to ij, is equal to twice the deformation gradient
times te to rs.
And those two equations are given down here.
If we now take variations on the left hand side and on the
right hand side, and if you recognize that's a variation,
is respect to the configuration at time t plus
delta t, it is clear that the deformation gradient terms
don't change.
And, therefore taking the variation on the left hand
side, that variation is simply applied to the ters, with
these being constant terms.
Similarly here, and this then, completes the derivation of
all the kinematic relationships that I had in
the earlier table and that we want to use now.
In addition, we have, also, the important relationship
between the second Piola-Kirchhoff stress and the
Cauchy stress.
And that relationship is shown here.
Of course, we introduced that one in an earlier lecture.
Finally, we have a relationship between the
tangent material tensor, corresponding to the total
Lagrangian formulation and corresponding to the updated
Lagrangian formulation.
Notice zero here, t there.
And that relationship is given here.
It's given by a fourth order tensor transformation, which
I'd like to now discuss with you further.
This is an important relationship that we want to
spend a bit off time on.
The relationship is derived as follows--
the increment in the second Piola-Kirchhoff stress is
clearly given by the tangent material relation times
increment in the Green-Lagrange strain.
And we're looking at differential increments here.
The increment in the second Piola-Kirchhoff stress, refer
to time t, is given by the right hand side here, where
this increment in Green-Lagrange strain is
refer to time t.
We also have this relationship between the stress increments.
This is the same relationship that holds between they total
second Piola-Kirchhoff stress at time t and the Cauchy
stress at time t.
And we have this relationship, which we just derived.
Now, using these four equations, we directly obtain
by substitution, that this left hand side, substituted
for d0 Sij, must be equal to this right hand side, where we
have substituted this relationship
for d0 epsilon rs.
And this equation, then, directly yields this equation,
where the coefficient here must be what we are looking
for, namely tCabpq.
This is the tangent material tensor to be used in the
updated Lagrangian formulation,
referred to time t.
The tangent material relation is once more summarized here.
And this is what we need to use in the updated Lagrangian
formulation.
Notice it's a fourth order tensor transformation on this
value here, on this tensor here.
Now we can compare the updated and total Lagrangian
formulations.
And the question that we ask is under what condition is
this left hand side, which we use in the total Lagrangian
formulation equal to this term here, which we use in the
updated Lagrangian formulation.
Well, we simply substitute here for the stress, and for
the strain, our earlier relationships.
We multiply out, and immediately, we obtain the
right hand side.
Notice ij, of course, are dummy indices, which can be
substituted by m and n.
It's interesting to upset here that this transformation is a
kinematic transformation, which is actually buried in
the proper use of the updated Lagrangian formulation.
This transformation is a stress transformation that has
to be enforced for any Cauchy stress that we would be
calculating in the computer program.
Let us look at the second term.
Here, we have, on the left hand side, the term of the
total Lagrangian formulation.
And we ask whether this is equal to the term that I
showed you on the right hand side, corresponding to the
updated Lagrangian formulation.
Well, if we substitute our definitions for the second
Piola-Kirchhoff stress here, and what we know to be a true
kinematic relations, here, and we multiply out, and also use,
of course, this relationship here, we directly obtain what
we know we have to obtain for the updated Lagrangian
formulation.
In other words, the left hand side is equal to
the right hand side.
Notice, again, the kinematic transformation here and the
stress transformation here.
Finally, we also have to look at this term, the term in
which the material tensor enters.
We ask again if this left hand side equal to the right hand
side using the proper transformations.
And if we substitute, as shown here, we directly observe yes,
if the term of the total Lagrangian formulation is
equal to the term of the updated Lagrangian
formulation, provided, of course, we use the proper
transformations.
And here, in particular, the constitutive transformation
that we discussed just a bit earlier.
So we conclude then that the updated Lagrangian terms are
identically equal to the total Lagrangian terms, provided we
follow the transformation rules that I have summarized.
This means that in the finite element analysis, if we use
the same finite element interpolation functions in the
updated Lagrangian formulation, as in the total
Lagrangian formulation--
of course we have to use the same interpolation functions
and the same finite element assumptions--
then, we can see directly that, for the total Lagrangian
formulation with this equivalent equation and the
updated Lagrangian formulation with this equivalent equation,
that these matrices are identically equal.
And that's a very important observation, a very important
observation.
It means, then, going back to the question that I asked
earlier, regarding the use of Program 1 and Program 2, it
means that, to summarize, if we use Program 2, then Program
2 gives the same results as Program 1, provided the Cauchy
stresses are calculated from this relationship.
What does that mean?
It means that the second Piola-Kirchhoff stresses of
course are given by the relationships that were
determined by physical laboratory results.
Remember, we know how old the second Piola-Kirchhoff stress
is defined as a function of deformations, of the
deformations in the material.
That was our assumption because that went into the use
of Program 1.
And we are asking now only how can we use Program 2, which
only contains the updated Lagrangian formulation.
So we know how St0Smn is defined.
And if we know this quantity, we have to make this
transformation to calculate the Cauchy stress
in the Program 2.
Similarly, we also remember, postulated that we know this
tensor from the physical laboratory test results.
We know this tensor.
And what I'm saying now is that we have to transform this
tensor, as shown here, to obtain this tensor.
And it is this one that would go then into all our Program 2
computations.
If, in Program 2, we use this Cauchy stress, defined by the
second Piola-Kirchhoff stress, as shown here.
And this tangent material relation, you find by this
tensor, which we know, then Program 2 will give
identically the same results as Program 1.
The kinematic relations that we talked earlier about are
buried in the proper implementation of the updated
Lagrangian formulation in Program 2.
Conversely, if we say that the material relationship for
Program 2 is given, say, in other words, the Cauchy stress
is given and also the tangent material relationship is given
from laboratory test results.
Then you can surely show that Program 1 with the total
Lagrangian formulation will give identically the same
results, provided these transformations are performed
in the program.
Now, the Cauchy stress is given, the tangent material
relationship is given, and what we have to do in the
program, in every solution step, in every iteration, is
to transform this Cauchy stress to the second
Piola-Kirchhoff stress that's going to be used in Program 1.
And we have to transform this tensor here, as shown here, to
this tensor.
And this tensor here, 0Cijs, is going to be
used in Program 1.
Therefore we can really conclude that's the choice of
whether to use a total Lagrangian formulation or the
updated Lagrangian formulation is based only merely on how
effective computationally one formulation is over the other.
And then we note now that the B matrix, the strain
displacement matrix in the U.L. formulation contains less
entries than the strain displacement matrix in the
T.L. formulation.
This means, of course, that the product, B transpose CB,
which is used in the stiffness matrix, is cheaper to
evaluate, less expensive to evaluate, in the updated
Lagrangian formulation.
Across So there's a plus here for the updated Lagrangian
formulation.
However, if the stress-strain law is available in terms of
the second Piola-Kirchhoff stress, then it is most
natural to use a total Lagrangian formulations.
And this is the case when we analyze rubber type materials,
for example we use the Mooney-Rivlin material law,
which we discussed very briefly in the previous
lecture, and, in particular, when we want to analyze
inelastic response and we want to allow for large
displacements and large rotations.
but we have small strain conditions.
I'd like to refer you now back to the previous lecture, where
we made a big point out of the fact that the second
Piola-Kirchhoff stress, the Green-Lagrange strains, are
the components of these two tensors are invariant under
rigid body rotation, and that that means that we can
directly use material relationships that are
applicable to infinitesimal displacement and strains, in
other words, the engineering stress and
engineering strain variables.
You can use these material relations directly in a large
displacement, large rotation analysis, provided we use the
total Lagrangian formulation.
And I explained that, I spent a bit of time on that, in the
previous lecture.
Please refer back to that information.
Let us now look at a very special case, namely, the case
of elasticity.
We discussed in the previous lecture already that one way
to proceed is to use this relationship here in the total
Lagrangian formulation, where we have the second
Piola-Kirchhoff stress on the left hand side, the
Green-Lagrange strain on the right hand side, multiplying
the constitutive relation, corresponding to the total
Lagrangian formulation.
Now if we use this equation in the relation that we just
established, or that we talked about, namely giving us of a
Cauchy stress, in terms of the second Piola-Kirchhoff stress,
we obtain this equation here.
And of course, we also have to use in the updated Lagrangian
formulation, this
transformation, as we just discussed.
Now notice that when we compare these two right hand
sides, we see the deformation gradient entering twice here,
whereas four times here.
We would like to recast, now, this equation on the Cauchy
stress, so as to have the deformation gradient here
entering in the same form as we have it entering here.
And that is achieved by the definition of the Almansi
strain tensor.
Here, now, we have the Cauchy stress in terms of the Almansi
strain tensor, multiplying a new constitutive tensor.
And that new constitutive tensor is given down here.
Notice the right hand side here contains the constitutive
tensor of the total Lagrangian formulation, t0 here.
And we are multiplying this tensor four times by the
deformation gradient components.
This is exactly what we try to achieve.
And we achieve it by using the Almansi strain tensor, which
is defined as follows.
Here we have the definition of the Almansi strain tensor, in
terms of the Green-Lagrange strain tensor.
And here, we have the inverse deformation
gradient entering twice.
The same definition is also given here,
but in matrix form.
And if we substitute for the inverse deformation gradient,
in terms of displacements, and multiply components out, we
obtain directly this relationship here for the
Almansi strain tensor, where, notice our notation is this t
and that t are those two t's here.
We are taking the displacement from time 0 to time t, and we
differentiating those displacements with respect to
the current coordinates.
And that's important.
There's also a minus here, which is a bit of a surprise,
because in the Green-Lagrange strain definition, of course,
we have a plus here, but different terms, of course,
different terms.
Anyway, this is the definition of the Almansi strain tensor.
And this tensor is very useful in the way I just described.
The Almansi strain tensor is a symmetric tensor.
The ij components are equal to the ji components.
The components are not invariant under rigid body
rotation of the material.
This is a property, an important property, of the
Green-Lagrange strain tensor, but it is not a property of
the Almansi strain tensor.
The Almansi strain tensor is really, overall, not a very
useful measure, strain measure.
But we wanted to introduce it here briefly.
And it is quite useful in the one particular case, namely,
in the analysis of isotropic materials undergoing large
displacements and large rotations, but
generally small strains.
Let's look at one simple example here that gives us a
bit of insight, what's the Almansi strain tensor looks,
what the strain components look like.
Here we have a simple four node element that is being
pulled out into the red configuration.
Notice the original length is 0L.
The pullout is t delta.
Of course, t delta could also be negative, in which case we
are pushing in here.
And the current length is tL.
Notice the Green-Lagrange strain is given here, where we
have a plus sign here and where we have 0L down here.
The Almansi strain 1, 1 component, of course, in this
particular case, has this relationship here.
Notice t [? sigma ?] over tL, versus 0L in the
Green-Lagrange strain.
And a minus sign here, whereas we have a plus sign there.
If we plot these strain tensor components as a function of t
delta over the original lengths of the element, we
find this blue curve for the Green-Lagrange strain.
That is very easy to see that this is the
Green-Lagrange strain.
Notice here we have 1/2 minus 1/2.
That minus 1/2 comes from this relationship.
This is minus one at this point.
And that is plus 1/2.
So we get a minus 1/2 here.
And at this point here, when delta over L0 is one, we have
3/2, verified by that formula.
The engineering strain, of course, is
just a straight line.
The Almansi strain looks as shown here by the red curve.
Notice that when delta is compressive, going into this
direction in the picture, in other words, when we are
compressing that element, of course, tL becomes smaller and
smaller, meaning this variable down here becomes smaller and
smaller, and our Almansi strain component very rapidly
becomes very large negative.
It turns out that the use of the Almansi strain tensor,
corresponding with this material relationship, t t
Caijrs, which I defined on the earlier view graph, is quite
effective when we want to analyze with the U.L.
Formulation situations that involve a linear isotropic
material and large displacements, large
rotations, but small strains.
In this case, we can use directly that t t Caijrs is
given by the right hand side here.
Notice here we have the Lame constants, lambda and mu,
which we already used in the previous lecture, and the
chronica deltas.
The t 0 Cijrs tensor is given as here on
the right hand side.
Again, Lame constants and chronica delta entries.
We would use here the same Lame constants in both of
these relationships.
And we also use that the incremental, the tangent
material relationship, is the same as the total
stress-strain relationship in both cases.
If we use reformulating, if we use this right hand side for
that tensor and make this tensor equal to that tensor,
and if we use the same right hand side as up there, down
here, for the t 0 Cijrs and the 0Cijrs, same numbers, same
Young's modulus, same Poisson ratio, then for large
displacement, large rotation, but small strain analysis, we
virtually obtain identically the same results.
I'd like to just demonstrate to you with a very simple
problem analysis what I mean here.
Let us turn to just one slide in which we show a solution
response of an arch that was analyzed using the updated and
total Lagrangian formulations.
Here you now see the slide that shows the arch.
It is subjected to a point load at it's apex.
And we measure the displacement at the apex.
The displacement is w0, half of the arch was modeled with
12 eight-node elements.
And as you can see, the TL and UL solutions for this arch are
practically, identically the same.
We can't see any difference to the accuracy that we have
plotted the response.
Notice that in these TL and UL solutions, we used exactly the
same Young's modulus and Poisson ratio.
The reason that practically the same response is
calculated using the updated Lagrangian formulation and the
total Lagrangian formulation lies in that the constitutive
transformations that would have to be applied to obtain
the exact response, exactly the same response with a two
formations, really reduce here to mere rotations.
And that is the fact because, for this type of problem,
large displacement, large rotation, but small strain
problem, the mass density remains constant and the
deformation gradient, which can always be written as a
product all of a rotation matrix, an orthogonal matrix,
times another symmetric matrix.
This is, of course, here is a polar decomposition of the
deformation gradient, that for this particular case, large
displacement, large rotation, but small strains, this
deformation gradient is simply almost equal to the rotation
matrix only.
In other words, the stretch matrix is almost equal to the
identity matrix.
However, when we look at large strain problems, large strain
analysis, and we were to use this relationship here for the
Cauchy stress, in terms of the Almansi strain, and this
relationship here, giving us a second Piola-Kirchhoff stress
in terms of the Green-Lagrange strain, with these two
constitutive tensors, given as shown here by
the same Lame constants.
In other words, Poisson ratio and Young's modulus, being
identically the same, entering here.
More specifically, you would put in here a Young's modulus,
say, of 30 million pounds per square inch and the Poisson
ratio of 0.3.
These are the exact numbers that would go into here, with
these, of course, being the chronica deltas.
If we were to use this relationship in the total
Lagrangian formulation and that relationship in the
updated Lagrangian formulation for large strains, then we
would get very different results.
Let us look at the simple problem that we already solved
earlier ones, or that we considered earlier ones in the
previous lecture, using the total Lagrangian formulation.
I'd like to now consider the same problem using the updated
Lagrangian formulation with this material relationship.
Cauchy stress given in terms of the Almansi strain, and
this E curl, which we used already before in the last
lecture in this total Lagrangian formulation.
Notice this E curl here is given on the right hand side
here in terms of Young's modulus and Poisson's ratio.
The problem, very briefly reviewed, is that we are
looking at a bar with constant cross sectional area A bar.
And we will pull this bar out and also compress it in and
look at the force required to pull the bar
out or compress it.
And we want to now look at this problem using this
material relationship here, Cauchy stress in terms of
Almansi strain with E curl constant.
Here we have, for this one dimensional problem, the
Almansi strain, given as shown up here.
Notice this differentiation here of tU1 with respect to
the coordinate 1 at time t.
It's nothing else than tL minus 0L over tL.
Here we have that same quantity squared.
Of course, there's a minus 1/2 in front.
And the result is this.
The Cauchy stress is directly obtained in terms of the
physical force applied to the bar divided by the cross
sectional area, which is constant.
If we use that current length is given in terms of the
original length plus the extension and this material
relationship here, we obtain directly this force
displacement response.
Force plotted vertically up, displacement plotted along
here, t sigma is the displacement.
And notice that the force displacement response is
highly nonlinear--
here is, by the way, the actual expression--
highly nonlinear, and looks quite different from what we
calculated earlier using the total Lagrangian formulation
with the same E curl, same E curl used earlier in the total
Lagrangian formulation, in which case the response really
looked something like this.
So we get a totally different description of the material
response using these two formulation when the material
is subjected to large strains.
Let us now also look at one example that demonstrates this
feature, the features that I just discussed a bit more.
Here we have a frame subjected to a tip load
over at this end.
A is the tip load.
The frame has a thickness h, L length here, L length there,
thickness h here, as well.
The width of the frame structure is b.
And the geometric data are given here.
And the material data are given here.
Notice that h/L is 1/50.
We modeled this structural using 51 two-dimensional eight
node elements, plane strain elements.
Notice here is a typical element shown,
an eight node element.
There are 25 elements along here.
There's one element there for the corner and another 25
elements in the column.
Now we want to ask the question what happens in the
elastic analysis when we subject this structure, in
other words, to the load r and we used once a TL and once a
UL formulation, but with the same material constants.
In other words, the material constants E and nu that I gave
on the earlier view graph are the same for the TL solution
and the UL solution.
And we do not make any transformation, therefore, on
the material relationships.
We simply use the total Lagrangian formulation with E
and nu given, these two constants plugged in.
And these constants are the same
throughout the TL solution.
We perform this way the TL solution.
And then, afterwards, we do the UL solution, again, simply
putting E and nu into the analysis and keeping it
constant throughout that analysis.
For large displacement, large rotation, but small strain
conditions, the TL and UL formulation will give similar
results, similar because we don't make these
transformations that I discussed earlier in the
lecture in order to obtain exactly the same results.
But for large displacement, large rotation, small strain
conditions, similar results are obtained.
Of course, we have shown that already, by the solution that
I showed you on the slide, the analysis of the arch subjected
to an apex load.
But we will see that now again here in this example.
For large displacement, large rotation, and large strain
conditions, the TL and UL formulations will gave quite
different results.
Well, let's look now at the results that we obtained for
this problem.
Here we have plotted the force vertically, meganewtons, and
the vertical displacement of the tip in terms of meter.
The 2-D elements, TL and UL formulations, give this result
here, the black curve, solid curve.
And we also wanted to solve this problem once using beam
elements using the total Lagrangian formulation.
In fact, we used four node isoparametric beam elements.
We have not discussed these yet.
We will discuss the formulation of these elements
in a later lecture.
And you can see that the beam elements give a slightly
different response solution.
Of course, the assumptions, the kinematic assumptions, in
the beam elements are different.
And that explains the difference in the response
calculated.
But the TL and UL formulations, using the eight
node isoparametric plane, the strain elements, we obtain
basically the same response.
The formations are very large for this frame.
Notice here we have the undeformed frame.
There's the load.
We are pressing down here.
These are the deformations at a load of one meganewton.
And this is how the frame looks at full load, five
meganewton--
very large deformations.
In fact, the deformation are so large for the frame, that
in an actual practical problem, probably the frame,
of course, would have undergone inelastic
deformations.
However, we look at this problem as a numerical
experiment just to demonstrate what is happening and what we
have been discussing in the lecture.
If you look at the maximum deformations once more, a
little bit closer, we find that the vertical tip
displacement in the TL formulation for this problem
is given as 15.289 meters, and in the UL formulation, as
15.282 meters.
Notice that there is only a change in the fifth digit.
The displacements and rotations have certainly been
very large in the analysis of this frame, but the strains
are still quite small.
If you look at the strain at the base of the frame, maybe
we have the maximum strain, maximal moments there, we find
that the moment is approximately as shown here.
And the strain, using strength of material formulas, is
approximately given by 3%.
Now up to 2%, you certainly would consider it a small
strain problem.
So here we are just at the limit of going over into a lot
strain region near the base of the column.
This is all I wanted to present to you, discuss with
you, in this lecture.
Thank you very much for your attention.