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We have been discussing issues related to
model reduction and substructuring, so we
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will continue with this topic in this lecture.
So we can quickly recall what we have been
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discussing,
we are considering a large finite element
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model governed by equation of this kind, we
are restricting our attention to linear time-invariant
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systems, the idea is that we want to reduce
this model to a lower order model, R is for
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reduced here through a transformation X(t)
is some sai into XM(t), that means we will
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partition this degree of freedom X(t) into
a set of master degrees of freedom, and slave
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degrees of freedom and the state vector X(t)
is related to the master degrees of freedom
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through this sai matrix. So once we find that
we can substitute this into the governing
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equation, and we get this reduced, mass, damping
and stiffness matrices and the reduce forces
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governing the master degrees of freedom.
There are different model reduction schemes
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they differ in their definition of this matrix
sai. In static condensation technique the
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slave degrees of freedom are taken to be related
to the master degrees of freedom through relations
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which are strictly valid only for static,
under static conditions, so we derived the
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relation between X and XM using static equilibrium
equations and this is a sai matrix. In dynamic
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condensation we again partition the degrees
of freedom into master and slave, in addition
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we specify frequency at which the condensation
is done, so the sai matrix is given as here.
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Now if one computes the natural frequencies
and mode shapes of the reduced system the
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set of, you know, natural frequencies from
this model need not agree with any of the
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natural frequencies of the larger model, that
is certainly true in case of static condensation
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method. In dynamic condensation method if
this omega is chosen such that it coincides
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with one of the natural frequencies of the
larger system then one of the frequencies
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in the reduced model will match with that
frequency.
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Now in a third method that is system equivalent
reduction expansion process the sai matrix
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is constructed from the modal matrix of the
larger system, so the modal matrix can be,
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I mean not all modes need to be evaluated
first P modes, suppose if we include the modal
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matrix will be a rectangular matrix, now the
transformation matrix there is synthesized
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from the partitioning of the available modal
information and this is the sai matrix that
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is obtained here. Here again we need to partition
the degrees of freedom to master and slave,
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in addition we need to also specify the modes
of the larger model which needs to be replicated
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in the reduced model, SEREP has the feature
that it exactly reproduces Eigen solutions
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of the larger model, a set of Eigen solutions
of a larger model are retained in the smaller
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model, so these are the 3 techniques that
we discussed.
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Now we move on to the next topic in this that
is known as coupling techniques or the substructuring
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techniques. The basic idea here is again we
deal with large FE models and we want to devise
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some way of reducing the size of the problem
while an computing the response, the idea
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here is that often a large finite element
model may be needed to take into account geometric
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complexities of the structure, and a very
detailed modeling may perhaps be not needed
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if one is interested in capturing the behavior
of the system, a simpler model can capture
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the spatial variation of the system behavior,
but a finer model is needed to capture all
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the geometric details, so in which case there
is no point in working with a very large model,
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a smaller model would suffice.
The other situation where coupling techniques
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would be needed is when a built up structure
is made up of different components and different
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engineers are involved in development of different
components, so every engineer therefore if
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possible would develop a computational model
as well as a experimental model for each of
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these substructures, so we need to synthesize
the models developed by individuals to predict
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the behavior of the global structure, the
testing on a global structure may or may not
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be possible, but finite element model indeed
can be constructed, so how to, what in such
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situation, so there are basically the coupling
techniques can be classified into two categories
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spatial coupling method and modal coupling
method.
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In modal coupling method there are two further
classification fixed interface and free interface,
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the main distinguishing feature of these two
methods is that in modal coupling method the
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coupling matrices are derived in terms of
Eigen solutions of the substructures, where
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as in spatial coupling method the coupling
matrices are derived in terms of the substructure
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structural matrices, mass, stiffness, damping
matrices, so there is no Eigen solutions that
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one need to compute if one is working with
spatial coupling method. So we will see some
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of these details and a good reference for
a discussion on this is a volume edited by
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Maia and Silva, the details are given here.
So we'll start with the discussion on spatial
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coupling method, will restrict our discussion
by enlarge to free vibration analysis, suppose
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this is a built-up structure and I assume
that this built-up structure is made up of
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2 substructures A and B, so this built up
structure is made up of A+ so called, I mean
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so this plus operation is combining these
2 systems. Now if we consider substructure
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A, the degrees of freedom in this model can
further be classified as interior degrees
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of freedom and coupling degrees of freedom,
coupling degrees of freedom occurs here where
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this node and this node are to be coupled
to produce a built up structure, so the degrees
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of freedom at this node constitute UC of A,
and other degrees of freedom in the interior
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are the interior degrees of freedom. Similarly
for system B, I have UI of B and UC of B,
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clearly you can see that UC of A and UC of
B need to be equal for compatibility relations
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to be you know obeyed.
Now let's consider the equation for subsystem
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A, so this subsystem A, the other degrees
of freedom as I already said are partitioned
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into interior degrees of freedom and coupling
degrees of freedom, so this partitioning of
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degrees of freedom induces a partition on
structural matrices and this is depicted here
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KIIA, I stands for interior, C stands for
coupling, it's not the I-th element of mass
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matrix that is not what is meant here, so
the equation of motion is obtained in this
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form, we assume as I already said there are
no external forces but nevertheless I should
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write FCA because these are the coupling forces
that we need to consider later on.
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Now if NC is the size of UC that is coupling
degrees of freedom, then UIA will have NA
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- NC degrees of freedom, the size of this
vector will be that. Similar equation for
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subsystem B with NB degrees of freedom can
also be written which is as shown here. Now
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the interior degrees of freedom include all
the unknown degrees of freedom or could be
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the master DOF's obtained after a model reduction,
so that also could be that means while arriving
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at this one may use a condensation technique
inbuilt into that if needed to be.
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Now how do we get the equation for the coupled
system? So the coupled system equation is
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obtained by considering the compatibility
of displacements at the interface which demanded
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UCA must be equal to UCB, and similarly equilibrium
of forces at the interface requires that some
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this FCA + FCB must be equal to 0, so if we
use these conditions I get this method, so
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this is the built-up structural matrix, I
mean for the built-up structure, this is a
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mass matrix, and the stiffness matrix. Now
the built-up structure can now be analyzed
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using these assembled matrices, the size of
assembled structural matrices will be NA +
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NB - NC, that is cross this, I mean this is
the size of a vector displacement vector,
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now this method is well suited if subsystems
are studied using a finite element analysis
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if you have access to all these matrices you
can easily construct this matrices for the
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built-up structure.
Now I have illustrated this with respect to
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2 subsystems, but it can be extended to a
built-up structure having more than two subsystems
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then the extension is fairly straightforward.
Now I come to the other class of methods that
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is the so called modal coupling method, we
will start with fixed interface method that
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is also known as component mode synthesis
is one of the well-known methods, so we will
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again start with the equation for subsystem
A, and we will partition the degrees of freedom
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as interior and coupling degrees of freedom
and this equation is obtained.
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Now I will denote this NA - NC as NIA, NI
superscript A, now we begin by assuming that
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at the interfaces the degrees of freedom are
made 0 that means the structure is fixed at
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the coupling point, okay. Then the governing
equation therefore UC will be 0, so the governing
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equation will be UIA double dot + this KIIA
UIA = 0, so this is 0 and this is 0 and this
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force is 0. Now we can do Eigen solution,
Eigenvalue analysis associated with this equation
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and determine the system natural frequencies
and mode shapes, okay. Now how do we, at the
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interface structure of course is not fixed
so we need to correct for that, so in any
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case the using
the modal matrix of the fixed interface system
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I obtained this representation in terms of
these modal coordinates denoted as PKA(t),
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the subscript K here denotes that I am not
retaining all the modes, I am retaining only
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K modes, so thereby I am already achieving
a model reduction, okay.
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Now we release the degrees of freedom at the
interface, this analysis has been done with
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degrees of freedom fixed, now once I release
that UC of A(t) is not equal to 0, now we
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make an assumption now that UC of A(t) is
related to interior degrees of freedom through
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relations that are strictly valid under static
condensation, that is we use static condensation
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to eliminate integral degrees of freedom in
terms of coupling degrees of freedom, so by
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that I get this is the equation, I get integral
degrees of freedom in terms of coupling degrees
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of freedom using this relation.
Now what I suggest, what we do is the solution
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is taken to be the sum of, this is a solution
that we assume, the unknowns that we take
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are the modes corresponding to the fixed interface
degrees of freedom, a fixed interface system
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and the coupling degrees of freedom obtained
through a condensation, and this is a transformation
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matrix, is that okay? Now this is how we represent
the assumed displacement, now we return to
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the governing equation and make this
substitution, so once I substitute this I
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will pre multiply by sai KA transpose and
I get this equation, and this leads to the
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definition of reduced, mass, and reduced structural
matrices, so if we now consider this matrix
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transpose M into this and analyze this a bit
we get terms involving phi IKA transpose into
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MIIA into phi IKA which I know is a diagonal
matrix, because this is a modal matrix of
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fixed interface system whose mass matrix is
MII of A, so making some of these simplifications
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we get the mass matrix for the substructure
to be this.
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Similarly this is the mass matrix, if we can
work through this I get I mean if we multiply
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all this and work through this I get the elements
of the mass matrix for subsystem A in this
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form. Now similarly the stiffness matrix we
can work through, and here if we see here
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the Eigenvectors here this phi is the Eigenvector
and it is orthogonal, and it is mass normalized,
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therefore phi transpose K phi is the square
of the natural frequency, so at the matrix
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level we get phi IKA
transpose KIIA into phi IKA to be a this lambda
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matrix, which is the diagonal matrix of natural
frequencies of subsystem A in that fixed interface
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form, so these are the matrices that I get
for substructure A, so this is the equation
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for substructure A at this stage in terms
of, let me recall
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once again the coupling degrees of freedom
and PKA(t) are the modes emerging from fixed
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interface model for the substructure, so these
together constitute the state vector.
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Similarly for system B, subsystem B I can
derive a similar equation, okay. Now I have
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to generate the equation for the coupled system
these are equations for subsystem A and B
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derived independent of each other. Now I will
impose this compatibility relations and this
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equilibrium equation, and I get the combined
equation to be of this form where, for the
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combined system the state vector consists
of normal modes of subsystem A, normal modes
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of subsystem B in the fixed interface format
and the coupling degrees of freedom. The coupling
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degrees of freedom are equal by this compatibility
requirement, so this is the state vector,
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and this is the mass matrix, and this is the
stiffness matrix, stiffness matrix itself
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is in terms of the subsystem natural frequencies
in the fixed interface form and the matrix
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at the, these matrices which we have derived.
Now this is an equation that we need to analyze
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to produce the Eigen solutions for the built-up
structure.
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Now we can make some observations, this method
requires a knowledge of structural matrices
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of substructures, it is not suited for experimental
studies since in experimental studies such
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matrices would not be easily available, what
would be available will be a set of natural
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frequencies and mode shapes if you do an experiment.
Now also creating fixed interfaces may not
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be feasible in an experimental substructure,
okay so you can't weld a member and so on
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and so forth, so this method is suited for
studies where substructures are studied using
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finite element method, but as I said one of
the motivations for studying substructuring
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methods at least in the current scenario where
computational power is fairly generously available,
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the reduction of the model in terms of degrees
of freedom may or may not be that crucial
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but it becomes indeed crucial if some of the
substructures are studied experimentally and
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they need to be combined with other substructures
which may be modeled mathematically, okay
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and also this kind of reduction in computational
efforts become relevant if you are doing uncertainty
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analysis where this, running of these programs
forms a part of a Monte Carlo simulation run,
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where you need to repeatedly run this programs
for, you know, nominally identical values
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of model parameters, so these methods remain
relevant in such situations.
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Now we'll move on to the next method known
as free interface modal coupling method, here
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the objective is to produce a solution strategy
where we bank on the Eigen solutions of substructures
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to construct the coupling matrices and the
modal for the coupled system, so how does
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that work? So here again let us consider the
equation for subsystem A, I write it in this
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form, and as before we will partition the
degrees of freedom and to interior and coupling
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degrees of freedom and that induces partitioning
of structural matrices and the forcing vector
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as shown here. Now the sizes are as shown
here, we can perform now the Eigenvalue analysis
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corresponding to this system, I am not putting
UCA to be 0 now, I am including it in the
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model. Now I will perform the Eigenvalue analysis
and determine NA x NA modal matrix and diagonal
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matrix of Eigenvalues, okay, and I will make
this modal matrix to be mass normalized so
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that phi transpose M phi is I, and phi transpose
K phi becomes lambda for the subsystem A.
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Now if I were to consider KA term expansion
for the response, this is a full NA by NA
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matrix, but I may not include all the modes
in a given calculation so out of capital N
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modes that are available I may retain K modes,
so I will say that I am going to write UA(t)
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as phi KA PKA(t), where PKA(t) are the generalized
coordinates. The subscript K here denotes
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the fact that I am retaining only K terms
in the expansion, and KA means I'm talking
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about subsystem A.
Now I will now partition, U has been partitioned
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as interior and coupling degrees of freedom,
and just as it has induced partitioning of
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stiffness and mass matrices it also induces
a partitioning of modal matrix that is written
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here. Now we substitute this in, this equation
and use the orthogonality relation I get the
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equation for subsystem A in this form.
Similarly for subsystem B using the same logic
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I will get this equation, so just to recall
I mean just to emphasize this capital Lambda
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is the natural frequencies of subsystem B,
and the interface is kept free, and this is
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the corresponding modal matrix. Now the combined
equation for the 2 systems is written in this
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form. Now we need to now impose the conditions
that UCA(t) is UCB(t), so how do I get that?
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That is not implicit here, so this PKA and
PKB contain information on coupling degrees
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of freedom, in some sense this equation need
to be now constrained by this additional requirement,
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how do we do that? So UCA(t) is given by this
in terms of degrees of freedom, generalized
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degrees of freedom PKA(t) and the modal matrix,
so these should be equal, so this produces
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00:21:14,630 --> 00:21:22,280
a constraint equation as shown here SP = 0,
where S is the modal matrix corresponding
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00:21:22,280 --> 00:21:32,390
to the coupling degrees of freedom for subsystem
A and similar entity for subsystem B.
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00:21:32,390 --> 00:21:41,100
Now what I will do is, this combined vector
P, I will now partition into PD and PI, this
198
00:21:41,100 --> 00:21:48,909
partitioning is done from a mathematical perspective,
and I will select D degrees of freedom so
199
00:21:48,909 --> 00:21:54,809
that S, this matrix S gets partitioned into
SD and SI, and SD remains as a matrix that
200
00:21:54,809 --> 00:22:03,340
can be inverted, it is a nonsingular matrix
that can be inverted, consequently I can retain,
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00:22:03,340 --> 00:22:13,530
I mean write PD as in terms of PI as - SD
inverse SI PI, so this is a crucial step here.
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00:22:13,530 --> 00:22:20,230
Now once I do this I
can write the vector P which is PD PI in this
203
00:22:20,230 --> 00:22:28,409
form retaining only PI, and this matrix I
denote as sai, that is sai is - SD inverse
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00:22:28,409 --> 00:22:36,000
SI by I. Now noting that this equation is
satisfied I get now the coupled equation in
205
00:22:36,000 --> 00:22:47,700
this form, this is purely in terms of PI which
degrees of freedom which I am retaining, so
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00:22:47,700 --> 00:22:54,090
I will pre multiply by sai transpose and carry
out this operation, and I get the equation
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00:22:54,090 --> 00:22:59,930
for the built up system M is sai transpose
sai, K is sai transpose this into sai, Q is
208
00:22:59,930 --> 00:23:14,900
PI, FQ is this.
Now since no forces are taken to act at the
209
00:23:14,900 --> 00:23:19,600
interface I take the equation to be MQ double
dot +
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00:23:19,600 --> 00:23:26,690
KQ = 0, so to summarize I got the mass matrix
as sai transpose sai, K as sai transpose this,
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00:23:26,690 --> 00:23:35,450
and sai itself as this, and this S etcetera
is as shown here. Now what needs to be noted
212
00:23:35,450 --> 00:23:44,870
here is that these matrices M and K are now
derived purely in terms of the Eigen solutions
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00:23:44,870 --> 00:23:53,090
of substructure A and B in the free interface
form, you don't need the structural matrices
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00:23:53,090 --> 00:24:03,830
for A and B while implementing this method,
the Eigen solutions for one or more of the
215
00:24:03,830 --> 00:24:08,190
substructure can be experimentally established
suppose A can be an finite element model,
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00:24:08,190 --> 00:24:12,630
B can be an experimental model, or both the
A and B can be experimental models, of course
217
00:24:12,630 --> 00:24:18,450
A and B can also be finite element models,
so the knowledge of structural matrices for
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00:24:18,450 --> 00:24:24,190
the substructures is not directly needed,
if you have it so well and good, but it is
219
00:24:24,190 --> 00:24:33,200
not needed, so this is the method known as
component mode synthesis.
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00:24:33,200 --> 00:24:43,490
Now suppose we are considering a situation
where 2 substructures are coupled elastically,
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00:24:43,490 --> 00:24:48,940
that means there is one more coupling element,
the equation for coupled system needs to be
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00:24:48,940 --> 00:24:55,870
modified now to take into account this feature,
so if you do free interface method the coupling
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00:24:55,870 --> 00:25:01,740
element will not be easily represented. Now
the compatibility condition in this case is
224
00:25:01,740 --> 00:25:12,570
no longer applicable, by this what I mean,
suppose I have a system, this is one substructure,
225
00:25:12,570 --> 00:25:19,520
and this is another substructure, and this
is a coupling element, this is A, and this
226
00:25:19,520 --> 00:25:31,549
is B, so I am talking about how to deal with
this elastic coupling, this UCA(t) will not
227
00:25:31,549 --> 00:25:37,260
be equal to UB(t) because there is an elastic
coupling between the two, okay, so it's not
228
00:25:37,260 --> 00:25:41,470
rigidly connected, so this displacement can
be different from this.
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00:25:41,470 --> 00:25:48,950
Now on the other hand this equation remains
valid, so what I do now is I will define a
230
00:25:48,950 --> 00:25:56,350
coupling matrix KCPL, let that represent the
stiffness matrix such that at the coupling
231
00:25:56,350 --> 00:26:02,360
the displacements and the forces are related
through this coupling matrix, with KCPL being
232
00:26:02,360 --> 00:26:12,420
this equal to this. Now the equation for coupled
system needs to be modified to take into account
233
00:26:12,420 --> 00:26:15,460
this coupling,
the compatibility condition is no longer applicable.
234
00:26:15,460 --> 00:26:23,260
Now you consider this and we get this, now
UA(t) as before I will express in terms of
235
00:26:23,260 --> 00:26:30,429
this, and UB(t) in terms of this, so this
is a representation I have. Now this forcing
236
00:26:30,429 --> 00:26:36,890
is related through to these degrees of freedom,
through this relation, now consequently if
237
00:26:36,890 --> 00:26:39,640
you substitute these relations into the governing
equation, I
238
00:26:39,640 --> 00:26:48,370
get equation of this form, and the final equation
of motion, this is for system A and B together,
239
00:26:48,370 --> 00:26:53,950
and the final equation for the built-up system
is obtained like this where the stiffness
240
00:26:53,950 --> 00:27:00,830
matrix for the coupled system is modified
through these additional matrices, so this
241
00:27:00,830 --> 00:27:07,110
helps in dealing with situations where there
is a elastic coupling between subsystems.
242
00:27:07,110 --> 00:27:14,460
Now there are few other strategies where,
I mean strategies become possible when in
243
00:27:14,460 --> 00:27:19,700
certain class of problems, for example if
the structure has certain planes of symmetry,
244
00:27:19,700 --> 00:27:26,440
for example if you have a 1 span beam like
this, can I construct the solution of this
245
00:27:26,440 --> 00:27:34,020
structure by analyzing a smaller problem,
okay, so instead of solving the larger problem
246
00:27:34,020 --> 00:27:41,020
A, we will solve 2 problems B and C, and synthesize
the solution of this problem, so what I do
247
00:27:41,020 --> 00:27:46,220
is I cut the structure here and I introduce
this boundary condition, this is a roller
248
00:27:46,220 --> 00:27:57,320
and a hinge. So now if you look at mode
shapes of this simply supported beam all the
249
00:27:57,320 --> 00:28:07,860
modes 1, 3, 5 etcetera are symmetric about
X = L/2, so these modes can be simulated by
250
00:28:07,860 --> 00:28:15,240
considering this situation. The displacement
here you can see this is 0 is not 0, okay
251
00:28:15,240 --> 00:28:19,740
you followed I mean, right this is first mode,
this is third mode, this is for fifth mode,
252
00:28:19,740 --> 00:28:26,620
and so on and so forth. So this can be simulated
this, symmetric modes can be simulated like
253
00:28:26,620 --> 00:28:31,899
this, the even modes that is 2, 4, 6 etcetera
are anti-symmetric, this is L/2
254
00:28:31,899 --> 00:28:37,510
you can see here this mode shape is this and
this is anti-symmetric, these anti-symmetric
255
00:28:37,510 --> 00:28:44,030
modes can be synthesized by considering this
problem, so what I can do is I can analyze
256
00:28:44,030 --> 00:28:51,539
2 small problems okay and find out mode shapes
in for this and this, and by suitably using
257
00:28:51,539 --> 00:28:58,690
planes of symmetry I can construct the mode
shape for this built up system, all the symmetric
258
00:28:58,690 --> 00:29:02,720
modes can be obtained from this, anti-symmetric
modes can be obtained from this, and complete
259
00:29:02,720 --> 00:29:07,770
system modes can be thus constructed without
solving a larger problem.
260
00:29:07,770 --> 00:29:15,480
Now this is again illustrated here, suppose
you have a multi-story building frame, this
261
00:29:15,480 --> 00:29:19,970
frame may require say 500 degrees of freedom
to analyze the problem, so what I will do
262
00:29:19,970 --> 00:29:27,070
is, I will utilize the plane of symmetry which
is this red line, and analyze 2 problems okay,
263
00:29:27,070 --> 00:29:31,029
so this will have 250 degrees of freedom,
this is 250 degrees of freedom, so 2 such
264
00:29:31,029 --> 00:29:36,600
smaller problems will be solved and I will
synthesize the solution for the built up system,
265
00:29:36,600 --> 00:29:42,740
okay that means instead of solving a larger
problem A we solve 2 smaller problems B and
266
00:29:42,740 --> 00:29:50,149
C, and synthesize a solution for the larger
problem, so this is another way of utilizing
267
00:29:50,149 --> 00:29:54,970
certain problem features to achieve model
reduction, of course this is clearly possible
268
00:29:54,970 --> 00:29:59,970
only when such symmetries exist in the given
structure.
269
00:29:59,970 --> 00:30:07,250
Now suppose the symmetric structure is loaded
asymmetrically, what happens, okay so that
270
00:30:07,250 --> 00:30:13,140
is not a problem, for example if you consider
a simply supported beam loaded symmetrically
271
00:30:13,140 --> 00:30:20,530
by a force P(t), I will now consider 2 problems
and this plus, so if you add these 2 you get
272
00:30:20,530 --> 00:30:27,580
solution for this problem. Now each of these
problems can be handled by solving these 2
273
00:30:27,580 --> 00:30:36,120
problems, okay, so this is you know advantage
of using symmetric, so in principle can be
274
00:30:36,120 --> 00:30:42,250
used even when if structure is of course symmetric
about X = L/2 in terms of boundary condition,
275
00:30:42,250 --> 00:30:48,080
geometry etcetera, but it is asymmetrically
loaded, okay, so still we can utilize symmetry
276
00:30:48,080 --> 00:30:57,390
by following these arguments/
We will now return to the earlier, one of
277
00:30:57,390 --> 00:31:04,720
the themes that we have been pursuing that
is development of structural matrices, so
278
00:31:04,720 --> 00:31:10,090
we considered a few lectures before, what
we achieved
279
00:31:10,090 --> 00:31:18,740
was we developed the mass and stiffness matrices
for this 3D generalized, 3-dimensional beam
280
00:31:18,740 --> 00:31:25,150
with the 2 nodes and 6 degrees of freedom
per mode, okay, so how did we do that? We
281
00:31:25,150 --> 00:31:31,450
considered this structure to you know the
displacement fields that we assumed was axial
282
00:31:31,450 --> 00:31:38,710
deformation, deformation in plane, deformation
out of plane, and the rotation about the longitudinal
283
00:31:38,710 --> 00:31:43,950
axis, so we accounted for all the energies,
so this is energy due to axial
284
00:31:43,950 --> 00:31:49,220
deformation, due to twisting, this is bending
about Z, bending about Y, and similarly we
285
00:31:49,220 --> 00:31:56,149
computed the kinetic energy due to axial deformation,
twisting, bending about Z, and bending about
286
00:31:56,149 --> 00:32:02,460
Y, and using this we constructed the Lagrangian
and we interpolated these field variable in
287
00:32:02,460 --> 00:32:08,630
terms of their nodal values for U and theta
we use linear interpolation functions, and
288
00:32:08,630 --> 00:32:14,570
for V and W use cubic interpolation functions,
and we derived the 12/12 structural matrices.
289
00:32:14,570 --> 00:32:21,570
Now we wish to now carry this exercise further,
beyond 3D element now we can think of continuum
290
00:32:21,570 --> 00:32:30,290
problems, so in continuum problems we can
think of plane, 2 dimensional problems first
291
00:32:30,290 --> 00:32:34,360
of all, so in 2 dimensional problems we have
plane stress, plane strain, and axisymmetric
292
00:32:34,360 --> 00:32:41,510
problems, so then we can consider 3 dimensional
solids and then plate bending elements, and
293
00:32:41,510 --> 00:32:52,149
some discussion on shell elements, so
the general template for achieving this is
294
00:32:52,149 --> 00:32:58,159
basically to write the appropriate equations
for strain energy and kinetic energy, identify
295
00:32:58,159 --> 00:33:05,440
the field variables interpolate, and you know
take an element, identify the nodes, and degrees
296
00:33:05,440 --> 00:33:14,260
of freedom, and develop the, represent the
field variables using polynomials in terms
297
00:33:14,260 --> 00:33:19,570
of the nodal degrees of freedom, substitute
into the Lagrange's equation and derive the
298
00:33:19,570 --> 00:33:26,750
equation for the nodal degrees of freedom,
and then question of coordinate transformation,
299
00:33:26,750 --> 00:33:31,870
assembling, imposition of boundary conditions,
calculating external nodal forces, and thus
300
00:33:31,870 --> 00:33:37,660
deriving the final equilibrium equation will
have to be done. Many of these steps we have
301
00:33:37,660 --> 00:33:45,230
already covered, the step beyond formulation
of element level structural matrices has already
302
00:33:45,230 --> 00:33:51,130
been covered, okay, the problem of coordinate
transformation, assembling of matrices, imposition
303
00:33:51,130 --> 00:33:55,840
of boundary conditions, computation of nodal
forces and assembling all that has been earlier
304
00:33:55,840 --> 00:34:05,769
done, so to be able to extend that framework
to more complicated problems we need to now
305
00:34:05,769 --> 00:34:11,730
expand our repertory of elements, so with
that in mind in the next few lectures we will
306
00:34:11,730 --> 00:34:18,089
start discussing about some of these elements,
and we'll begin by talking about 2-dimensional
307
00:34:18,089 --> 00:34:25,490
element, namely plane stress and plane strain
elements.
308
00:34:25,490 --> 00:34:37,960
Now in all this I'm assuming the structure
behaves linearly, and material is isotropic
309
00:34:37,960 --> 00:34:45,070
and elastic, elastic, isotropic we are also
going to assume homogeneity within an element,
310
00:34:45,070 --> 00:34:49,399
so the governing equations we will quickly
recall from 3 dimensional linear elasticity,
311
00:34:49,399 --> 00:34:55,060
the state of stress at any point X1, X2, X3
is at given by this tensor, and the state
312
00:34:55,060 --> 00:35:01,750
of strain is given by this tensor, and the
displacement fields U1, U2, U3 all these quantities
313
00:35:01,750 --> 00:35:06,660
are functions of X1, X2, X3, and time, so
a number of independent variables is 4, that
314
00:35:06,660 --> 00:35:14,650
is X1, X2, X3 and T, and number of dependent
variables is 15, there are 6 stress components,
315
00:35:14,650 --> 00:35:20,250
6 strain components, and 3 displacements,
so how do we tackle this problem. So upon
316
00:35:20,250 --> 00:35:26,780
coordinate transformation the stresses obey
this rule of transformation, therefore second
317
00:35:26,780 --> 00:35:32,670
order tensors the displacement is a vector
it follows this rule of transformation, where
318
00:35:32,670 --> 00:35:39,480
T is the coordinate transformation matrix.
Now we have 3 equilibrium equations which
319
00:35:39,480 --> 00:35:49,580
are given here, and this symmetry rho IJ = rho
JI is assumed, that means we assume that there
320
00:35:49,580 --> 00:35:54,089
are no body moments acting on the system,
so these are the 3 equilibrium equations.
321
00:35:54,089 --> 00:36:00,520
Similarly we have the 6 strain displacement
relations as shown here, and as I said we
322
00:36:00,520 --> 00:36:04,530
are using basically linear models, so the
strain displacement
323
00:36:04,530 --> 00:36:15,130
relation would be linear and we assume material
is hookean and also isotropic, therefore the
324
00:36:15,130 --> 00:36:21,390
relationship between stress and strain there
will be 6 constitutive laws relating stress
325
00:36:21,390 --> 00:36:28,540
and strains, and they are given here, there
are 2 elastic constants E and nu which is
326
00:36:28,540 --> 00:36:34,100
Young's modulus and person's ratio or lemma
is constant and shear modulus, either we can
327
00:36:34,100 --> 00:36:41,970
express stress in terms of strains, or strain
in terms of stresses. We can also use index
328
00:36:41,970 --> 00:36:46,950
notations, this I'm not going to do in this
discussion, so for completeness I have mentioned
329
00:36:46,950 --> 00:36:53,280
here all these equations can be written in
short form in this way. There is an alternative
330
00:36:53,280 --> 00:36:57,980
notation where the stress instead
of being written as a matrix we write it as
331
00:36:57,980 --> 00:37:05,540
a vector 6 x 1 vector, this is the representation
for stress, this is a representation for strain,
332
00:37:05,540 --> 00:37:11,580
and this is the matrix D that I will be needing
in the formulation, and this 6 x 1 stress
333
00:37:11,580 --> 00:37:16,510
and strain matrices are related, for example
stress is related to strain through this 6/6
334
00:37:16,510 --> 00:37:23,800
matrix or strain is related to stress through
this C tilde which is this, okay, so this
335
00:37:23,800 --> 00:37:31,060
is as I said material that obeys Hooke's law
and which is isotropic.
336
00:37:31,060 --> 00:37:36,150
Now in terms of this matrix D the equilibrium
equation can be written in this form, and
337
00:37:36,150 --> 00:37:40,980
this is the strain displacement relations,
this is the constitutive law either these
338
00:37:40,980 --> 00:37:48,040
2 represent constitutive law, and by eliminating
using these relations I can write for sigma
339
00:37:48,040 --> 00:37:57,950
C into epsilon I get this equation, and for
epsilon I can write in terms of DU so I will
340
00:37:57,950 --> 00:38:04,140
get this equation, and this is the governing
equation for displacement in terms of operator
341
00:38:04,140 --> 00:38:12,670
D and matrix C, so this is the equation that
we'll be solving.
342
00:38:12,670 --> 00:38:17,579
We also need to discuss about the energies
in a 3 dimensional situation, what is the
343
00:38:17,579 --> 00:38:22,250
expression for energy? Now this can be explained
by considering an infinitesimal element as
344
00:38:22,250 --> 00:38:28,430
shown here, and suppose if it is loaded you
know if you consider this and only the axial
345
00:38:28,430 --> 00:38:36,290
stresses, the force on this phase is sigma
XX into DY into DZ, and this force will deform
346
00:38:36,290 --> 00:38:41,430
the object, and therefore this force would
have done some work on that deformation, and
347
00:38:41,430 --> 00:38:45,800
the question is what happens to that work
done? It is stored as strain energy in the
348
00:38:45,800 --> 00:38:56,650
system, so this is given by this integral
sigma XX into epsilon XX DX, this is the force
349
00:38:56,650 --> 00:39:02,500
and the elongation will be epsilon XX into
DX, therefore this into this integrated or
350
00:39:02,500 --> 00:39:06,770
the entire volume will give us the total strain
energy, so this is total strain energy. Kinetic
351
00:39:06,770 --> 00:39:14,560
energy is given in terms of displacement U,
velocity U dot V dot and W dot as shown here.
352
00:39:14,560 --> 00:39:24,540
Now we would like to simplify the problem
of 3-dimensional elasticity, so as I said
353
00:39:24,540 --> 00:39:35,400
here we have 15 dependent variables, and we
have 15 equations that is 3 equilibrium equations,
354
00:39:35,400 --> 00:39:41,250
6 strain displacement relations, and 6 constitutive
laws, so there are 15 equations, and 15 unknowns,
355
00:39:41,250 --> 00:39:48,160
now not all these 15 equations need to be
solved for all situations, in some situations
356
00:39:48,160 --> 00:39:56,020
we can simplify. There are 2 aspects to this
simplification, there are 4 independent variables
357
00:39:56,020 --> 00:40:01,589
X, Y, Z, and T, and there are 15 dependent
variables, we can cut down on number of dependent
358
00:40:01,589 --> 00:40:07,300
variables and also we can cut down on the
dependence of this variables on X, Y, Z, and
359
00:40:07,300 --> 00:40:13,589
T by you know suitable simplification, if
there is no dynamics the T becomes irrelevant
360
00:40:13,589 --> 00:40:20,060
so it will be independent variables will be,
3 instead of 4, but can we use certain features
361
00:40:20,060 --> 00:40:25,740
of the problem and reduce these quantities
further, so one such approach is the so-called
362
00:40:25,740 --> 00:40:35,470
plane stress model. So to consider a plane
stress model in XY plane, we consider a continuum
363
00:40:35,470 --> 00:40:41,470
as shown here and the basic problem is, the
red lines show the surface traction, there
364
00:40:41,470 --> 00:40:47,079
is a body force acting on this, and we want
to analyze this continuum would be supported
365
00:40:47,079 --> 00:40:54,070
in some manner, and all that will be specified
given the geometry of this continuum the way
366
00:40:54,070 --> 00:40:58,520
it is supported, and the surface tractions,
and the body forces, and the constitutive
367
00:40:58,520 --> 00:41:05,119
law of the material which makes this continuum,
what are the stresses, strains, and displacements,
368
00:41:05,119 --> 00:41:10,250
that is a problem we need to solve.
Now the specification of geometry for plane
369
00:41:10,250 --> 00:41:16,650
stress model we restrict our attention to
prismatic body, okay, so this is what is meant
370
00:41:16,650 --> 00:41:20,410
prismatic body along the Z axis the cross
sectional area, the cross sectional property
371
00:41:20,410 --> 00:41:25,880
do not change, and lateral dimensions are
much greater than thickness, so this thickness
372
00:41:25,880 --> 00:41:30,589
is smaller compared to the lateral dimensions,
this is the restriction on geometry.
373
00:41:30,589 --> 00:41:37,230
How about loads? The loads are only in the
XY plane, and there is no body force in the
374
00:41:37,230 --> 00:41:42,890
Z direction, the loads can be surface tractions
and body forces but they are restricted only
375
00:41:42,890 --> 00:41:50,000
to lie in the XY plane, so the surface Z equal
to this thickness I have taken it as 2H, so
376
00:41:50,000 --> 00:41:58,730
Z = H, and Z = -H is free from surface tractions,
so this top surface is free from surface tractions,
377
00:41:58,730 --> 00:42:04,560
so for these class of problems we can make
a simplification to the 3-dimensional elasticity
378
00:42:04,560 --> 00:42:10,690
problem and that simplification is known as
plane stress approximation. How does it work?
379
00:42:10,690 --> 00:42:15,730
Now let's consider
the surface Z = +H and - H and as we said
380
00:42:15,730 --> 00:42:19,670
there are no surface tractions on that, therefore
the stresses acting on those planes namely
381
00:42:19,670 --> 00:42:27,900
sigma ZZ, sigma ZX, and sigma ZY must be 0,
at Z = plus minus H, for in for all X and
382
00:42:27,900 --> 00:42:36,220
Y. And on omega that means on this surface
sigma XX, sigma YY and Sigma XY are independent
383
00:42:36,220 --> 00:42:40,210
of Z, because loading doesn't change with
respect to Z, the loading is uniform across
384
00:42:40,210 --> 00:42:46,680
thickness, so that is independent of Z, in
developing plane stress model what we do is
385
00:42:46,680 --> 00:42:52,569
we assume that these conditions prevail not
just at the surface but in the interior also
386
00:42:52,569 --> 00:42:57,630
that means we interpolate these features into
the interior by that what I mean strictly
387
00:42:57,630 --> 00:43:06,020
speaking sigma ZZ is 0 only on Z = plus minus
H, but I assume that sigma ZZ is 0 throughout,
388
00:43:06,020 --> 00:43:11,310
and sigma ZX similarly is 0 only on these
2 outer surfaces, but I assume that it is
389
00:43:11,310 --> 00:43:19,079
0 for all X, Y, Z. Similarly sigma XX and
sigma YY and sigma XY these forces do not
390
00:43:19,079 --> 00:43:24,180
change with respect to Z on omega, but what
I assume is they don't change anywhere in
391
00:43:24,180 --> 00:43:31,470
the interior as well, so this is the approximation.
Now therefore at the end of this I have now
392
00:43:31,470 --> 00:43:40,100
certain quantities have become 0, and certain
quantities have become independent of Z. Now
393
00:43:40,100 --> 00:43:44,051
let's see whether equilibrium equation is
satisfied, so you go to the equilibrium equation
394
00:43:44,051 --> 00:43:50,040
right there 3 equilibrium equations you can
see here the quantities that are written in
395
00:43:50,040 --> 00:44:00,851
the red are zeros as per our model, this is
0, so this drops off, this drops off, sigma
396
00:44:00,851 --> 00:44:05,550
XZ is 0 this drops off, YZ is 0 this drops
off, ZZ is 0 this drops off,
397
00:44:05,550 --> 00:44:11,470
and Z is 0 there is no body force that drops
off so that is 0, so the remaining equations
398
00:44:11,470 --> 00:44:15,940
are the equilibrium equations which I need
to consider further.
399
00:44:15,940 --> 00:44:23,500
In the constitutive law again I have assumed
stresses now I have to figure out what these
400
00:44:23,500 --> 00:44:30,010
strains are, so this is a relation between
stress and strain but these quantities shown
401
00:44:30,010 --> 00:44:39,600
in red are 0, so once I expand I get epsilon
XX and epsilon ZZ to be this, and epsilon
402
00:44:39,600 --> 00:44:47,740
XY is given this, but epsilon YZ and XZ become
0, so 2 of the strain components becomes 0,
403
00:44:47,740 --> 00:44:54,140
because certain stresses are 0. So at the
end of it the stress matrix with all nonzero
404
00:44:54,140 --> 00:44:59,140
elements and dependencies is
shown here, the original stress matrix is
405
00:44:59,140 --> 00:45:07,590
function of X, Y, Z it has 6 independent components,
now this is 0 and these two are 0, so these
406
00:45:07,590 --> 00:45:15,270
are 0 as per our model, and these quantities
are independent of Z, the nonzero strain components
407
00:45:15,270 --> 00:45:24,829
are shown here, so epsilon XZ is 0, YZ is
0, but epsilon ZZ will be 0. So now how many,
408
00:45:24,829 --> 00:45:28,900
come from stress to strain, now how do I get
displacement? I have to use strain displacement
409
00:45:28,900 --> 00:45:33,740
relation, so the strain displacement relation
epsilon XX is this, epsilon YY is this, epsilon
410
00:45:33,740 --> 00:45:39,690
ZZ is this, and epsilon XY is related to this.
Now epsilon YZ I have got it to be 0, and
411
00:45:39,690 --> 00:45:45,150
epsilon XZ I have got it to be 0, therefore
these quantities need to be equal to 0, but
412
00:45:45,150 --> 00:45:49,330
these will not be able to honor and we simply
ignore these relations in further development,
413
00:45:49,330 --> 00:45:58,339
and that is where the approximations coming.
So now there are 10 unknowns that we are considering
414
00:45:58,339 --> 00:46:06,060
sigma XX, sigma YY, sigma XY and epsilon XX,
epsilon YY, epsilon ZZ, and the U, V, W are
415
00:46:06,060 --> 00:46:11,050
the unknown displacement, so what are the
equations? I have 2 equilibrium equations,
416
00:46:11,050 --> 00:46:16,430
4 stress-strain relations, and 4 strain displacement
relations there are 10 equations, so this
417
00:46:16,430 --> 00:46:22,260
is the simplification that we have achieved.
What has happened? We have achieved a reduction
418
00:46:22,260 --> 00:46:33,240
in number of unknowns from 15 to 10, reduction
in number of independent spatial coordinates
419
00:46:33,240 --> 00:46:44,819
so this X ,Y, Z becomes only XY and this because
of the approximations in treatment of strain
420
00:46:44,819 --> 00:46:50,170
displacement relations is not an exact model,
suppose if you substitute this into 3 dimensional
421
00:46:50,170 --> 00:46:54,450
equations of elasticity by that I mean suppose
if you analyze the plane stress problem, and
422
00:46:54,450 --> 00:47:00,210
the solution that you obtain you substitute
into the 3 dimensional equations of elasticity
423
00:47:00,210 --> 00:47:06,220
you will be able to satisfy equilibrium equations,
constitutive laws, but you will have difficulty
424
00:47:06,220 --> 00:47:10,170
in satisfying certain compatibility relation,
not all 6 compatibility equations will be
425
00:47:10,170 --> 00:47:15,350
satisfied, so that is one of the limitations
of plane stress model.
426
00:47:15,350 --> 00:47:28,651
Now in
the next lecture what we will do is, we will
427
00:47:28,651 --> 00:47:46,920
consider another form of simplification that
is known as plane strain mode, here the objective
428
00:47:46,920 --> 00:47:51,950
is again prismatic one of the example that
we can give is that of a you know gravity
429
00:47:51,950 --> 00:48:03,020
dam subjected to hydrostatic force, so if
this dam is in a valley like this we can assume
430
00:48:03,020 --> 00:48:11,800
that in this portion the dam cross section
is prismatic, and the load doesn't change
431
00:48:11,800 --> 00:48:18,550
suppose this is the Z direction the load does
not change in the Z direction. Now if at Z
432
00:48:18,550 --> 00:48:29,850
= 0, and Z = L if we assume that the dam is,
the displacement W is 0 that means it is held
433
00:48:29,850 --> 00:48:35,940
fixed, then in plane strain model we deal
with this type of situations where we again
434
00:48:35,940 --> 00:48:42,309
consider prismatic objects where the thickness,
this is much larger than the lateral dimension
435
00:48:42,309 --> 00:48:46,250
it is the opposite of plane stress model,
where in the plane stress model thickness
436
00:48:46,250 --> 00:48:53,990
was small in relation to the lateral dimension,
whereas here it will be the opposite. Now
437
00:48:53,990 --> 00:49:02,980
the way we proceed is as follows now Z is
0 at W is 0 at Z = 0, and W is 0 at Z = L,
438
00:49:02,980 --> 00:49:09,130
now if you consider a mid plane the object
is prismatic therefore it is symmetric about
439
00:49:09,130 --> 00:49:16,230
this mid plane, and the way the boundary conditions
are applied on displacement that is also symmetric
440
00:49:16,230 --> 00:49:22,000
therefore if W is 0 here and W is 0 here,
W must be 0 here, because load hasn't changed,
441
00:49:22,000 --> 00:49:26,640
geometry has not changed, boundary conditions
have not changed, so this is 0.
442
00:49:26,640 --> 00:49:32,380
Now next what I do is I consider a subsection
which is half of this dam so I'll consider
443
00:49:32,380 --> 00:49:42,190
the quarter point. Now again the same logic
at Z = 0 and Z = L/2, W is 0. Now the object
444
00:49:42,190 --> 00:49:49,079
is again symmetric about X = L/4, Z = L/4,
boundary conditions are the same but loading
445
00:49:49,079 --> 00:49:55,550
is the same etcetera, etcetera, I get W = 0,
so what happens is in this model by repeatedly
446
00:49:55,550 --> 00:50:08,079
using this argument we postulate that W(X,Y,Z)
is 0. And next we assume that U(X,Y,Z) is
447
00:50:08,079 --> 00:50:25,339
function of U(X,Y) and similarly V(X,Y,Z)
is V(XY) so this is a displacement field we
448
00:50:25,339 --> 00:50:31,390
will assume, W is 0 U is function of X and
Y, V is a function of X and Y, we develop
449
00:50:31,390 --> 00:50:36,450
this model by considering
from this we'll go to now from displacement
450
00:50:36,450 --> 00:50:46,780
we go to strain, by using strain displacement
relations and from this I use constitutive
451
00:50:46,780 --> 00:50:52,480
relations and come to stresses, so this model
is known as plane strain model.
452
00:50:52,480 --> 00:50:58,609
So in the next lecture we will consider the
mathematical theory behind plane strain model
453
00:50:58,609 --> 00:51:03,579
and after having covered the two back ground
to the theory of these two models, we will
454
00:51:03,579 --> 00:51:10,990
start developing finite element models for
plane stress and plane strain continue, so
455
00:51:10,990 --> 00:51:12,640
we will take up that problem in the next class.