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In this lecture we are going to go further with the definition of higher order shape
functions for two-dimensional problems. We were working with triangular elements. How
do I construct higher order approximation, that is, an order greater than two for a triangular
element?
Here are the corners of the master element. Given the order, the shape functions should
become the one dimensional shape function, when I project them on a given edge of the
triangle. If this is node 1, node 2, node 3 and if I
am talking of the first edge connected to nodes 1 and 2; and if I am talking of P equal
to 3, a cubic approximation, that approximation or those functions should become the one dimensional
cubic shape functions on this edge. The one-dimensional cubic shape functions, in terms of definition
- the Lagrangian functions, correspond to these four points on this edge, that is, these
four equally spaced points. This point will correspond to the coordinate one-third, zero,
this will correspond to two-third, zero and this is one, zero.
Similarly, on this edge, I should have the four equally spaced points given by these
locations. Here the coordinate will be psi is equal to two-third, eta is equal to one-third,
while this one will be psi equal to one-third, eta is equal to two-third because psi plus
eta has to be one. On this edge I will have two more points, which are given by eta is
equal to two-third and eta is equal to one-third. I have made 1, 2, 3, 4, 5, 6, 7, 8, 9 points.
For P equal to 3 from the Pascal triangle we had drawn earlier, we needed (P+1) into
(P+2) by 2 (number of monomials) for completeness. For P equal to 3, this is equal to 10. So
nine points means nine definitions of Lagrangian shape functions. We need one more to give
us ten independent basis functions. Let us say that I connect these points by
straight lines. If I have drawn everything correctly, they should pass through like this.
I have connected these points by a straight line, these two points by a straight line
and then I draw straight lines parallel to the three edges. If I look at these straight
lines, they intersect at an interior point. This interior point will have a coordinate.
It lies on the line eta is equal to one-third and on the line psi is equal to one-third.
So it will have a coordinate one-third, one-third. This is nothing but the centroid. This point
is another node. This becomes the tenth node. Our shape functions are now defined. Now I
am going to color the nodes with respect to these ten nodes and I will follow some numbering scheme. Let
us say, as far as the definition of shape function is concerned, this is 1, this is
2, this is 3, this is 4, this is 5, this is 6, 7, 8, 9 and 10. I need ten cubic basis
functions in order to have a complete set, which can completely represent any cubic polynomial.
How do I now go about defining the shape functions? Let us say I would like to define N1 as a
function of psi and eta. This function N1 is with respect to this node such that N1
is 1 at the nodes with coordinates zero, zero and zero at all other nodes.
How can N1 be a cubic polynomial in terms of psi and eta and vanish at all other nodes
barring node 1? I will put N1 as psi. If I take the N1 to vanish on this line, which
is nothing but the line psi plus eta is equal to one-third, make it vanish on this line
which is the line psi plus eta is equal to two-third and on this line which is psi plus
eta is equal to one. So if I choose N1 to vanish on these three lines, then automatically
N1 vanishes on all other nodes barring the first node. So I will say, it will have to
vanish on the line psi plus eta equal to one-third. So psi plus eta is minus one-third. It has
to vanish on the line psi plus eta equal to two-third, so psi plus eta minus two-third
and it has to vanish on the line psi plus eta equal to one. So it is psi plus eta minus
1. This is going to give the N1 such that N1 hat at the point zero, zero, which is the
first point is equal to 1. This is equal to c into minus one-third into minus two-third into
minus 1. C implies c is equal to minus nine by two. So it is quite easy to find the definition
of the first shape function. Similarly, if I want to find N2, N2 is a function,
which is one at this point and vanishes at all other points. It simply means that it
has to vanish along this line, vanish along this line, vanish along this line and vanish
along this line. N2 is now written in terms of equation of this line. Equation of this
line is psi is equal to two by three; equation of this line is psi is equal to one by three;
and equation of this line is psi is equal to zero.
N2 hat, which is a function of psi and eta is, let us say, some constant D. It has to
vanish on the line psi is equal to zero, so psi minus 0, into (on the line psi equal to
one-third) psi minus one-third, into (on the line psi is equal to two-third) psi minus
two-third.
N3 vanishes on this line, vanishes on this line and vanishes on this line. Equations
of these lines are: eta is equal to zero, eta is equal to one-third, eta is equal to
two-third. We should have a value one at the node three.
N3 will be some constant E into eta minus 0 into eta minus one by three into eta minus two by three.
If I have to find N4, N4 has to be one at this point and zero at all other points. If
N4 has to be taken to be zero at all other points, N4 has to be zero on this line. It
has to be zero on this line and it has to be zero on this line. Equation of this line
is psi is equal to zero. Equation of this line is psi plus eta is equal to minus one.
Equation of this line is psi plus eta is equal to two-third.
By the same token, if I want to construct N8, N8 will be 1 at the node 8 and it should
be zero at all other nodes, which means that it has to be zero. If I take it to be zero
on this line, zero on this line, which takes care of all these nodes and zero on this line,
then N4 is taken care of. Similarly, I can define N10. N10 has to vanish
on the three edges of the triangle. If I have to talk of N10, N10 will be zero along this
line, which is eta is equal to 0, 0 along this line which is psi is equal to 0, 0 along
this line which is psi plus eta is equal to 1.
N10 is the function of psi and eta is equal to F into psi minus 0 into eta minus 0 into
psi plus eta minus 1. I can construct this way, all the shape functions that we need
in the master element. If I want to now go to P = 4, P = 4 will require, by our definition
of (P+1) (P+2)/2 monomials, to define a fourth order polynomial. This will require 15 shape
functions to be defined.
Let me give some notations. These corner degrees of freedom are called the vertex degrees of
freedom. These degrees of freedom on the edge are called the side degrees of freedom and
this interior degree of freedom, which is ten, is called the interior degree of freedom.
Ten is non-zero in the given element. It is going to be zero outside this element. Ten
is called an internal bubble function. These edge functions are going to be non-zero only
in the two elements, which share this edge. So these are called side bubble functions.
The vertex functions are non-zero in all the elements, which have this vertex as a common
vertex.
Let us now see how to do the P=4 case.
We will make these vertices. When I project the shape functions on the edges of the triangle,
they should become equivalent to the one-dimensional fourth order shape functions (defined in the
one-D case) on this edge. They are given by specifying five equally spaced points on the
line P+1. So I have these three points here. Then I will have these three points here and
similarly, three points here on these edges. If we count all these points, I have 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. As far as the vertex and edge functions are concerned,
there are only twelve side functions. So I need to have three more functions, which are
now internal bubbles. We connect these points with straight lines on opposite edges. I have
constructed these additional internal points in this grid. I define the shape functions
for all these points. There are exactly three internal points and fifteen points with respect
to which I define the shape functions. Here I have fifteen nodes for the definition of
shape functions. I follow the same procedure as before to define the shape functions.
For example, if I have to find the shape function corresponding to this one, let us give the
numbering 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15. Let us say I am interested
in the eleventh one. The eleventh one should be such that it is one at this point and zero
at all other nodes. How can I choose four lines such that the eleventh shape functions
vanishes on all other points? I can choose this line, I can choose this line, I can choose
this line and I can choose this line. So the shape function should vanish on this line,
in this line and this line and this line. Looking at the equations of these lines and
multiplying them together, I will get the fourth order shape function corresponding
to the eleventh point. I can keep on constructing the higher order approximations to any order
that I wish by using this kind of a structure. It can be done in the master element and immediately
used. We are now, in principle, in a position to
construct any triangular Lagrangian basis function of any order that we wish to obtain.
And the beauty of these basis functions is that we have exactly the number of functions
that we need to have a complete definition of the polynomial of the given order.
Now we are going to go to another family of shape functions or basis functions, which
are very popular and are probably used more than triangles. Let us look at rectangular
elements. Instead of meshing the simple domain that I had taken earlier with triangles, I
have meshed it with rectangles such that the size here is h1 and the size here is h2.
Let us take a generic element here. This is element 1, 2, 3, and 4. So the rectangular
element will have four corner nodes or four corner vertices. Everything has to be done
or defined with respect to these corner vertices. The simplest thing we can do is to take a
generic rectangular element. For simplicity, I am taking the first one. I will give a coordinate
axis here, x and y, such that the first node has point locations (0, 0). This is the first
node, second node, third node, fourth, fifth, sixth, seventh, eight, and ninth.
I take these four nodes and I will have 1 for the element, 2 for the element, 3 for
the element, and 4 for the element. So I am taking the first element and by 1, 2, 3 and
4 here I mean the local numbering, just like we did for the triangles. The coordinates
here are going to be (0, 0), (h1, 0), (h1, h2) and (0, h2). I am doing a counter clockwise
numbering. What are the simplest basis functions or simplest element shape functions that we
can define? They have to be defined with respect to these four vertices. Let us
take this element apart because these angles are 900.
Let us imagine I have taken these two edges apart - this edge and this edge. Let us say
this edge is with the nodes 1 and 2 and this edge is with the nodes 1 and 4. On each of
these edges, as we had said for triangles also, the shape functions should be such for
the two-D domain that the projection on the edge becomes a one-D shape function.
Let us take the same principle here. We have to define the shape functions in such a way
that the projections on the edge become the one-D shape function. On this edge it will
become this one-D shape function. Let us say these two linear are N1 of x and N2 of x.
Similarly, here I will have this function. This will be N1 of y and this will be N2 of
y. I have defined the linear on each of the edges. What is N1 of x? As a function of x
on this edge, it is quite easy to define the shape function. It should vanish at this point.
It should be equal to 1 here. I will define it as (h1-x) divided by h1, which is 1 minus
x by h1. The second one, N2 of x as a function of x (because this is a one-dimensional shape
functions and along this edge there will only be functions of x), this will be equal to
nothing but x divided by h1.
Similarly, N1 of y should be 1 here, 0 here at the point y equal to h2. By the same token,
it becomes 1 minus y by h2 and N2 of y becomes y by h2. This is the simple definition.
Using these one-D shape functions that we have defined along this line and along this
line, we take the product of these functions to get all the functions. If I take the product
of N1 of x, that is, the function on this line corresponding to this node and function
on this line corresponding to this node, which is N1 of y, by the definition of N1x and N1y,
this function is 1 at this node and vanishes at this node, which means it vanishes at this
node also. It satisfies all our constraints that the function should have a value 1 at
this node and 0 at all other nodes. Similarly, I go to this node. This node lies as a second
node on this line, so I take N2 of x. On this line, this is the first node and so I take
N1of y and by definition, N1of y is going to vanish here and here; N2of x is going to
vanish here. So it vanishes at all other points and gives me a value 1 at the second. N3 of
x lies as a second node of this line. So N2of x and it lies as a second node of this line
and N2of x into N2 of y should do the job and the fourth one should be N1 of x into
N2 of y.
We are going to define our shape functions in terms of this: N1, which is a function
in the element x and y in the element 1, equal to N1 of x into N1 of y which is the function
y. N2 in the element 1 has a function of x and y is equal to N2 of x into N1 of y. N3
in an element 1 is equal to N2 of x which is a function of x into N2 of y which is the
function of y. N4 in element 1 which is the function of x and y is equal to N1 of x and
N2of y. I have defined these four shape functions with respect to the four vertices of the rectangle
and this is the minimum we can do, because these functions have to be defined by the
logic that we have been following with respect to these four vertices. By definition, these
are products of functions in the x direction and functions in the y direction. Since, they
are defined as products of one-D functions defined in the x and y directions, they are
said to be Tensor Product Family of shape functions. Let us look at some features of it. N1 of
x is nothing but 1 minus x by h1 and N1 of y is 1 minus y over h2. So if I take this
product, this is going to be 1minus x by h1 minus y by h2 plus xy by h1h2. This part is
linear in x and y. This part is bilinear, that is, it is not linear, it is more than
linear in x and y. The basis functions or shape functions seem to represent more than
the set of functions we need to represent completely or to define a linear. That is,
they contain more terms than just the linear.
We have the Pascal triangle here: 1, x, y, x2, xy, y2, x3, x2y, xy2, y3, (I will write
up to the 4th order) x4, x3y, x2y2, xy3 and y4.
If I expand all of these functions N1 N2 N3 N4, I will find the same feature that they
will contain the linear part in terms of 1, x and y plus they will also contain the product
of x and y.
If I come to the Pascal triangle, what happens is, if I look at the representation of these
functions, they will not only represent the linear, they will also contain the xy part.
That is more than what is really required to define the linear polynomials.
Similarly, if I define the quadratic shape functions, I will simply add intermediate points on each of these edges.
We connect these points just like we had done for the triangles. I will get an additional
node in the middle. There are nine such functions that are products: three functions on this
edge, three functions on this edge; three into three and I have nine and these nine
functions are nicely given here in terms of these nodal values. If these nine functions
are to be given in terms of the quadratic, I will define the quadratic shape functions
on the edge and multiply for the two edges to get the quadratic shape functions for the
element. Here the feature is that I get P+1 functions
in one direction into P+1 functions in the other direction. That is, I will get P+1 squared
functions. For the linear I had 2 into 2 which is 4, for the quadratic I will have 3 into
3 which is 9, for the cubic I will have 4 into 4 which is 16. I am increasing the number
of functions as I am increasing the approximation.
If we look at the quadratics by the same formula that we have followed, that is, the same line
of approach, the quadratics should represent this much to be complete, but it turns out
that they not only represent this much, they go and represent this whole set. That is,
the quadratic definition will include all these functions. It will go beyond the quadratic
and have these extra functions. These additional functions are also in the representation above
the requirement of completeness. This is a feature of the Tensor Product Family.
It is not always that we will have such a nice rectangular domain. Let us take this
kind of a domain. If I try to mesh this domain with four nodded entities, this mesh will
have quadrilaterals. It will be a mesh of quadrilaterals, not of rectangles. When I
have a mesh of quadrilaterals, then we do not have the luxury of having these two perpendicular
edges for which we define these individual functions and take the product. I cannot do
anything at the physical level, that is, at the level of the physical element.
For that we map this physical element to a master element. So we will have to define
the master element now for these quadrilaterals. We define the master element in such a way
that I can use this tensor product representation to get the shape functions in the master element.
Let us say this is the physical element, which is a generic quadrilateral, which has n nodes
(x1, y1), (x2, y2), (x3, y3) and (x4, y4). Let us say, I have the xy coordinate system
somewhere. We are going to take it to a master element, which is defined as this. This is
psi. This is eta. This is node 1 in the master element, node 2 in the master element, node
3 and node 4. This is the generic element tau. This is the master element tau hat and
this is the node 1 in the physical element, node 2, node 3 and node 4.
Node 1 will have coordinates (-1, -1). Node 2 will have coordinates (1, -1). Node 3 will
have coordinates (1, 1) and node 4 will have (-1, 1). The four nodes in the master element
-- 1, 2, 3, 4, map to, let us say, 1 hat, 2 hat, 3 hat and 4 hat nodes with these coordinates.
The master element is now a square and is called a master square.
How do I define the shape functions with the master element? The master element has edges
that are perpendicular to each other. So I will take these edges out with the nodes 1
hat, 2 hat, 1 hat and 4 hat and then on these edges, I can redo the whole job as I had done
earlier. I will define these linear functions on this edge, which is in the direction psi,
on this edge, which is in the direction eta. I will define this linear again.
This is N1 psi. This is N2 psi. This is N1 eta. This is N2 eta. This way I can define
these one-D shape functions on the psi edge and the eta edge. In terms of the products
of these one-D shape functions, the shape function for the master element can be defined.
If I have to define the bilinear or the linear approximation, it is a tensor product and
it is called Bi-P approximation. Which means, it is P in one direction and P in the other
direction. It is a product of these two Ps.
When P is equal to 1, it is called a bi-linear approximation. N1 hat, which is a function
of psi and eta, will be nothing but N1 psi into N1 eta. N2 hat as a function of psi and
eta is equal to N2 psi, N1 eta. N3 hat, which is a function of psi and eta, is N2 psi, N2
eta. N4 hat psi and eta is equal to N1 psi into eta. We can define the bi-linear shape
functions here. It will turn out to be 1/4th of (1 minus psi) into (1 minus eta). This
one will be 1/4th of (1 plus psi) into (1 minus eta). This one will be 1/4th of (1 plus
psi) into (1 plus eta) and this will be 1/4th of (1 minus psi) into (1 plus eta). So this
way, I can define the bi-linear shape functions for the master element. Similarly, we can
define the quadratic shape functions by quadratic shape functions.
Let us look at biquadratics. As we had done earlier, this is the psi edge. This is the
eta edge. I will add these additional nodes on the edge at the mid side and I will define
these one-D quadratic functions on the edge. This becomes N1 psi. This becomes N2 psi.
This becomes N3 psi. On the eta edge, I will similarly define N1 eta, N2 eta and N3 eta.
I have three functions on each edge. So I have these nine functions defined on the master
element with respect to these nine points. I will call this point 1, this is 2, this
is 3, this is 4, and this is 5, 6, 7, 8 and 9. Depending on which line the point lies
on, I will find the corresponding shape function. For example, N1 hat as a function of psi and
eta should be equal to (it has the first node of the psi side, first node of the eta side)
N1 psi into N1 eta. Similarly, N2 hat, lies on the third node of the psi side and the
first node of the eta side. So it will become N3 psi into N1 eta. To define N3 hat: (N3
hat lies on the third node of the psi side and the third node of the eta side) N3 psi
into N3 eta. N4 hat is equal to N1 psi into N3 eta. Similarly N5 will be equal to (see
it is a second node for the psi side and first node for the eta side) N2 psi into N1 eta.
To get N9 hat: N9 hat is equal to (it lies on the second node of the psi side and second
node of the eta side) N2 psi into N2 eta. I am able to define all the nine basis functions
with respect to these definitions of N1 N2 N3 and so on, depending on P on the edge.
Similarly, I can construct bi cubic approximation. If I look at the master element, I will have
these corner vertices. On this edge I will have two more equally spaced points, just
like we had done in the triangles. Connect these by lines. We will get four more points
lying in the interior and we will get a grid of sixteen points. Bi Cubic will have sixteen
functions. We can continue this to whatever order we want.
Constructing this basis functions in the generic quadrilateral element has to be done at the
master element level and we follow the same principle that we had followed for the rectangular
element and we get the shape functions. Question is will it give me a continuous approximation?
Answer is yes, because, if I have the next element setting here and if I have two elements
like this, the shape functions of both these elements should become equivalent to the one-D
shape function on the edge. Both these functions on this edge will match from either side because
they have to become equivalent to the same one-D shape functions. The approximation will
match because the functions match. Continuity is not a problem here in this definition.
We have not discussed how we map this quadrilateral to the master element. What is the kind of
mapping that gives me this? This is a domain with straight edges. For the mapping of the
geometry, in this case, as long as I do not have curved elements, we are going to use
bilinear map. That is, x at any point psi and eta in the master element will be sum of i is 1 to 4, xi Mi (I am deliberately
writing it as Mi) as a function of a psi and eta, y as a function of psi and eta is sum
i is 1 to 4, yi Mi hat as a function of psi and eta. Where are these xis, yis? This is
the (xi, yi), this is (x1, y1), (x2, y2), (x3, y3) and (x4, y4).
Why have I written Mi? Because I want to bring out clearly that mapping need not always be
of the same order as the approximation. Here, Irrespective of whether I am using a cubic
approximation or a fourth order, that is, bi cubic or bi quadratic, the mapping is always
going to be bilinear for this kind of a domain, where the quadrilateral is a straight edged
quadrilateral. So the mapping should always be done with respect to the bilinear shape
functions, while the approximation can be done with respect to any shape function that
I wish. These Mis are nothing but the bilinear shape functions.
We see that as far as our computations are concerned, we finally want to put everything
in a computer program. As far as the computations are concerned, we need quantities like del
x divided by del psi, del x divided by del eta, del y divided by del psi and del y divided
by del eta because, we have converted, we have mapped all our functions from the physical
domain to the master domain. So why not do all the integration and other procedures which
are required to get the stiffness matrices and the load vectors in the master element?
These quantities are required. How do I find these quantities?
X will be equal to x1 into ¼ of (1 minus psi) into (1 minus eta) plus x2 into ¼ of
(1 plus psi) into (1 minus eta) plus x3 into ¼ of (1 plus psi) into (1 plus eta) plus
x4 into ¼ of (1 minus psi) into (1 plus eta). y is also written similarly, in terms of the
y. So these are our Ms - M1, M2, M3 and M4. Del x divided by del psi becomes equal to
minus x1 into (1 minus eta) by 4 plus x2 into (1 minus eta) by 4 plus x3 into (1 plus eta)
by 4 minus x4 into (1 plus eta) by 4. So this one will be equal to, if I write it in terms
of psi and eta, 1/4th of (collect all the constant parts) (x2 minus x1 plus x3 minus
x4) plus eta into (x1 minus x2 plus x3 minus x4). Similarly, del x divided by del eta is
equal to minus x1 into (1 minus psi) by 4, minus x2 into (1 plus psi) by 4, plus x3 into
(1 plus psi) by 4, plus x4 into (1 minus psi) by 4. If I collect, it will be 1 by 4 into
((x3 plus x4 minus x1 minus x2) plus psi into (x1 minus x2 plus x3 minus x4)).
I can get the quantities corresponding to y by replacing this by y and replacing this
by y1, y2 and so on. Similarly, I replace this one by y and these by y1, y2 and so on.
So del x divided by del eta, del x divided by del psi, del y divided by del eta, del
y divided by del eta can be now obtained. Once I have obtained these quantities, I use
them to get the matrix of the transformation.
I would like to write dx, dy is equal to (del x divided by del psi, del x divided by del
eta, del y divided by del psi, del y divided by del eta) into (d psi, d eta).
If I go back and look at this one, I can write this expression as A1 plus A2 eta and this
expression becomes B1 plus B2 psi.
This one is going to be C1 plus C2 eta and this one is going to be D1 plus D2 psi. Now
I have to compute the Jacobian, because Jacobian is needed for the transformation of the integrals
from the physical domain to the master domain. This will be equal to (del x divided by del
psi into del y divided by del eta) minus (del x divided by del eta into del y divided by
del psi). Del x divided by del psi is A1 plus A2 eta. This one is D1 plus D2 psi, where
the D1, D2 are again obtained from the previous expression. Del x divided by del eta is B1
plus B2 psi and the second one is going to be del y divided by del psi, that is, C1 plus
C2 eta. In the product, it will be A1 bar plus A2
bar psi plus A3 bar eta plus A4 bar psi eta, where the bars are obtained like this for
us: A1 bar is A1D1 minus B1C1 and so on. The bottom line is that the Jacobian is no longer
a constant. Even with this bilinear map, this is no longer a constant because it is now
a function of psi eta and the product psi eta. While in the triangular element, when
we did the linear map, the Jacobian turned out to be a constant.
When we have to do the numerical integration, we have to account for this part of the Jacobian.
Jacobian is also a bilinear in terms of the psi and eta and because it is a bilinear in
terms of psi and eta, we would like as such, that the Jacobian should always be greater
than zero, because that is what is physical and that positive area maps to a positive
area. It is not that the area can become negative or a given area cannot map to a point. An
area will map to another area, but this will be a positive number, the ratio may be less
than one or greater than one but nevertheless, it is a positive number and the ratio of the
two areas is nothing but the Jacobian. When we say Jacobian is zero, it means the area
maps to a point, to which we not aligned, we cannot have. It is unphysical. We would
like to avoid elements, geometries of quadrilaterals for which this Jacobian can become a negative
or zero at a point. Who stops this from becoming negative or zero? I can have some combinations
of psi and eta given this A1, A2, A3, and A4 that are the four nodal coordinates, and
will define A1 bar, A2 bar, A3 bar A4 bar in such a way that the Jacobian could be negative
at a point.
It is quite easy to show that this can only happen, when (let us take this as the initial
physical element) I have 1, 2, 3, 4, these four points and I decide to start moving point
3 inwards. It will turn out that point 3 could lie on a triangle, degenerate triangle. If
I go beyond this, that is, if point 3 moves to this point here, then the Jacobian would
be negative at some points. Geometries of quadrilaterals for which one of the angles
becomes an obtuse angle are not allowed. One has to be very careful when making the
mesh and ensure one does not get this kind of a quadrilateral domain. We can always convert
it into two quadrilaterals for which the quadrilateral is convex, that is, the angle is certainly
lesser than or equal to 180 degrees. We do not want quadrilaterals for which angles are
large. We should not have these kinds of quadrilaterals -- very large or very small. In general their
performance will be pretty bad. So while making the mesh we have to ensure that these quadrilaterals
are as good as possible. The angles are not too large and the angles are not too small.
In this case, we will have a point where Jacobian is less than or equal to 0 for some psi and
eta. This is a very important point that has to be kept in mind, when we are using quadrilateral
elements.
As far as the definition of the quadrilateral is concerned, these rectangular elements,
we have over-done the job. That is, we have in the Pascal triangle, accounted for more
terms. The question is can we redefine our basis functions or the shape functions over
the quadrilateral in such way that we do not over-do the job. That is, can we cut out some
of these extra terms?
The answer is yes. That comes by using what we call the 'serendipity family of functions'.
Completeness is guaranteed and these extra terms are cut down.
So in the next class, as far as the approximation is concerned, we are going to discuss the
serendipity family of shape functions. Then we are going to now consolidate everything
in terms of the finite element approximation in an element. We will then move on to look
at how we are going to use this in a computational regime. That is, how numerical integrations
are done for the quadrilateral or for the triangular element.