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In this lecture, we will be looking at higher order elements for one-dimensional problems.
If you recall, in the last lectures, we used two node element, linear element and we also
developed the governing… We developed element equations for general one-dimensional problems
using two node linear element. And we have also seen applications like heat conduction
problems and also a column buckling problem. And in the last class, or in the last lectures,
when we have seen this column buckling problem, we solve this problem using four, two node
linear elements. And we have seen the critical buckling load, that we obtained has about
5 percent error when we use four linear element.
So, one way to improve the accuracy of the solution is to go for higher order element.
So, in today’s lecture, or in the next two lectures, we will be looking at higher order
elements. So, finite element formulation presented so far has been on linear trial solution over
an element, that is, we discussed about two node linear element.
The main advantage of these linear elements is that they are very simple to formulate
as you have seen, but they have some drawback. Their drawback is that, usually a large number
of linear elements are required to get a reasonably accurate solution. So, one way of improving
the solution, which you have seen in the last lecture for column buckling problem, is to
use more number of linear elements, but as we increase the number of elements, the equation
system also gets increased, because we will be having more number of nodes. So, one way
of improving solution is to go for higher order elements, so that is what we will be
doing in this class.
So, this lecture presents finite elements for one-dimensional problems based on quadratic
and higher order trial solutions. So, we have seen in the last lectures, we have seen linear,
so next one is a quadratic or you can go for cubic and quartic order elements. So, in this
lecture, we will see a procedure for doing that. And when we use higher order elements,
the element formulation becomes little bit more complicated, however, few of these elements
are required for a given solution accuracy. So, even though so formulation is more complicated,
we can use less number of elements, so the size of equation system that we will be solving
will be smaller than what we usually get using linear elements. We will be solving same column
buckling problem that we are solved using four linear elements at the end or once we
complete the element formulation using quadratic elements.
And these higher order elements are especially useful if more accurate solution derivatives
are desired. If your recall, we are after the solution, but not after the derivative
of solution. But, if you some body is interested in derivative of solution or accurate values
of let us say strain, so then higher order elements are very much useful in that case,
if more accurate solution derivatives are desired.
Now, let us look at what are the various general steps involved in the formulation of finite
element equations. So, first step is to choose the element shape, it depends on the kind
of problem, line elements for one-dimensional problems. For two and three-dimensional problems,
we will be seeing in a little bit later, there are different shapes possible, rectangular
or quadrilateral elements, or if this 3D, you can have tetrahedron or a quadrilateral
or a cubic element. And next step is to choose the nodal degrees
of freedom. How many nodes you want for an element? This depends on the order of boundary
value problem, which dictates what the essential boundary conditions are. And if you recall,
if the boundary value problem is of order 2 p, those boundary conditions of order 0
to p minus 1 are essential, and those boundary conditions of order p to 2 p minus 1 are natural
boundary condition. So, depending on the order of the boundary value problem which dictates
what the essential boundary conditions are, the number of degrees of freedom for each
element will be decided based on that. The nodal variable chosen correspond to the essential
boundary conditions, because if you see a fourth order differential equation, which
corresponds to beam bending problem, there we have zeroth order and first order equations
as essential boundary conditions, whereas third and fourth order equations as natural
boundary conditions, so we need at least first order continuity.
And the solution or we need derivatives also for imposing the essential boundary condition,
so that is what this step says, choose the nodal degrees of freedom depending on the
problem that we are solving. So, after these two steps, we need to obtain suitable shape
functions for the chosen element that is express all trial solution in terms of nodal parameters
of that particular element. And once we have trial solution in terms of nodal parameters
through shape functions, we can substitute a trial solution into an equivalent functional
if you are using variational method or if you are using Galerkin criteria, or in Galerkin
criteria if you are using weighted residual method.
And once we substitute a trial solution into equivalent functional or Galerkin criteria,
then we can do all the manipulations like carrying out integrations. And in case of
Galerkin methods, the above steps gives or give the finite element equations, whereas
for variational method, we need to apply stationarity condition that is differentiate the functional,
which is now, which now is a simple function of nodal parameters with respect to the nodal
degrees of freedom, to obtain the element equations.
Basically, this all steps you have already experienced, and we have solved problems in
earlier lectures. So, in today’s lecture, we will be looking at Lagrange interpolation
formula for constructing finite element trial solutions that is shape functions for second
order boundary value problem. And also, a basic concept of isoparametric mapping we
will be looking at it, where shape functions are used for transforming the physical element
geometry into a parent element simple geometry, so we will be looking at isoparametric mapping.
A quadratic element for general one-dimensional boundary value problem is going to be presented,
and we will be developing finite element equations using quadratic element for general one-dimensional
boundary value problem using Galerkin method. So, once we have the element equations for
a quadratic element, then we can solve any problem like we did in the earlier lectures,
like we can solve heat conduction problem or under various boundary conditions, or we
can solve using even the column buckling problem by just comparing or making a comparative
table of the corresponding variables in the general one-dimensional boundary value problem
and that specific problem. So, now, for isoparametric element, it is
usually difficult to express element equations in an explicit form, because of need to integrate
complicated functions, so we will be seeing this. And also for that, because the explicit
expressions are possible only if we can carry out integrations accurately, or we can carry
out integrations without adopting any numerical techniques, but sometimes the integrant, when
it becomes complicated, we need to adapt numerical integration.
So, we will also be looking at numerical integrations schemes: one of the numerical integration
schemes is Galerkin’s numerical integration procedure, so that also we will be looking
as a part of study of these higher order elements.
So, now, let us look at how to derive the shape functions for higher order elements
or second order boundary, second order problems. So, one of the key steps in derivation of
element equations is development of shape functions that express trial solution over
an element in terms of nodal parameters. For any problem, the shape functions can be
derived starting from a suitable polynomial as we did for linear elements. When we are
deriving the shape functions for linear elements, what we did is we assume trial solution to
be u is equal to a naught plus a 1 x, and we solved for a naught a 1, or we expressed
a naught a 1 in terms of nodal parameters, and back substituted at this a naught a 1,
and grouped the terms containing nodal values, and then we got the shape function, similar
procedure can be adapted.
For second order problems, in which only the solution and not its derivative is nodal parameter,
it is possible to write shape functions more easily using Lagrange interpolation formula.
So, the procedure which you have seen earlier that is starting with a suitable polynomial,
it becomes more cumbersome when you go for deriving for higher order element, so the
other alternative is to use Lagrange interpolation formula.
So, one point that you need to keep in mind is Lagrange interpolation formula can all
only be used, when the solution, not its derivative, is the nodal parameter. So, note that Lagrange
interpolation formula will not be applicable to, for example, a fourth order problem, which
is like a beam bending problem, were we have both solution as well as derivative as the
nodal parameters. The reason is simple; reason is being that fourth order - for a fourth
order problem, nodal parameters must include the solution as well as its first derivative.
So, let us look now first starting with a polynomial how to derive shape functions at
for a quadratic element and derivation of shape functions starting from a polynomial.
And this method starts by choosing a suitable polynomial; how to choose the polynomial?
The number of coefficients in the polynomial is equal to the number of degrees of freedoms
of element. If you recall, for linear element - the shape
function derivation of linear element, we started with a naught plus a 1 x, because
there are only two nodes, so a naught, a 1 are the coefficients - two coefficients. So,
here, if we want to derive shape function for quadratic element, we need to start with
a polynomial having three coefficients that is a naught plus a 1 x plus a 2 x square.
These coefficients are then expressed in terms of nodal degrees of freedom resulting in the
shape function.
Shape functions for a quadratic element, the derivation is shown. A quadratic polynomial
trial solution is assumed, u is equal to a naught plus a 1 x plus a 2 x square. Why,
because since there are three unknown parameters, each element must have three nodes, so that
these parameters can be expressed in terms of unknown solutions at the nodes.
For one-dimensional problem, nodes are required at the ends of the element to ensure continuity
of solution across element boundaries. Furthermore essential boundary conditions can be imposed
directly, because one of the nodes will be at the end, where boundary conditions may
be specified or prescribed. So, for a quadratic element, the third node can be placed anywhere
over the element, but the placement of middle node is going to influence the solution that
will be seen once we develop the shape functions.
So, a typical quadratic element is shown in the figure here. Having three nodes, node
1, 2, 3; node 1 at x 1, node 2 at x 2, node 3 at x 3 and the corresponding nodal parameters
at node 1 u 1, u 2 at node 2 and u 3 at node 3. The unknown solutions at nodes are identified
as u 1, u 2 and u 3. From the assumed polynomial, u is equal to a naught plus a 1 x plus a 2
x square. By substituting x is equal to x 1, x is equal to x 2, x is equal to x 3, we
are going to get three equations.
The three equations that we get by substituting x is equal to x 1, x is equal to x 2, x is
equal to x 3 are given here. And using these three equations, we can solve for a naught,
a 1, a 2. Solution of this system of equations can be written like this. Once we solve for
a naught, a 1 and a 2, here the constant or the parameter c is defined like this.
So, solving three equations for a naught, a 1, a 2, we got the corresponding a naught,
a 1, a 2. And now substituting these values of coefficients a naught, a 1, a 2 back into
the trial solution, we get this one. And this equation can be rearranged such a way that the terms containing u 1,
u 2, u 3 are grouped together. Grouping terms with same nodal unknown together, we can write
the previous equation in this form u is equal N1 u1, N2 u2 and N3 u 3.
Where N1 is given by this, substituting the value of c and simplifying, N 1 is equal to
x minus x 2 times x minus x 3 divided by x 1 minus x 2 times x 1 minus x 3. And we see
this N 1, N 1 by substituting x is equal to x 1, we notice that N 1 is equal to 1 by substituting
x is equal to x 2 or x is equal to x 3, N 1 is equal to 0.
So, N 1 is going to be 1 at x is equal to x 1; N 1 is going to be 0 when x is not equal
to x 1. Similarly, N 2 and N 3 are given by this, so these N 1, N 2, N 3 are called shape
functions, shape functions for quadratic element. And instead of going through all this cumbersome
procedures starting with a polynomial, as you can notice that when we took this quadratic
element and compare to the linear element, the procedure or the effort is very cumbersome.
So, instead of going through this procedure, there is an alternative way, which is Lagrange
interpolation formula for shape functions. Using this formula, we can write shape functions
for a second order boundary value problem up to any order of trial solution.
So, now, let us look at Lagrange interpolation formula for shape functions. For second order
boundary value problems, a general formula called Lagrange interpolation formula is available
for writing shape functions directly without going through the detailed calculations, which
we have seen in the last, in the previous slide or previous slides.
Remember once again that this formula is not, this formula is not applicable for 4th order
and higher order boundary value problem, because Lagrange interpolation formula is only applicable
for the case when only the solution is nodal parameter, not its derivative; so that you
need to keep in mind.
So, according to the Lagrange interpolation formula, the i th shape function for an n
node line element, for second order boundary value problem can be written like this. Where
n is the total number of nodes in the element, so using this formula we can write shape function
expressions for any order element quadratic, cubic, quartic, 5th order, 6th order, or 100th
order element. n is total number of elements; if it is linear
element, n is equal to 2; quadratic, n is equal to 3; cubic, n is equal to 4; quartic,
n is equal to 5 and so on. And here, in this expression, note that the numerator is a product
of terms x minus x 1, x minus x 2, dot, dot, x minus x n. And if you see the numerator
carefully, the only missing term from this series is one associated with the i th node,
x minus x i is going to be, is not going to be there, that is the term that is missing.
And in the denominator, denominator is a product of terms x i, because we are writing shape
functions for i th node, x i minus x 1, x i minus x 2, dot, dot, x i minus x n, with
i th node that is x i minus x i is missing that is i th node term is missing.
So, using this Lagrange interpolation formula, we can write shape functions for any order
element. And some of the important characteristics of this shape functions N i is equal to 1
at x is equal to x i, N i is equal to 0 where x is not equal to x i or x is equal to x j,
j is not equal to i, because of these characteristics, the i th shape function is considered to be
associated with node i of element. And if you see the shape functions expressions for
N i, N i is a polynomial of degree n minus 1.
So, the shape function for a linear element is going to be of order 1, shape function
for a quadratic element is going to be order of 2, shape function for a cubic element is
going to be order of 3 and so on. And this nodes 1, 2, 3 need not be evenly spaced, the
nodes do not have to be evenly spaced, however placement of interior nodes influence the
shape functions, which will be seen once we have solved some examples.
So, here we have seen two ways of deriving shape function, one is starting with a polynomial
- suitable polynomial, or you can use Lagrange interpolation formula, but using Lagrange
interpolation formula, writing shape function expressions for any order element is much
much easier.
So, now, let us take some examples and apply this formula - Lagrange interpolation formula.
Using Lagrange interpolation formula, write down trial solution for a two node linear
element with nodes x 1 coordinate is given, x 2 is given for a two node linear element.
And also show a plot of trial solution or solution if the nodal values are u 1 is equal
to 4 and u 2 is equal to 1. So, trial solution is given by u is equal
to N 1 u 1 plus N 2 u 2, which can be arranged in matrix and vector form like this, where
N 1 and N 2 can be obtained using Lagrange interpolation formula. One, you can check
these expressions for N 1 N 2 with the expressions that we already looked at, when we derived
starting with a polynomial a naught plus a 1 x, we can notice that these N 1 N 2 expressions
that we got using Lagrange interpolation formula are identical to that what you already have,
using starting with a linear polynomial. So, these are N 1 N 2 values, so the trial
solution is u 1 times N 1; N 1 is substituted u 2 times - plus u 2 times N 2, N 2 is also
substituted. And once we simplify this, it turns out that this is equal to 4 minus 1.5
times x. So, u - approximate solution as a function of x is obtained, so now we are ready
to plot this as a function of x and we can also plot N 1 N 2 as a function of x.
So, the plot showing N 1 N 2 and the approximate solution are given in the figure. And note
that N 1 N 2 are linear shape functions, and since u is a linear function, with slope - negative
slope, so that is how the variation of u with respect x is going to be as shown in the plot.
So, this is how you can apply Lagrange interpolation formula to get the shape functions and also
to interpolate the nodal values. Now, let us apply Lagrange interpolation formula for
a quadratic element, now let us taken another example. Use Lagrange interpolation formula,
write down trial solution for a three node quadratic element with nodal coordinates given
x 1 x 2 x 3 and also show a plot if nodal values are given u 1 u 2 u 3.
So, first step is we need to write shape function expressions N 1 N 2 N 3 by substituting x
1 x 2 x 3 values, because N 1 N 2 N 3 are required for obtaining the approximate solution,
which is u is equal to N 1 u 1 plus N 2 u 2 plus N 3 u 3, which can be put in a matrix
and vector form like what is shown there on the slide.
So, now, N 1 N 2 N 3 after substituting x 1 x 2 x 3 values and simplifying, we will
obtain what is shown there. Now, using this N 1 N 2 N 3 values, approximate value of u
or approximate expression for approximation of u can be obtained by substituting N 1 N
2 N 3 and simplifying. And u 1 u 2 u 3 numerical values are given, so simplifying that we are
going to get u is equal to 4 minus x over 3 minus x square over 12. You can see here,
this trial solution is quadratic and also each of the shape functions is quadratic.
So, now, we can plot this N 1 N 2 N 3 and also the approximate solution. And you can
see by looking at this, you can note by looking at these figures or these plots that the location
of interior node that is x 2 influences behavior of these functions N 1 N 2 N 3, moving the
interior node closer to an end, will make functions change sharp linear at the ends.
Since shape functions represent trial solution, finite element solution in a region can be
made to change more rapidly by moving interior nodes closer to the end nodes in that region.
So, by playing with this location of interior node, we can change the way the trial solution
is varying. Note the location of interior node influences behavior of these shape functions,
and moving to the interior node closer to an end, will make functions change sharp linear
that end.
Since shape functions represent trial solution, finite element solution in a region can be
made to change more rapidly by moving interior nodes closer to end nodes in that region.
So, depending on the requirement, we can do this. Now, let us go to the next concept -isoparametric
mapping. The basic idea of isoparametric mapping is that shape functions are used for transforming
physical element geometry into parent element with a simple geometry.
So, the physical element, the actual element, we are going to map on to a parent element.
And especially elements with curved boundaries are mapped onto straight boundaries. Shape
function derivation and integration are performed over the parent element. And this concept
of isoparametric mapping is useful for one-dimensional problems; however, it is crucial for development
of practical elements for two and three-dimensional problems, which will be seen in the later
lectures.
A typical three node element and its parent element are shown here. So, the actual element
is having three nodes, located at node 1 at x 1, node 2 at x 2, node 3 at x 3. And this
actual element is mapped on to a parent element, in s coordinate system, with node 1 at s is
equal to minus 1, node 2 at s is equal to 0, node three at s is equal to 1.
So, the portion x 1 whatever is there between x 1 and x 2, it is mapped to 0 between minus
1 to 0 in s coordinate system. And whatever is there between s 2, x 2 and x 3 that is
mapped on to 0 and 1 in s coordinate system.
For one-dimensional elements, the isoparametric mapping essentially defines a local coordinate
system. For one-dimensional problems, the parent element is a line element of length
2 units with s going from minus 1 to 1. And also, all the nodes on the parent element
are place symmetrically. In isoparametric mapping concept, the relationship between
the parent element coordinate s and the actual element coordinate x is expressed using parent
element shape functions as follows. If you recall for linear element, for a two
node linear element, earlier we have derived the relationship between x coordinate and
s coordinate using the formula for linear interpolation, but we can also derive that
relation between x coordinate and s coordinate using shape functions of the parent element,
like using the expression that is given here, x is equal to small i taking values 1 to n,
where small n is the number of nodes, n i is the shape function, x y is the location
of these nodes. So, this formula is applicable for any order
element, where i takes values over the number of nodes, where N i are the parent element
shape function, x i are the nodal coordinates of the actual element.
So, this application of this formula, these concepts are illustrated by considering a
three node quadratic element as an example. Now, let us take the three node quadratic
element or typical three node element like this. This is same as what you have seen a
few minutes back, it is reproduced here.
So, now, we need to write the shape functions of the parent element, because we require
the shape functions of the parent element to get the relationship between x coordinate
and s coordinate system. And we can get write the or we can get the expressions for shape
functions of the parent element using Lagrange interpolation formula. So, the shape functions
for the parent element can be written using a Lagrange interpolation formula.
Applying Lagrange interpolation formula, we get N 1 is equal to half s times s minus 1,
N 2 is equal to 1 minus s square, N 3 is equal to half s into s plus 1. So, we got the shape
functions for parent elements for node 1, 2 and 3. So, in isoparametric mapping, the
relation between x and s coordinates is expressed using parent element shape functions, by this
relation x is equal to, here there are three nodes, so i takes values 1 2 3, so x is equal
to N i or N 1 x 1 plus N 2 x 2 plus N 3 x 3. Substituting N 1 N 2 N 3, the shape functions
of the parent element, we get this expression, where x 1 x 2 x 3 are the locations of the
nodes in the actual element. And by substituting s is equal to minus 1, you can notice here
by substituting s is equal to minus 1 in the relationship between x and s, we get x is
equal to x 1, by substituting s is equal to 0, we get x is equal to x 2, by substituting
s is equal to 1, we get x is equal to x 3, which you can easily verify.
And also the original geometry, which is from x 1 to x 3 is mapped onto a line in terms
of s with nodes at minus 1, 0 and 1.
Furthermore note that x is expressed in terms of s. For derivation of finite element equations,
it will be more convenient if we have s in terms of x. However, in general, it is difficult
to get the inverse mapping. You will understand in a while why this inverse mapping is required,
because once when we have derivatives like this, therefore derivative of form du dx must
be evaluated using chain rule as follows, du dx is equal to du ds times ds dx using
chain rule. Note that from the isoparametric mapping,
the expression that we have is x in terms of s, so it is easy to calculate dx over ds.
So, note that from isoparametric mapping dx ds over dx cannot be calculated directly.
So, first, we need to calculate dx over ds, which is called also called Jacobian. And
for a linear element, it turns out that this Jacobian is equal to L over 2, L being the
length of the element. And for higher order elements, we need to first calculate dx over
ds, because we already have expression or the relation between x and s. So, we can differentiate
it on both sides, we get dx over ds.
And once we have this, we can invert this to get ds over dx, but this is possible only
if dx over ds is not equal to 0. This is, this need for inverting Jacobian puts restriction
on when we can use isoparametric mapping. In particular, if Jacobian is 0 anywhere in
the domain or anywhere between minus 1 to 1, its inverse does not exist at that point
and the mapping does not make sense. Because, if Jacobian is 0, inverse of Jacobian
goes to infinity, so Jacobian becomes singular. Thus in isoparametric elements, we must avoid
situations where Jacobian is singular. It is always not necessary that we need to avoid,
but there are circumstances or there are specific cases where we required the singularity. And
we will use this property to our advantage when we solve such kind of problems, where
singularity is required for regular problems, where singularity is not required, we will
try to avoid singularity Jacobian. For simple cases, it is possible to explicitly state
conditions, which will make mapping invalid.
So, here, when we say mapping is invalid, we mean when Jacobian become singular, then
mapping is invalid, so we can specify or we can come up with some conditions such a way
that Jacobian singularity can be avoided. So, we will be deriving those conditions now.
For a quadratic element, we have the relationship between x and s, by taking derivative of that,
derivative of x with respect s, we get Jacobian. So, that is what is shown there.
And we note that or this can be rewritten like this, and noting that the length of element
is or the x 3 minus x 1 is equal to length of element. If length of element l is denoted
using l, then x 3 minus x 1 or x 3 is equal to x 1 plus l, because x 3 minus x 1 is equal
to l. So, substituting this, that is x 3 in terms of x 1 and l, we can rewrite that Jacobian
in the form shown in the slide. For isoparametric mapping to be valid J must
be greater than 0 over the domain 0 to 1, J must be greater than 0 or the domain s go
from minus 1 to 1, not 0 to 1. And you can see here J is the linear function of s.
Since, J is linear function of s, if I make sure that J is positive at s is equal to minus
1, and J is positive at s is equal to 1, then J is going to be positive everywhere between
minus 1 and 1. And that kind of check is only possible if J is possible, because here J
is linear function of s. For isoparametric mapping to be valid J must be greater than
0 over the domain s going from minus 1 to 1.
Since, J is linear function of s, it is enough to check the values of J at s is equal to
minus 1 and s is equal to 1. Clearly if J is greater than 0 at these points, it will
be positive for the entire domain as going from minus 1 to 1.
So, that is the condition we are going to apply to get the required conditions, for
to avoid singularity in Jacobian. So, substituting the condition that J evaluated at s is equal
to minus 1 should be greater than 0, we get that, after simplification we get x 2 should
be greater than x 1 plus l over 4. And the other condition is J evaluated at s is equal
to 1 should also be greater than 0, and applying this condition and simplifying, we get that
x 2 should be less than x 1 plus 3 l over 4. So, the so the middle node that is x 2
should lie or x 2 should satisfy these conditions, thus isoparametric mapping for a quadratic
element is valid as long as middle node is placed such that x 1 plus 3 l over 4 or x
2 lies between x 1 plus 3l over 4 and x 1 plus l over 4.
And we can verify this if we evaluate what happens, and if we satisfy this, what happens?
We can check that using an example. Write down an expression for isoparametric mapping
for three node element shown in figure below and plot the mapping and its Jacobian. The
actual element nodal coordinates are given, isoparamatric mapping gives us relationship
between x and s, substituting the shape functions of the parent element, and substituting the
nodal values of the actual element x 1 x 2 x 3, we obtain this relation.
Now, taking derivative of this, we get this relation J is equal to s plus 3 over 2. And
it can be easily verified that between s going from minus 1 to 1, J is always positive. Jacobian
and mapping from s to s x to s are plotted in the figure below. And you can see, since
Jacobian is positive everywhere, it is mapping is valid. In s coordinate system, all points
is originally between x 1 and x 2, lie between minus s is equal to minus 1 to 0, and those
all and all those points between x 2 and x 3 lie between 0 and 1, and so the mapping
is proper or mapping is valid.
Now, let us take a special case or slightly modify this element such a way that node 2, which is at
location x is equal to 2, is moved to x is equal to 1.5, and see how Jacobian and mapping
looks like.
So, now, we will take that example. Repeat the previous example, if middle node is moved
to x is equal to 1.25, and lets repeat the entire process, isoparametric mapping becomes
this by substituting the parent element shape functions and actual element nodal coordinates,
and taking derivative of it, of this, we get Jacobian, Jacobian is equal to 5 over 2 s
plus 3 over 2.
So, now, we are ready to plot x as a function of s, J as a function of s. And now, let us
see those plots, and before doing that note that J is equal to 0 at s is equal to minus
3 over 5. In fact, for s going from minus 1 to minus 3 over 5 J is negative, and for
s going from minus 3 over 5 to 1, j is positive, so that you can easily verify by looking at
these plots.
And it can be seen from figure here that some values - that for some values of s, or some
values of s result in x being less than 1, which is obviously nonsense, because the actual
element is between x equal to 1 and 4. So, the mapping, there is some problem with the
mapping, so mapping is not good. So, these two examples give us or illustrate isoparametric
mapping. And how to check the mapping and also what are the conditions for placing the
middle node in case of a quadratic element or a three node finite element?
So, we will continue in the next class looking at the derivation of the finite element equations,
for a quadratic element, for general one-dimensional boundary value problem and its applications.