Virtually every phenomenon in the nature, whether biological, geological or mechanical

can be described with the aid of laws of physics in terms of algebraic, differential or

integral equations relating various quantities of interest.

Studying a physical phenomenon involved two major tasks

( i). Mathematical formulation of the physical process

(ii) Numerical analysis of the mathematical model

The mathematical formulation of a physical process needs background in

the related subjects. The formulation results in mathematical statement, often differential

equations. The derivation of the governing equation for most of the problem is not a

difficult task. But their solution by exact methods of analysis is not at all a easy task. In

such cases approximate method of analysis are used to find the solution. Among these

methods, functional approximation method, finite difference method, and finite element

method are most frequently used.

Functional approximation:-

In this method, a set of independent functions satisfying the boundary

condition are chosen. A linear combination of finite number of them is taken to specify

the field variable at any point. The unknown parameters that combine the functions are

found out.

Examples for this method of solution are

( i). Collocation methods

(ii). Least square methods

(iii).Sub domain methods

(iv).Galerkin methods

Finite element method :-

In this method a given domain (area) is divided into a number of sub domains

(sub area), called element or finite elements and an approximate solution to the problem

is developed over each of these finite elements.

The sub division of a whole geometry in to part involves two advantages

(i). It allows accurate representation of complex geometries.

(ii). It enables accurate representation of the solution within each element to bring out

local effects (e.g. Stress concentration).

Practical applications :-

(i). Determining the system distribution in a pressure vessel with oddly shaped holes

and number of stiffness which are subjected to mechanical , thermal and

Aerodynamic loads.

(ii).Finding the concentration of pollutant in sea water or in the atmosphere

(iii). Simulating the weather in an attempt to understand and to predict the formation

of thunderstorms

Steps involved in FEM:-

(1). Divide the whole geometry in to parts.

(2). Over each part, find an approximate solution.

(3). Assemble them to obtain the solution to the whole geometry

The process of dividing a complex geometry into a number of sub domains is called

discretization or meshing.

Example to understand the basic idea of FEM

To determine the circumference of a circle

Ancient mathematician estimated that the value of circumference of the circle by

approximating the circle by line segment whose lengths can be measured.

The approximate value of P can be obtained by summing the lengths of the line elements.

This ancient method lays the building for the development of FEM techniques

FEM Procedure

(1). Finite element discretization:-

The domain, the circumference of the circle represented as a collection of finite

number n sub domain. This process is called discretization of domain.

Each sub domain is called an element. The collection of element is called as finite

element mesh.

R

Uniform

Mesh

Non-

Uniform

Mesh

An element is connected with another one element at selected points called nodes.

The line segment can be of same length or of different length. When the lengths of

the line segment are same for all elements, then the resulting mesh is called as uniform

mesh.

If the lengths of line segments are not uniform then the resulting mesh is called as

non-uniform mesh.

(2).Element equation

An element is isolated form mesh and its required property (length) is

computed by same approximate means

he/2

From the geometry,

Sin (2

e) =

R

he

he = 2Rsin (2

e) (1)

R is the radius of the circle and e is the subtended angle by the element.

Equation (1) is called as element equation.

he e e/2

(3). Assembly of element equation and solution

The approximate value of the circumference of the circle is obtained by putting

together the element properties in a meaningful manner. This process is called as the

assembly of element equation. In our case, the perimeter or circumference of the circle is

equal to the sum of the lengths of the individual elements.

Pn =

n

e 1

he

Here, Pn represents an approximation to the actual perimeter P. If the mesh size is

uniform, then

e = n2

Here n is the number of elements.

Pn = n2R sin ( n22

)

Pn = n2R sin ( n

)

(4). Convergence and Error estimate

In our case, the exact solution is known to us i.e. P =2R

So, we can estimate the error in the approximation. Also, we can show that the

approximate solution Pn converges to the exact solution P in the limits n

For our case, the error in the approximation is equal to

En = P- Pn

=2R- n2R sin ( n

)

=2R (- n sin ( n

))

Proof for convergence

En 0 when n

En = P- Pn, 0= P- Pn, P= Pn

Pn = n2R sin ( n

) put x = n

1

Pn = x

R2 sin ( x )

As n x 0

Lim Pn = Lim x

R2 sin ( x )

n x

= Lim (2Rcos x ) (Apply L Hospital Rule )

x

=2R =P

From the above proof, it is clear that as the number of elements increases, the

approximation improves i.e. the error in the approximation decreases.

APPROXIMATE DETERMINE OF CENTROID / CENTRE OF MASS

Here, the body is conveniently divided in to several elements of simple shape for

which the centre of mass can be computed readily.

In our case, the complex geometry is divided into a finite number of rectangular strips.

Each elements are having he as the width and be as the height. The area of an elemental

strip is Ae= he be. here the area Ae is an approximation of the true area of the element.

Because, be is the estimated average height of the elements. The co-ordinates of the

centroid of the region is given by

X =

Ae

exAe & Y =

Ae

eyAe where

ex and ey are the co-ordinate of the centroid of the e th element with respect to the co-

ordinate system used for the whole body

The accuracy of approximation will be improved by increasing the number of strips ie

by decreasing the width; we can increase the number of elements

Instead of rectangular shaped elements, we use any geometry as elements that can be

approximate the given domain to a satisfactory level of accuracy.

For example, a trapezoidal element will require two height

to compute its area.

Ae = he (be + be+1)

. .

ex

ey Y

X

be+1 be

he

FINITE DIFFERENCE METHOD

In the functional approximation technique, the assumed trial function must be

continuous and should satisfy all the prescribed boundary condition. No simple

guidelines are available to select such functions. Hence except in simple situation, this

approach could not be used to solve practical problems

In this finite difference method, the given geometry is bounded with in a solid region

and the region is discretized by nodal points as shown

Here, the field variable is represented by the discrete values of the variable at the nodal

points. The governing differential equation and the boundary condition are converted into

finite difference form. The finite difference form of the governing D.E and B.C are then

applied over each of the nodes in turn, this will yield a set of algebraic equations. The

resulting equations are then solved for the nodal values of the field variables.

In this method, the finite difference form of the D.E can be obtained easily for meshing

regular shapes but it becomes tedious to derive it for irregular shapes. Move over the

governing D.E itself will become move complex in the case materials other than

isotropic.

Determine the approximate solution of the D.E

x

dd

u2

2

- u + x2 = 0; u (0) = 0; u

l (1) = 1

Let the trial function be

u(x) = a0 + a1x + a2x2 + a3x

3

u(0) = 0 = a0 ( by applying B.C 1 )

dx

du = a1 + 2a2x

2+ 3a3x

2

ul(1) = 1 a1 + 2a2+ 3a3= 1

a1 = 1- 2a2- 3a3 ( A )

u(x) = (1- 2a2- 3a3) x + a2x2 + a3x

3

= x + a2(x2-2x) + a3(x

3+3x) ( 1 )

dx

du = 1 + a2(2x-2) + a3(3x

2+3)

2

2

dx

ud = 2a2 + 6a3x ( 2 )

Sub (1) and (2) in given D.E

-2a2 - 6a3x - x a2(x2-2x) - a3(x

3+3x) + x

2 = 0

x2 - x + a2(2x -x

2-2) + a3(3x -x

3-6x) = 0

x2 - x + a2(2x -x2-2) + a3(-x

3-3x) = 0 is the residue equation

Collocation method

Collocation points are to be selected with in the limits. Here 1/3, 2/3 are to be

taken as collocation points.

R (3

1) = (

3

1)

2 -

3

1 + a2(2*

3

1 -(

3

1)

2-2) + a3(-(

3

1)

3-3*

3

1) = 0

= 1.4444a2 + 1.037a3 = -0.222 ( 3 )

R (3

2) = (

3

2)

2 -

3

2 + a2(2*

3

2 -(

3

2)

2-2) + a3(-(

3

2)

3-3*

3

2) = 0

= 1.1111a2 + 2.2963a3 = -0.222 (4)

(3) *- 0.7694 = -1.1111a2 0.7978a3 = .01708

1.4985a3 = -0.0512

a3 = -0.03416

Sub a3 = -0.03416 in (3) a2 = -0.1292

Sub a2, a3 in (A)

a1 = 1+ 2*0.1292+ 3*0.03416

a1 = 1.3609

u (3

1) = 1.3609*

3

1 - 0.1292x (

3

1)

2 - 0.03416 (

3

1)

3

u (3

1) = 0.4405

u (3

2) = 0.8397

Least square method

1

0

R * Coeff of a2 =0

1

0

{( x2 - x )+ a2(2x -x

2-2) + a3(-x

3-3x) }(2x -x

2-2)dx =0

1

0

2x2- x

4-2x

2-2x

2+ x

3+2x+ a2(4x

2-2x

3-4x-2x

3+ x

4+2x

2-4x+2x

2+4) +

a3(-2x4+ x

5+2x

3-6x

2+3x

3+6x)dx=0

1

0

- x4-2x

2+ x

3+2x+ a2(8x

2-4x

3-8x+ x

4+4) + a3(-2x

4+ x

5+5x

3-6x

2+6x)dx=0

2

2 2x-

3

2 3x+

4

4x-

5

5x+ a2(8

3

3x-

2

8 2x+4x-

4

4 4x+

5

5x)+ a3(

2

6 2x-

3

6 3x+

4

5 4x-

5

2 5x+

6

6x)

1

0] = 0

1-3

2+

4

1-

5

1+ a2(

3

8-4+4-1+

5

1) + a3(3-2+

4

5 -

5

2+

6

1)=0

0.3833+1.867a2+2.0167 a3=0

1

0

R * Coeff of a3 =0

1

0

{(x2 - x) + a2 (2x -x2-2) + a3 (-x

3-3x)}(-x

3-3x) dx = 0

1

0

-x5- 3x

3 + x

4 + 3x

2 +

a2(-2x

4-6x

2+x

5+3x

3+2x

3+6x)+a3(x

6+3x

4+3x

4+9x

2) dx = 0

6

6x-

4

3 4x+

5

5x+

3

3 3x+ a2(

2

6 2x-

3

6 3x+

4

5 4x-

5

2 5x+

6

6x)+ a3(

3

9 3x+

5

3 5x-

5

3 5x+

7

7x)

1

0] = 0