Lagrange Interpolation Polynomials 2-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Nodes, Quadratic, One-dimensional, Element, Shape, Functions, Polynomials, Lagrange, interpolation

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Assignment on
Chapter#4
4.18- The nodes of a quadratic one-dimensional element are located at x = 0. x= 1/2, and
x= 1. Express the shape functions using Lagrange interpolation polynomials.
()
()
1 2 3
23
11
1 2 1 3
13
22
2 1 2 3
12
33
3 1 3 2
1 2 3
1
()
()
( )( )
() ( )( )
( )( )
() ( )( )
( )( )
() ( )( )
0, ;
2
( )( ) 2
2
(0 )(0 )
2
e
e
xN
x N N N
where
x x x x
N L x x x x x
x x x x
N L x x x x x
x x x x
N L x x x x x
l
Substituting x x and x l we get
l
x x l
Nll
l














2
22
32
( )( )
2
( 0)( ) 4 ()
( 0)( )
22
( 0)( ) 2
2()
2
( 0)( )
2
l
x x l
x x l
N x x l
ll l
l
l
xx l
N x x
ll
ll





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Assignment on

Chapter#

4.18- The nodes of a quadratic one-dimensional element are located at x = 0. x= 1/2, and

x= 1. Express the shape functions using Lagrange interpolation polynomials.

( )

( ) 1 2 3

2 3 1 1 1 2 1 3

1 3 2 2 2 1 2 3

1 2 3 3 3 1 3 2

1 2 3

e

e

x N

x N N N

where

x x x x N L x x x x x

x x x x N L x x x x x

x x x x N L x x x x x

l Substituting x x and x l we get

l x x l

l

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