Deriving Shape Functions for 2,3,4,5 Noded Line Element by Lagrange Functions and Verified

(1)

Deriving Shape Functions for 2,3,4,5 Noded

Line Element by Lagrange Functions and

Verified

P. Reddaiah#1

#_{Professor of Mathematics, Global College of Engineering and Technology, kadapa, Andhra Pradesh, India.}

Abstract — In this paper, I derived shape functions for 2,3,4,5 noded line element by lagrange functions and also I verified two verification conditions for shape functions. First verification condition is sum of all the shape functions is equal to one and second verification condition is each shape function has a value of one at its own node and zero at the other nodes. For computational purpose I used Mathematica 9 Software [2].

Keywords — Line element, Lagrange functions, Shape functions.

I. INTRODUCTION

Lagrange polynomials can be straight way used as shape functions for one dimensional problems. Line element with 2,3,4,5,6,7,8,9,10 nodes is shown in figures.1,2,3,4,5,6,7,8,9. In finite element method, the complex domain like a complicated structure is discretized into many number of finite elements and the solution like nodal displacements, element stresses are evaluated in the various locations of the structure. For achieving the nodal displacements in all parts of the element, we have to depend on mathematical relation namely shape functions.

II.GEOMETRICALDESCREPTION

Line element with 2,3,4,5 nodes is shown in figures.1,2,3,4.

Figure.1: 2 noded line element

III. DERIVING SHAPE FUNCTIONS FOR 2,3,4,5 NODED LINE ELEMENT

(i) The typical 2 noded element is shown in Figure.1.

1 2

Shape function for node 1 is N and for node 2 is N

1 2

: 1

x

  

function for node 1&2

Lagrange







2



1

1 2

: x x

N

x x

 









1



2

2 1

: x x

N

x x

 



1 2 N

(2)

1 2 1

2

Output

x





1 2

2 noded element shape functions are

1 1

N and N

2 2

For

x x



 

 

Condition : Sum of all the shape functions is equal to one

st

Verification

I

1

2

1 2

1 N :

2 1 N :

2

[N N ]

1

x

FullSimplifly

Output

 

 



Condition: Each shape function has a value of one at its own node and zero at the other nodes.

nd II

1 2

Node 1 x = -1 then we get

N 1, N 0

At

 

1 2

Node 2 x = 1 then we get

N 0, N 1

At

 

(ii) The typical 3 noded element is shown in Figure.2.

1 2

3

Shape function for node 1 is N , for node 2 is N and for

node 3 is N 1

2 3

: 1

: 0 : 1

x

   

function for node 1,2&3

Lagrange









2



3



1

1 2 1 3

: x x x x

N

x x x x

 



 









1



3



2

2 1 2 3

: x x x x

N

x x x x

 



 









1



2



3

3 1 3 2

: x x x x

N

x x x x

 



 

1 2 3

N

1

( 1 )

2

( 1 )(1 )

1

(1 )

2

Output

x x

x x

 

   



1 2

3

3 noded element shape functions are 1

N ( 1 ) , N ( 1 )(1 )

2 1

and N (1 )

2

For

x x x x

x x



       

 

st

Verification

I

1 2

3

1 2 3

1

N : ( 1 ) N : ( 1 )(1 )

2 1

N : (1 )

2

FullSimplify[N + N +N ]

1

x x x x

x x

Output

       

 

nd II

1

2 3

Node 1 x = -1 then we get N 1,

N 0, N 0

At 

 

1

2 3

Node 2 x = 0 then we get N 0,

N 1, N 0

At 

 

1 2 3

Node 3 x = 1 then we get

N 0, N 0, N 1

At

  

(iii) The typical 4 noded element is shown in Figure.3.

1 2

(3)

1 2 3 4 : 1 1 : 3 1 : 3 : 1 x x x x      

function for node 1,2,3&4

Lagrange









2



3



4



1

1 2 1 3 1 4

: x x x x x x

N

x x x x x x

  



  









1



3



4



2

2 1 2 3 2 4

: x x x x x x

N

x x x x x x

  



  









1



2



4



3

3 1 3 2 3 4

: x x x x x x

N

x x x x x x

  



  









1



2



3



4

4 1 4 2 4 3

: x x x x x x

N

x x x x x x

       1 2 3 4 N N N N





















9 1 1

( 1 )

16 3 3

27 1 1 1 16 3 27 1 1 1 16 3

9 1 1

1

16 3 3

Output

x x x

      _  _  _        _  _         _  _    _{ }  _  _      





















1 2 3 4

9 1 1

N ( 1 ) ,

16 3 3

27 1

N 1 1 ,

16 3

27 1

N 1 1 and

16 3

9 1 1

N 1

16 3 3

For

x x x

        _  _  _         _  _         _  _        _  _  _    

st Verification I





















1 2 3 4

9 1 1

N : ( 1 )

16 3 3

27 1

N : 1 1

16 3

27 1

N : 1 1 and

16 3

9 1 1

N : 1

16 3 3

x x x

       _  _  _         _  _         _  _        _  _  _    

1 2 3 4

FullSimplify[N + N +N N ]

1

Output



nd II

1 2 3 4

N 1, N 0, N 0, 0

At

N

   

1 2 3 4

1

Node 2 x = - then we get 3

N 0, N 1, N 0, N 0

At

   

1 2 3 4

1

Node 3, x = then we get 3

N 0, N 0, N 1, N 0

At

   

1 2 3 4

Node 4, x = 1 then we get

N 0, N 0, N 0, N 1

At

   

(iv) The typical 5 noded element is shown in Figure.4.

1

2 3

4 5

Shape function for node 1 is N , for node 2 is N , for node 3 is N , for node 4 is N and for node 5 is N .

(4)

function for node 1,2,3,4&5.

Lagrange









2



3



4



5



1

1 2 1 3 1 4 1 5

: x x x x x x x x

N

x x x x x x x x

   



   











1 3



4



5



2

2 1 2 3 2 4 2 5

: x x x x x x x x

N

x x x x x x x x

   



   









1



2



4



5



3

3 1 3 2 3 4 3 5

: x x x x x x x x

N

x x x x x x x x

   



   











1 2



3



5



4

4 1 4 2 4 3 4 5

: x x x x x x x x

N

x x x x x x x x

   



   









1



2



3



4



5

5 1 5 2 5 3 5 4

: x x x x x x x x

N

x x x x x x x x

         1 2 3 4 5 N N N N N

8 1 1

( 1 )

15 4 4

Output

x  x x  x

  _  _{ }  _

   





128 1

( 1 ) 1

15 x 4 x x x

 

   _  _ 

 





1 1

16( 1 ) 1

4 4

x  x x x

  _  _  _ 

  





128 1

( 1 ) 1

15 x x 4 x x

 

   _  _ 

 

8 1 1

(1 )

15 4 x x 4 x x

_{ }   _  _        













1 2 3 4 5

8 1 1

N ( 1 ) ,

15 4 4

128 1

N ( 1 ) 1 ,

15 4

1 1

N 16( 1 ) 1 ,

4 4

128 1

N ( 1 ) 1

15 4

8 1

and N

15 4

For

x x x x

x x x x

x x x x

x         _  _{ }  _           _  _          _  _  _           _  _       _  _   1 (1 ) 4

x_ x_ x

 

st Verification I













1 2 3 4 5 1

8 1 1

N : ( 1 ) ,

15 4 4

128 1

N : ( 1 ) 1 ,

15 4

1 1

N : 16( 1 ) 1 ,

4 4

128 1

N : ( 1 ) 1

15 4

8 1 1

and N : (1 )

15 4 4

FullSimplify[N

x x x x

x x x x

x x x x

       _  _{ }  _           _  _          _  _  _           _  _         _  _{ }  _     

2 3 4 5

+ N +N N N ]

1

Output

 

nd II

1 2 3 4

5

N 1, N 0, N 0, 0,

0 At N N     

1 2 3 4

5

1

Node 2 x = - then we get 4

N 0, N 1, N 0, N 0,

N 0

At

   



1 2 3 4

5

N 0, N 0, N 1, N 0,

N 0

At

   



1 2 3 4

5

1

Node 4, x = then we get 4

N 0, N 0, N 0, N 1,

N 0

At

   



1 2 3 4

5

N 0, N 0, N 0, N 0,

N 1

At

   



IV. CONCLUSIONS

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2. Verified sum of all the shape functions is equal to one

3. Verified each shape function has a value of one at its own node and zero at the other nodes.

REFERENCES

[1] S.S. Bhavikatti, Finite Element Analysis, New Age International (P) Limited, Publishers, 2

nd

Edition, 2010. [2] Mathematica 9 Software, Wolfram Research, Version

number 9.0.0.0, 1988-2012.

[3] J.N. Reddy, An introduction to Finite Element Method, 2nd