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

(1)

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

Line Element by Polynomial 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 polynomial functions, by taking natural coordinate system 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, Polynomial functions,

Shape functions.

I. INTRODUCTION

Initially shape functions were derived interms of Cartesian coordinates. Polynomial functions were used for this. After natural coordinates were identified and its advantage was noticed researchers started deriving shape functions interms of natural coordinates. By this approach more elements could be developed [1]. By increasing more number of nodes we can minimize error in Finite Element Method [3]. In this computation one of the important observation is one dimensional polynomial changes to Lagrange interpolation formula.

II. GEOMETRICAL DESCREPTION

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

(2)

1 2 typical bar element in the natural coordinate x varying from -1 to 1 is shown in Figure.1.In figure.1 nodal unknowns are displacements u and u along x-axis. For this element we have to select po

The

lynomial with only two constants to represent displacement at any point in the elements.Since there are only two nodal values, a linear polynomial is to be selected. Let displacement at any point P(x) be

1 2

u  x

 

₁

 

1 2

u x _

    

₁ 1 ₁ ₁

1

2 2 2

u x

  

 

1 ₁ :

1 ₂

x A

x



: [ ] // Form

b Inverse A Matrix

B

2 1

1 2 1 2

1 1

1 2 1 2

Output

x x

x x x x



   



   

 

 

 

2 1

1 2 1 2

:

1 1

1 2 1 2

x x

x x x x

d

x x x x



   

 

   

 

 

 





: 1

: .

e x

f e d

 

[ ]//

FullSimplify f MatrixForm

    

2 1

1 2 1 2

: x x x x

a

x x x x

    

  _ _ 

 

a



1 1



, 2 2

Output

x x

 

 

 

 

1 2 noded element shape functions are N₁

2

1 and N2

2

x For

x



 

 

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

st

Verification I

1 1

: , N :

1 2

2 2

x x

N    

1 2

[ ]

1

FullSimplify N N 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

typical bar element in the natural coordinate x varying from -1 to 1 is shown in Figure.2. Since there are only

(3)

2

1 2 3

u  x x



₂



1

1 ₂

3

u x x

   

         

 

2 1 ₁ ₁

1 2 1

1

2 2 2 2

2

3 1 ₃ ₃ 3

x x

u

u x x

u

x x

   

 

  _ _ 

  _ _ 

  _ _ 

 

2 1 ₁ ₁

2 : 1 ₂ ₂

2 1 ₃ ₃

x x

A x x

x x



 

 

 





: [ ]; 2

: 1 ;

: . ;

b Inverse A

e x x

f e b

  

[ ]//





































2 3 1 3

1 2 1 3 1 2 2 3

1 2

1 3 2 3

Output

x x x x x x x x

x x x x

   



   

  

  

  

  

1: 1 2: 0 3: 1

x   x  x 

  

     

  

: 2 3 1 3

1 2 1 3 1 2 2 3

1 2

1 3 2 3

a x x x x x x x x

x x x x x x x x x x x x

x x x x

_     

   





  _

 _

   _



 

   



1 1



1 , 1 1 , 1

2 2

a

Output

x x x x x x

      

 

 

 







 

 

3 noded element shape functions are

1

: 1 x N : 1 1

1 2

2

1 N₃: 1

2

For

N x x x

x x



       

 

st

Verification I

1 2 3

[ ]

1

FullSimplify N N N 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.

typical bar element in the natural coordinate x varying from -1 to 1 is shown in Figure.3.Since there are only four nodal values, Hence a polynomial with 4 generalized coordinates

is to be selecte

The

d. Let displacement at any point P(x) be

2 3

1 2 3 4

(4)



2 3



1₂

1

3 4

u x x x

              

   

2 3

1 ₁ ₁ ₁

1 2 3 1

1 ₂ ₂ ₂

2 2

2 3

3 1 ₃ ₃ ₃ 3

2 3

4 4

1 ₄ ₄ ₄

x x x

u

x x x

u

u _x _x _x

u

x x x

         _ _    _ _    _ _    _ _    _ _ 

   

2 3

1 ₁ ₁ ₁

2 3

1 ₂ ₂ ₂

: ₂ ₃

1 ₃ ₃ ₃

2 3

1 ₄ ₄ ₄

x x x

A

x x x

              





: [ ]; 2 3

: 1 ;

: . ;

b Inverse A

e x x x

f e b

  

[ ] //









































































1 2 4

1 3 2 3 3 4

1 2 3

1 4 2 4 3 4

2 3 4

1 2 1 3 1 4

1 3 4

1 2 2 3 2 4

Output

x x x x x x

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

1 2 3 4

1 1

: 1 : : : 1

3 3

x   x   x  x 

















































































2 3 4

:

1 2 1 3 1 4

1 3 4

1 2 2 3 2 4

1 2 4

1 3 2 3 3 4

1 2 3

1 4 2 4 3 4

x x x x x x

a

x x x x x x

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









 





 

9 1 1

1 ,

16 3 3

27 1

1 1 ,

16 3

27 1

1 1 ,

16 3

Output

x x x

                     _ _ _              

 

9 1 1

1

16 3 3

x x x

    

  



  

   

For 





9 1 1

: 1

1

16 3 3

N    x  x x

  





 

27 1

: 1 1

2

16 3

N   x  x x

 





 

27 1

: 1 1

3

16 3

N    x  x x

 

 

9 1 1

: 1

4

16 3 3

N   x x x

  

Condition : Sum of all the shape functions

Verification

(5)

[ ₁ ₂ ₃ ₄]

1

FullSimplify N N N N

Output

  

nd

II

1

2 3 4

N 0, N 0, 0

At

N



  

1

2 3 4

1

Node 2 x = - then we get N 0, 3

N 1, N 0, N 0

At 

  

1

2 3 4

1

Node 3, x = then we get N 0, 3

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.

typical bar element in the natural coordinate x varying from -1 to 1 is shown in Figure.3. Since there are only four nodal values, Hence a polynomial with 4 generalized coordinates is to be selected.

The

Let displacement at any point P(x) be

2 3 4

5

1 2 3 4

u  x x  x  x





1 2 2 3 4

1 ₃

4 5

u x x x x

     

             

     

2 3 4

1 ₁ ₁ ₁ ₁

2 3 4

1 1

1 ₂ ₂ ₂ ₂

2 ₂ ₃ ₄ 2

1

3 3 3 3 3 3

2 3 4

4 ₁ 4

4 4 4 4

2 3 4

5 5

1 ₅ ₅ ₅ ₅

x x x x

u

x x x x

u

u x x x x

u

x x x x

u

x x x x

     

 

    

    

  _ _ 

  _ _ 

 

     

2 3 4

1 ₁ ₁ ₁ ₁

2 3 4

1 ₂ ₂ ₂ ₂

2 3 4

: 1 ₃ ₃ ₃ ₃

2 3 4

1 ₄ ₄ ₄ ₄

2 3 4

1 ₅ ₅ ₅ ₅

x x x x

A x x x x

x x x x



 

 

 





: [ ]; 2 3 4

: 1 ;

: . ;

b Inverse A

e x x x x

f e b

  

[ ]//





























































2 3 4 5

1 2 1 3 1 4 1 5

1 3 4 5

1 2 2 3 2 4 2 5

1 2 4 5

1 3 2 3 3 4 3 5

1 2 3 5

1 4 2 4 3 4 4 5

1 2 3 4

1 5 2 5 3 5 4 5

Output

x x x x x x x x

   



   



    

   



     

   



      

  

  

1 1

: 1 : : 0 : : 1

1 2 3 4 5

4 4

(6)





























































2 3 4 4 :

1 2 1 3 1 4 1 5

1 3 4 5

1 2 2 3 2 4 2 5

1 2 4 5

1 3 2 3 3 4 3 5

1 2 3 5

1 4 2 4 3 4 4 5

1 2 3 4

1 5 2 5 3 5 4 5

x x x x x x x x

a

x x x x x x x x

a

   



   



   



    

   



     

   



      

  

  









 

8 1 1

1 ,

15 4 4

128 1

1 1 ,

15 4

Output

x x x x

    

     

    

    

   

 

 

 

 





 





 

1 1

16 1 1

4 4

128 1

1 1 ,

15 4

x x x x

x x x x

     

    

  

  

  

 

 

 

 

8 1 1

1

15 4 4

x x x x

    

   



   

    

For 





8 1 1

: 1

1

15 4 4

N   x  x  x x

   





 

128 1

: 1 1

2

15 4

N    x  xx x

 





1 1

 

: 16 1 1

3

4 4

N   x  x x x

  





 

128 1

: 1 1

4

N    x  x x

 

 

8 1 1

: 1

5

15 4 4

N   x  x x x

   

Verification

st I

[ ₁ ₂ ₃ ₄ ₅]

1

FullSimplify N N N N N

Output

   

nd

II

1

2 3 4 5

N 0, N 0, 0, 0

At

N N



   

1

2 3 4 5

1

Node 2 x = - then we get N 0, 4

N 1, N 0, N 0, N 0

At 

   

1

2 3 4 5

N 0, N 1, N 0, N 0

At 

   

1

2 3 4 5

1

Node 4, x = then we get N 0, 4

N 0, N 0, N 1, N 0

At 

   

1

2 3 4 5

N 0, N 0, N 0, N 1

At 

   

V. CONCLUSIONS

1. Derived Shape functions for 2,3,4,5 noded line element by polynomial.

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

(7)

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

Edition, McGraw Hill International Editions, 1993. [4]. S.Md.Jalaludeen, Introduction of Finite Element Analysis,

Anuradha Publications, 2012.

[5]. P. Reddaiah, Deriving shape functions for 8-noded rectangular serendipity element in horizontal channel geometry and verified, International Journal of Mathematics Trends and Technology (IJMTT), Volume 50, Number 2 , October 2017.

[6]. P. Reddaiah, Deriving shape functions for 9-noded rectangular element by using lagrange functions in natural coordinate system and verified, International Journal of Mathematics Trends and Technology (IJMTT),Volume 51, Number 6, November 2017.

[7]. P. Reddaiah, Deriving shape functions for Hexahedral element by natural coordinate system and Verified,

International Journal of Mathematics Trends and Technology (IJMTT), Volume 51, Number 6, November 2017.

[8]. P. Reddaiah, Deriving shape functions for hexahedron element by lagrange functions and verified, International Journal of Mathematics Trends and Technology (IJMTT), Volume 51, Number 6, November 2017.

[9]. P. Reddaiah, Deriving shape functions for 2,3,4,5 noded line element by lagrange functions and verified, International Journal of Mathematics Trends and Technology (IJMTT), Volume 51, Number 6, November 2017.

[10]. P. Reddaiah and D.R.V. Prasada Rao, Deriving Vertices, Shape Functions for Elliptic Duct Geometry and Verified Two Verification Conditions, International Journal of Scientific & Engineering Research, Volume 8, Issue 5, May-2017.