Deriving Shape Functions and Verified
Hermite Functions for a Two Node Element
with Three Primary Variables at Each Node
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 a two node element with three primary variables
, , , w
W k
x
and also I verified three
verification conditions for shape functions. First verification condition at node 1 is
1 2 3 4 5
6
( ) 1 and 0, 0, 0, 0,
0
i N N N N N
N
3
2 1 4
5 6
( ) 1 and 0, 0, 0,
0, 0
N
N N N
ii
x x x x
N N
x x
2 2 2
3 1 2
2 2 2
2 2
2
5 6
4
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
iii
x x x
N N
N
x x x
Second Verification condition at node 2 is
4 1 2 3
5 6
( ) 1 and 0, 0, 0,
0, 0
iv N N N N
N N
5 1 2
3 4 6
( ) 1 and 0, 0,
0, 0, 0
N N N
v
x x x
N N N
x x x
2 2 2
6 1 2
2 2 2
2 2 2
3 4 5
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
vi
x x x
N N N
x x x
Third Verification condition is N1N4 1. For computational purpose I used Mathematica 9 Software [2].
Keywords — Primary variables, Hermite Functions, Two node element, Shape functions.
I. INTRODUCTION
Two-dimensional problems involve the specification of the secondary variables on discrete portions of the boundary of the domain. In such cases, one can use appropriate one-dimensional interpolation functions to compute the contributions of the specified secondary variables to the nodal force.
II. GEOMETRICAL DESCRIPTION
.1 Two noded Beam element
varying from 0 to h
Figure
2 2 two noded beam element shown in Figure.1 in which nodal unknowns are
W the displacement W and Slope =
x W and primary unknown k=
x
A
III. DERIVING SHAPE FUNCTIONS FOR A TWO NODE ELEMENT WITH THREE PRIMARY VARIABLES AT EACH NODE
Since the element in figure.1 has six degrees of freedom,We have to select the polynomial with only 6 constants. In this polynomial after substituting boundary conditions we get shape functions this we can take as first order Hermitian Polynomials as shape functions.
2 3
1 2 3 4
4 5
5 6
( )
+A x +A x (1)
W x A A xA x A x
1 2 3 4 5 6
W is the transverse displacement and A , , , ,A ,A are polynomial Coefficients
Where
A A A
If higher order (i.e., higher than cubic) approximation of w is desired, one must identify additional dependent (primary) unknowns at each of the two nodes. For example, if we a
2 2 dd
the primary unknown at each of the two nodes, there will be total of six conditions, and a fifth-order polynomial is required to interpolate the end conditions.
w as x
partially eq(1) w.r.t. 'x'
Differentiating
2
2 3 4
3 4
5 6
(1) 0 (1) (2 ) (3 )
(4 ) (5 )
W
A A x A x
x
A x A x
2 3 4
2 2 3 3 4 +4 5 5 6 (2)
W
A A x A x A x A x
x
partially eqation(2) w.r.t. 'x'
Differentiating
2
2
3 4 5
2
3 6
0 2 (1) 3 (2 ) 4 (3 )
5 (4 )
w
A A x A x
x
A x
2
2 3
3 4 5 6
2 2 6 12 20 (3)
W
A A x A x A x
x
2
1 1 2 1
the nodal conditions such that W
W=W , and =k at x= 0 x
Applying
W x
2
2 2 2 2
W
W=W and , =k
x
at x=h , where h is step length
W and
x
equations (1), (2) and (3) we get
in
1
W=W and x = 0
When
2 3
1 1 2 3 4
(1)W AA(0)A(0) A(0)
1 1 0 0 0
W A
1 1 (4)
W A
1 W
and x = 0 x
When
2
1 2 3 4
(2) A 2A(0) 3 A (0)
1 A2+0+0
1 A2 (5)
2 1
2 at x=0
W
When k
x
1 3
(3) k 2A (6)
2
W=W and x = h
When
2 3
2 1 2 3 4
4 5
5 6
(1) ( )
+A h +A h (7)
W A A h A h A h
2 W
and x = h x
When
2 3
2 2 3 4 5
4 6
(2) 2 3 +4A h
+5A h (8)
A A h A h
2 2
2 and x=h
W
When k
x
2 3
3 4 5 6
(3)2A 6A h12A h 20A h (9)
1 2 3 4 5 6
sin Mathematica 9 Software Solving (4),(5) ,(6),(7),(8) and (9) we get A , ,
, , ,
U g
A
A A A A
1 1 2 1
3 1 1 2
2 3 4 5
3 4 5 6
2 2 3 4
2 3 4
5 6 2
2
2 3 4 5
3
6 2 1 2 3 4
5 6
[ 0 & & 0
& & 2 * 0 & & *
* * * *
0 & & 2 * * 3*
* 4 * * 5* * 0
& & 2 * 6 * * 12 * *
20 * * 0,{ , , , ,
, }]
Input
Solve A W A
A k A A h
A h A h A h A h
W A A h A
h A h A h
A A A h A h
A h k A A A A
A A
1 1 2 1
{{ , ,
Output
A W A
1
3 4
2 2
3 1 2 20 1 20 2 12 1 8 2 , 3
2 ,
2
h k h k W W h h
h k
A A
5
2 2
3 1 2 2 30 1 30 2 16 1 14 2 4
2
,
h k h k W W h h
h A
6
2 2
12 12 6 6
1 2 1 2 1 2
}} 5
2
h k h k W W h h
h
A
1 2 3 4 5 6
A , , , ,A ,A in eq(1)
Substituting A A A
1 1 2 1
: :
A W
A
1 3
4
2 2
3 1 2 20 1 20 2 12 1 8 2 3
2 :
2
: h k h k W W h h
h k
A
A
5
2 2
3 1 2 2 30 1 30 2 16 1 14 2 4
2
: h k h k W W h h
h
A
6
2 2
12 12 6 6
1 2 1 2 1 2
5 2
: h k h k W W h h
h
A
2 3
1 2 3 4
4 5
5 6
( ) : * * *
* *
W x A A x A x A x
A x A x
[ ( )]
Expand W x
Output
2 3 4 5 3
1 1 1 1 2
2 3
4 5 3 4 5
2 2 1 1 1
1
2 3 3 4 5
3 3
2 2 2 2 2
10 15 6
2
x k x k x k x k x k
h h h h
x k x k x w x W x W
W
h h h h h
3 4 5 3
2 2 2 1
1
3 4 5 2
4 5 3 4 5
1 1 2 2 2
3 4 2 3 4
10 15 6 6
8 3 4 7 3
x W x W x W x
x
h h h h
x x x x x
h h h h h
2 3 4 5
1 1 1 1
2 3
3 4 5 3
2 2 2 1
1
2 3 3
3 3
( )
2 2 2 2
10
2 2
x k x k x k x k
W x
h h h
x k x k x k x w
W
h h h h
4 5 3 4
1 1 2 2
4 5 3 4
5 3 4 5
2 1 1 1
1
5 2 3 4
3 4 5
2 2 2
2 3 4
15 6 10 15
6 6 8 3
4 7 3
x W x W x W x W
h h h h
x W x x x
x
h h h h
x x x
h h h
1 33 44 553 4 5
1 2 3 4
2 3 4 5
1 2 3
3 4 5
2 3 4 5
3 4 5
2
2 2 3 4
3 4 5
2 2 3
10 15 6
1
6 8 3
3 3
2 2 2 2
10 15 6
4 7 3
+
(10)
2 2
x x x
W x W
h h h
x x x
x
h h h
x x x x
k
h h h
x x x
W
h h h
x x x
h h h
x x x
k
h h h
6
. .,
W=N1 1 2 1 3 1 N4 2 5 2 2
(11)
=N1 1 2 2 3 3 N4 4 5 5 6 6
i e
W N N k W N N k
N N N N
1 2 3 4 5 6
1 2 3 4 5 6
N , , , ,N ,N are shape functions for the beam elements and , , , , , are the nodal displacements
Where N N N
1 1
2 1
3 1
4 2
5 2
2 6
. ., { }=
W
k i e
W
k
(10) and (11) we get
Comparing
3 4 5
2
1 01 3 4 5
10 15 6
( ) 1 x x x (12)
N H x
h h h
3 4 5
2
2 11 2 3 4
6 8 3
( ) x x x (13)
N H x x
h h h
2 3 4 5
2
3 21 2 3
3 3
( ) (14)
2 2 2 2
x x x x
N H x
h h h
3 4 5
2
4 02 3 4 5
10 15 6
( ) x x x (15)
N H x
h h h
3 4 5
2 2
5 12 2 3 4
4 7 3
( ) x x x (16)
N H x
h h h
3 4 5
2
6 22( ) 2 3 (17)
2 2
x x x
N H x
h h h
Here
2 th
01
H ( ), 0 represents Zero order
derivative, 1 represents node number
one and power 2 represents second order
Hermitian function.
In x
2 11
H ( ), 1 represents first order
derivative, 1 represents node number
one and power 2 represents second order
Hermitian function.
In x
2 21
H ( ), 2 represents second order
derivative, 1 represents node number
one and power 2 represents second order
Hermitian function.
In x
2 th
02
H ( ), 0 represents Zero order
derivative, 2 represents node number
two and power 2 represents second order
Hermitian function.
In x
2 12
H ( ), 1 represents first order
derivative, 2 represents node number
two and power 2 represents second order
Hermitian function.
2 22
H ( ), 2 represents second order
derivative, 2 represents node number
two and power 2 represents second order
Hermitian function.
In x
IV. VERIFICATION
(I). 1st VERIFICATION CONDITION First verification condition at node 1 is
1 2 3 4
5 6
( ) 1 and 0, 0, 0,
0, 0
i N N N N
N N
3
2 1
5 6
4
( ) 1 and 0, 0,
0, 0, 0
N
N N
ii
x x x
N N
N
x x x
2 2 2
3 1 2
2 2 2
2 2
2
5 6
4
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
iii
x x x
N N
N
x x x
Node 1, x = 0
At
Input
3 4 5
2
1 01 3 4 5
10 15 6
( ) : 1 x x x
N H x
h h h
3 4 5
2
2 11 2 3 4
6 8 3
( ) : x x x
N H x x
h h h
2 3 4 5
2
3 21 2 3
3 3
( ) :
2 2 2 2
x x x x
N H x
h h h
3 4 5
2
4 02 3 4 5
10 15 6
( ) : x x x
N H x
h h h
3 4 5
2 2
5 12 2 3 4
4 7 3
( ) : x x x
N H x
h h h
3 4 5
2
6 22( ) : 2 3
2 2
x x x
N H x
h h h
: 0
x
1
N
2
N
3
N
4
N
5
N
6
N
Output
1 0 0 0 0 0
first derivatives for (12),(13),(14) , (15),(16), and (17)
Finding
Input
3 4 5
2
1 01 3 4 5
10 15 6
( ) : 1 x x x
N H x
h h h
3 4 5
2
2 11 2 3 4
6 8 3
( ) : x x x
N H x x
h h h
2 3 4 5
2
3 21 2 3
3 3
( ) :
2 2 2 2
x x x x
N H x
h h h
3 4 5
2
4 02 3 4 5
10 15 6
( ) : x x x
N H x
h h h
3 4 5
2 2
5 12 2 3 4
4 7 3
( ) : x x x
N H x
h h h
3 4 5
2
6 22( ) : 2 3
2 2
x x x
N H x
h h h
1 ( )
x N
2 ( )
x N
3 ( )
x N
4 ( )
x N
5 ( )
x N
6 ( )
x N
Output
2 3 4
3 4 5
30x 60x 30x
h h h
2 3 4
2 3 4
18 32 15
1 x x x
h h h
2 3 4
2 3
9 6 5
2 2
x x x
x
h h h
2 3 4
3 4 5
30x 60x 30x
h h h
2 3 4
2 3 4
12x 28x 15x
h h h
2 3 4
2 3
3 4 5
2 2
x x x
h h h
2 3 4
1 3 4 5
30 60 30
( ) (18)
x
x x x
N
h h h
2 3 4
2 2 3 4
18 32 15
( ) 1 (19)
x
x x x
N
h h h
2 3 4
3 2 3
9 6 5
( ) (20)
2 2
x
x x x
N x
h h h
2 3 4
4 3 4 5
30 60 30
( ) (21)
x
x x x
N
h h h
2 3 4
5 2 3 4
12 28 15
( ) (22)
x
x x x
N
h h h
2 3 4
6 2 3
3 4 5
( ) (23)
2 2
x
x x x
N
h h h
first derivative condition
at node 1, x = 0
Partial
Input
2 3 4
1
3 4 5
30 60 30
:
N x x x
x h h h
2 3 4
2
2 3 4
18 32 15
: 1
N x x x
x h h h
2 3 4
3
2 3
9 6 5
:
2 2
N x x x
x
x h h h
2 3 4
4
3 4 5
30 60 30
:
N x x x
x h h h
2 3 4
5
2 3 4
12 28 15
:
N x x x
x h h h
2 3 4
6
2 3
3 4 5
:
2 2
N x x x
x h h h
x = 0
1
N x
2
N x
3
N x
4
N x
5
N x
6
N x
Output
0 1 0 0 0 0
second derivatives for (12),
(13),(14) , (15),(16), and (17)
Finding
3 4 5
2
1 01 3 4 5
10 15 6
( ) : 1 x x x
N H x
h h h
3 4 5
2
2 11 2 3 4
6 8 3
( ) : x x x
N H x x
h h h
2 3 4 5
2
3 21 2 3
3 3
( ) :
2 2 2 2
x x x x
N H x
h h h
3 4 5
2
4 02 3 4 5
10 15 6
( ) : x x x
N H x
h h h
3 4 5
2 2
5 12 2 3 4
4 7 3
( ) : x x x
N H x
h h h
3 4 5
2
6 22( ) : 2 3
2 2
x x x
N H x
h h h
, 1
x x N
, 2
x x N
, 3
x x N
, 4
x x N
, 5
x x N
, 6
x x N
Output
2 3
3 4 5
60x 180x 120x
h h h
2 3
2 3 4
36x 96x 60x
h h h
2 3
2 3
9 18 10
1 x x x
h h h
2 3
3 4 5
60x 180x 120x
h h h
2 3
2 3 4
24x 84x 60x
h h h
2 3
2 3
3x 12x 10x
h h h
2 3, 1 3 4 5
60 180 120
x x
x x x
N
h h h
2 3, 2 2 3 4
36 96 60
x x
x x x
N
h h h
2 3, 3 2 3
9 18 10
1
x x
x x x
N
h h h
2 3, 4 3 4 5
60 180 120
x x
x x x
N
h h h
2 3, 5 2 3 4
24 84 60
x x
x x x
N
h h h
2 3, 6 2 3
3 12 10
x x
x x x
N
h h h
second derivative condition
at node 1, x = 0
Partial
2 2 3
1
2 3 4 5
60 180 120
:
N x x x
x h h h
2 2 3
2
2 2 3 4
36 96 60
:
N x x x
x h h h
2 2 3
3
2 2 3
9 18 10
: 1
N x x x
x h h h
2 2 3
4
2 3 4 5
60 180 120
:
N x x x
x h h h
2 2 3
5
2 2 3 4
24 84 60
:
N x x x
x h h h
2 2 3
6
2 2 3
3 12 10
:
N x x x
x h h h
: 0
x
2 1 2
N x
2 2 2
N x
2 3 2
N x
2 4 2
N x
2 5 2
N x
2 6 2
N x
0
0
1
0
0
0
Output
(II) 2nd VERIFICATION CONDITION Second Verification condition at node 2 is
4 1 2 3
5 6
( ) 1 and 0, 0, 0,
0, 0
iv N N N N
N N
5 1 2
3 4 6
( ) 1 and 0, 0,
0, 0, 0
N N N
v
x x x
N N N
x x x
2 2 2
6 1 2
2 2 2
2 2 2
3 4 5
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
vi
x x x
N N N
x x x
Node 2, x = h
At
3 4 5
2
1 01 3 4 5
10 15 6
( ) : 1 x x x
N H x
h h h
3 4 5
2
2 11 2 3 4
6 8 3
( ) : x x x
N H x x
h h h
2 3 4 5
2
3 21 2 3
3 3
( ) :
2 2 2 2
x x x x
N H x
h h h
3 4 5 2
4 02 3 4 5
10 15 6
( ) : x x x
N H x
h h h
3 4 5
2 2
5 12 2 3 4
4 7 3
( ) : x x x
N H x
h h h
3 4 5
2
6 22( ) : 2 3
2 2
x x x
N H x
h h h
:
x h
1
N
2
N
3
N
4
N
5
N
6
N
Output
0 0 0 1 0 0
first derivative condition at node 2, x = h
Partial
2 3 4
1
3 4 5
30 60 30
:
N x x x
x h h h
2 3 4
2
2 3 4
18 32 15
: 1
N x x x
x h h h
2 3 4
3
2 3
9 6 5
:
2 2
N x x x
x
x h h h
2 3 4
4
3 4 5
30 60 30
:
N x x x
x h h h
2 3 4
5
2 3 4
12 28 15
:
N x x x
x h h h
2 3 4
6
2 3
3 4 5
:
2 2
N x x x
x h h h
x = h
1
N x
2
N x
3
N x
4
N x
5
N x
6
N x
Output
second derivative condition
at node 2, x = h
Partial
2 2 3
1
2 3 4 5
60 180 120
:
N x x x
x h h h
2 2 3
2
2 2 3 4
36 96 60
:
N x x x
x h h h
2 2 3
3
2 2 3
9 18 10
: 1
N x x x
x h h h
2 2 3
4
2 3 4 5
60 180 120
:
N x x x
x h h h
2 2 3
5
2 2 3 4
24 84 60
:
N x x x
x h h h
2 2 3
6
2 2 3
3 12 10
:
N x x x
x h h h
:
x h
2 1 2
N x
2 2 2
N x
2 3 2
N x
2 4 2
N x
2 5 2
N x
2 6 2
N x
0
0
0
0
0
1
Output
Verification Condition
Third
1 4 3 verification condition is Nrd N 1
3 4 5
2
1 01 3 4 5
10 15 6
( ) : 1 x x x
N H x
h h h
3 4 5
2
4 02 3 4 5
10 15 6
( ) : x x x
N H x
h h h
1 4
[ ]
1
FullSimplify N N
Output
V. CONCLUSIONS 1. Derived Shape Functions for a
two node element with three
primary variables at each node.
2. Verified First verification condition at node 1
1 2 3 4
5 6
( ) 1 and 0, 0, 0,
0, 0
i N N N N
N N
3
2 1
5 6
4
( ) 1 and 0, 0,
0, 0, 0
N
N N
ii
x x x
N N
N
x x x
2 2 2
3 1 2
2 2 2
2 2
2
5 6
4
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
iii
x x x
N N
N
x x x
3. Verified Second Verification condition at node 2
4 1 2 3
5 6
( ) 1 and 0, 0, 0,
0, 0
iv N N N N
N N
5 1 2
3 4 6
( ) 1 and 0, 0,
0, 0, 0
N N N
v
x x x
N N N
x x x
2 2 2
6 1 2
2 2 2
2 2 2
3 4 5
2 2 2
( ) 1 and 0, 0,
0, 0, 0
N N N
vi
x x x
N N N
x x x
1 4
4. Verified Third verification condition N N 1.
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 forHexahedral 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.
