# MANE 4240 CIVL 4240 Introduction to Finite Elements

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```					    MANE 4240 & CIVL 4240
Introduction to Finite Elements

Prof. Suvranu De

Higher order elements

Lecture notes

Summary:

• Properties of shape functions
• Higher order elements in 1D
• Higher order triangular elements (using area coordinates)
• Higher order rectangular elements
Lagrange family
Serendipity family
Recall that the finite element shape functions need to satisfy the
following properties

1. Kronecker delta property
 1 at node i
Ni  
0 at all other nodes
Inside an element
u  N1u1  N 2 u 2  ...
At node 1, N1=1, N2=N3=…=0, hence

u node 1  u1

Facilitates the imposition of boundary conditions
2. Polynomial completeness

If   u  1   2 x   3 y
Then
N
i
i   1

N x
i
i   i   x

N y
i
i   i   y
Higher order elements in 1D

2-noded (linear) element:

x  x2
N1 
x1           x2                x1  x 2
x               x  x1
N2 
1             2                   x 2  x1

In “local” coordinate system (shifted to center of element)

ax
N1 
x             2a
ax
1            2                N2 
a   a                          2a
3-noded (quadratic) element:                x  x2 x  x3 
N1 
x1 x3      x2                        x1  x2 x1  x3 
x                 x  x1 x  x3 
N2 
1   3       2                         x2  x1 x2  x3 
x  x1 x  x2 
N3 
x3  x1 x3  x2 
In “local” coordinate system (shifted to center of element)
x1  a ; x 2  0 ; x3  a
x
xa  x 
N1  
1     3      2                      2a 2
a a                          xa  x 
N2 
2a 2
a2  x2
N3 
a2
x  x2 x  x3 x  x4 
N1 
4-noded (cubic) element:        x1  x2 x1  x3 x1  x4 
x  x1 x  x3 x  x4 
N2 
x1 x3 x4 x2               x2  x1 x2  x3 x2  x4 
x        x  x1 x  x2 x  x4 
N3 
1 3 4 2                    x3  x1 x3  x2 x3  x4 
x  x1 x  x2 x  x3 
N4 
x4  x1 x4  x2 x4  x3 
In “local” coordinate system (shifted to center of element)
9
N1          ( x  a)(x  a / 3)(x  a / 3)
2a/3 2a/3 2a/3         x        16a  3

9
N2         ( x  a)( x  a / 3)( x  a / 3)
1       3   4   2              16a  3
a       a
27
N3       3
(a  x)(a / 3  x)(a  x)
16a
27
N4        3
( x  a)(x  a / 3)( x  a / 3)
16a
Polynomial completeness

1                                        Convergence
rate (displacement)
x                           2 node; k=1; p=2
x2
3 node; k=2; p=3
x3                          4 node; k=3; p=4
x4

Recall that the convergence in displacements
u  uh 0  Ch p ; p  k 1
k=order of complete polynomial
Triangular elements

Area coordinates (L1, L2, L3)
1
Total area of the triangle A=A1+A2+A3

y               P A2                At any point P(x,y) inside
A3                     the triangle, we define
A1                         A1
3       L1 
A
2                                   A2
L2 
A
x
A3
L3 
A
Note: Only 2 of the three area coordinates are independent, since
L1+L2+L3=1
ai  bi x  ci y
Li 
2A

1 x 1             y1 
A  area of triangle  det 1 x 2             y2 
1
2                         
1 x 3
                  y3 

a1  x 2 y 3  x3 y 2   b1  y 2  y 3   c1  x3  x 2
a 2  x3 y1  x1 y 3    b2  y 3  y1    c 2  x1  x3
a3  x1 y 2  x 2 y1    b3  y1  y 2    c3  x 2  x1
Check that

L1  L2  L3  1
L1 x1  L2 x2  L3 x3  x
L1 y1  L2 y 2  L3 y3  y
1
L1= constant
P‟
y                P
A1
3
2
x

Lines parallel to the base of the triangle are lines of constant „L‟
L1= 1
1

y              P                      L1= 0
A1
3
2                                   L2= 0
L3= 1
x L =1
2
L3= 0

We will develop the shape functions of triangular elements in
terms of the area coordinates
For a 3-noded triangle

N1  L1
N 2  L2
N 3  L3
For a 6-noded triangle

L2= 0                                           L1= 1
1
L2= 1/2                                             L1= 1/2
6
4
y                                             L1= 0
L2= 1
3
5
2
x
L3= 0   L3= 1/2       L3= 1
How to write down the expression for N1?
Realize the N1 must be zero along edge 2-3 (i.e., L1=0) and at
nodes 4&6 (which lie on L1=1/2)

N1  cL1  0L1  1 / 2
Determine the constant „c‟ from the condition that N1=1 at
node 1 (i.e., L1=1)

N1 (at L1  1)  c11  1 / 2   1
c2
 N1  2 L1 L1  1 / 2 
N1  2 L1 ( L1  1 / 2)
N 2  2 L2 ( L2  1 / 2)
N 3  2 L3 ( L3  1 / 2)
N 4  4 L1 L2
N 5  4 L3 L2
N 6  4 L3 L1
For a 10-noded triangle

L2= 0                                      L1= 1
L2= 1/3               1                     L1= 2/3
9
L2= 2/3           4                         L1= 1/3
8
y               10                  L1= 0
L2= 1         5
3
7
2       6
x
L3= 0 L3= 1/3 L3= 2/3 L3= 1
9
N1      L1 ( L1  1 / 3)( L1  2 / 3)
2
9
N2       L2 ( L2  1 / 3)( L2  2 / 3)
2
9
N3      L3 ( L3  1 / 3)( L3  2 / 3)
2
27
N4        L1 L2 ( L1  1 / 3)
2
27
N5        L1 L2 ( L2  1 / 3)
2
27
N6        L2 L3 ( L2  1 / 3)
2
27
N7        L2 L3 ( L3  1 / 3)
2
:
N 10  27 L1 L2 L3
NOTES:
1. Polynomial completeness
1                            Convergence
rate (displacement)
x         y         3 node; k=1; p=2
x2       xy        y2    6 node; k=2; p=3
x3    x2 y        xy 2 y 3   10 node; k=3; p=4

2. Integration on triangular domain
k! m! n!
   L1 L2 L3 dA  2 A
k     m   n
1.
1              A                            (2  k  m  n)!
k! m!
2.              L1 L2 dS  l1 2
k   m
l1-2                        1 2 edge                      (1  k  m)!
y

3
2
x
3. Computation of derivatives of shape functions: use chain rule
e.g.,
N i N i L1 N i L2 N i L3
                
x   L1 x L2 x L3 x
L1   b1        L2 b2 L3 b3
But             ;              ;   
x 2 A           x   2 A x   2A

e.g., for the 6-noded triangle

N 4  4 L1 L2
N 4        b2        b1
       4 L1     4 L2
x         2A        2A
Rectangular elements
Lagrange family
Serendipity family

Lagrange family

4-noded rectangle       In local coordinate system
y
2               1               ( a  x )(b     y)
a     a             N1 
b                                     4 ab
( a  x )(b      y)
x   N2   
b                                    4 ab
( a  x )(b     y)
3               4        N3   
4 ab
( a  x )(b      y)
N4   
4 ab
Corner nodes
 x(a  x)   y (b  y )         x(a  x)   y (b  y ) 
y 5                N1                              N 2  
 2a   2b 
                         2a 2   2b 2 
            
2             2

2        a         1
a                     x(a  x)   y (b  y )            x(a  x)   y (b  y ) 
N 3                              N4                
b                                       2a 2         2b 2 
           2a 2  
     2b 2   
9      8
6                  x
b                             Midside nodes
7                      a 2  x 2   y (b  y )           x(a  x)   b  y 
2     2
3                 4         N5                               N 6                           
    a 2   2b 2 
                           2a 2   b 2 

 a 2  x 2   y (b  y )             x(a  x)   b  y 
2     2
N7                               N8                          
    a 2          2b 2             2a   b
2            2


Center node
 a 2  x2   b2  y 2 
N9        2         2    
 a  b 
NOTES:
1. Polynomial completeness
Convergence
1                          rate (displacement)
x         y            4 node; p=2
x2        xy       y2       9 node; p=3
x3    x2 y        xy 2 y 3
x4    x3 y         x 2 y 2 xy 3 y 4
x 5 x 4 y x 3 y 2 x 2 y 3 xy 4 y 5

Lagrange shape functions contain higher order terms but miss
out lower order terms
Serendipity family

4-noded same as Lagrange
8-noded rectangle: how to generate the shape functions?
First generate the shape functions of the
y 5                 midside nodes as appropriate products of
2     a     a     1
1D shape functions, e.g.,
b
8
x N   a  x   (b  y)  ; N   (a  x)   b  y 
2     2                              2   2
6
                       2a   b 2 
a 2   2b 
5                         8
b       7                                                              
3                 4 Then go to the corner nodes. At each corner
node, first assume a bilinear shape function as in
a 4-noded element and then modify:
ˆ  (a  x)(b  y )
“bilinear” shape fn at node 1: N1               4ab
actual shape fn at node 1:                  ˆ N N
N1  N1  5  8
2    2
8-noded rectangle
Midside nodes
y 5
2         a    a   1               a 2  x 2   (b  y )          (a  x)   b  y 
2   2
N5                         N 6   2a   b 2 
  2b                        
2
b                                  a                                                      
8
6                  x         a 2  x 2   (b  y )         (a  x)   b  y 
2   2
N7                         N 8   2a   b 2 
b          7                      a
2
  2b                                  
3                  4

Corner nodes

(a  x)(b  y ) N 5 N 8       (a  x)(b  y ) N 5 N 6
N1                         N2                        
4ab         2   2             4ab            2      2
(a  x)(b  y ) N6 N 7       (a  x)(b  y ) N8 N 7
N3                        N4                        
4ab         2   2            4ab           2      2
NOTES:
1. Polynomial completeness
Convergence
1                                              rate (displacement)
x                y                         4 node; p=2
x2            xy           y2                    8 node; p=3
x3     x2 y         xy 2 y 3                          12 node; p=4
x4       x3 y       x 2 y 2 xy 3 y 4
5        4      3       2        2       3         4       5
16 node; p=4
x        x y x y                 x y              xy       y

More even distribution of polynomial terms than Lagrange
shape functions but „p‟ cannot exceed 4!

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