# Higher Order Elements - Introduction to Finite Elements - Lecture Slides, Slides for Finite Element Method. Shree Ram Swarup College of Engineering & Management

## Finite Element Method

Description: The lecture slides of the Introduction to Finite Elements are very helpful and interesting the main points are:Higher Order Elements, Properties of Shape Functions, Triangular Elements, Area Coordinates, Lagrange Family, Rectangular Elements, Serendipity Family, Kronecker Delta Property, Polynomial Completeness, Boundary Conditions
Showing pages  1  -  4  of  7
1
MANE 4240 & CIVL 4240
Introduction to Finite Elements
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
=nodesotherallat
inodeat
Ni0
1
...
2211 ++= uNuNu
Inside an element
At node 1, N1=1, N2=N3=…=0, hence
1
1uu node =
Facilitates the imposition of boundary conditions
2. Polynomial completeness
yyN
xxN
N
i
ii
i
ii
ii
=
=
=
1
yxuIf 321
α
α
α
+
+
Then
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2
Higher order elements in 1D
2-noded (linear) element:
12
x2
x1
12
1
2
21
2
1
xx
xx
N
xx
xx
N
=
=
In “local” coordinate system (shifted to center of element)
12
x
aa a
xa
N
a
xa
N
2
2
2
1
+
=
=
x
12
x2
x1
()
(
)
()()
()()
()()
()()
()()
2313
21
3
3212
31
2
3121
32
1
xxxx
xxxx
N
xxxx
xxxx
N
xxxx
xxxx
N
=
=
=
In “local” coordinate system (shifted to center of element)
12
x
aa
()
()
2
22
3
2
2
2
1
2
2
a
xa
N
a
xax
N
a
xax
N
=
+
=
=
3
x3x
axxax
=
== 321 ;0;
3
4-noded (cubic) element:
12
x2
x1
()()()
()()()
()()()
()()()
()()()
()()()
()()()
()()()
342414
321
4
432313
421
3
423212
431
2
413121
432
1
xxxxxx
xxxxxx
N
xxxxxx
xxxxxx
N
xxxxxx
xxxxxx
N
xxxxxx
xxxxxx
N
=
=
=
=
In “local” coordinate system (shifted to center of element)
12
2a/3 x
aa
)3/)(3/)((
16
27
))(3/)((
16
27
)3/)(3/)((
16
9
)3/)(3/)((
16
9
3
4
3
3
3
2
3
1
axaxax
a
N
xaxaxa
a
N
axaxax
a
N
axaxax
a
N
+=
+=
++=
+=
3
x3x
x4
4
34
2a/3
2a/3
Polynomial completeness
#
4
3
2
1
x
x
x
x2 node; k=1; p=2
3 node; k=2; p=3
4 node; k=3; p=4
Convergence
rate (displacement)
1;
0+=kpChuu p
h
Recall that the convergence in displacements
k=order of complete polynomial
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3
Triangular elements
Area coordinates (L1, L2, L3)
1A=A1+A2+A3
Total area of the triangle
At any point P(x,y) inside
the triangle, we define
A
A
L
A
A
L
A
A
L
3
3
2
2
1
1
=
=
=
Note: Only 2 of the three area coordinates are independent, since
x
y
2
3
PA1
A2
A3
L1+L2+L3=1
A
ycxba
Liii
i2
+
+
=
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det
2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
===
===
===
==
Check that
yyLyLyL
xxLxLxL
LLL
=++
=++
=++
332211
332211
321 1
x
y
2
3
PA1
1
L1= constant
P’
Lines parallel to the base of the triangle are lines of constant ‘L’
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4
We will develop the shape functions of triangular elements in
terms of the area coordinates
x
y
2
3
PA1
1
L1= 0
L1= 1
L2= 0
L2= 1
L3= 0
L3= 1
For a 3-noded triangle
33
22
11
LN
LN
LN
=
=
x
y
2
3
1
L1= 0
L1= 1
L2= 0
L2= 1
L3= 0 L3= 1
For a 6-noded triangle
L1= 1/2
L2= 1/2
L3= 1/2
4
5
6
How to write down the expression for N1?
Realize the N1must be zero along edge 2-3 (i.e., L1=0) and at
nodes 4&6 (which lie on L1=1/2)
(
)
(
)
2/10 111
=
LLcN
Determine the constant ‘c’ from the condition that N1=1 at
node 1 (i.e., L1=1)
(
)
(
)
()
2/12
2
12/111)1(
111
11
=
=
=
=
=
LLN
c
cLatN
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