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Two DimensionsTwo Dimensions
Larry Caretto
Mechanical Engineering 501B
Seminar in Engineering
Analysis
April 27-29, 2009
2
Outline
- Review last lecture
- Quadratic basis functions in two
dimensions
- Border integral terms
- Triangular elements
- Natural coordinates (area coordinates)
- Linear basis functions
- Higher order basis functions
- Boundary terms
3
Review Quadrilateral Equation
∫ ∫ − −
1
1
1
1
Aki f (ξ ,η) d ξ d η
a J
J
xx yy
J
x y
J
xx yy
J
x y
f
k i
i k k
i k k
ξ ξ η ξ η ξ
η η η ξ η ξ
2
2 2
2 2
⎥^ −
∫ ∫ ∑∑ − − =^ =
n
k
n
j
k j k j
f dd f
1 1
1
1
1
1
- Select basis functions and n (^4)
Review Linear φ i Quadrilateral
(x
4
, y
4
(x 2 , y 2 )
(x
1
, y
1
(x 3 , y 3 )
3 4
1 2
Note:
i
( x
(j)
= δij
5
Review φ Derivatives
4 4 4
3 3 3
2 2 2
1 1 1
ξ
η
η ϕ
ξ
ξ η ϕ ϕ
ξ
η
η ϕ
ξ
ξ η ϕ ϕ
ξ
η
η ϕ
ξ
ξ η ϕ ϕ
ξ
η
η ϕ
ξ
ξ η ϕ ϕ
6
Review x and y Derivatives
(x 4 , y 4 )
(x
2
, y
2
(x 1 , y 1 )
(x 3 , y 3 )
4 1 3 2
4 1 3 2
2 1 3 4
2 1 3 4
y y y y y
x x x x x
y y y y y
x x x x x
ξ ξ
ξ ξ
η η
η η
η
η
ξ
ξ
η^ = –
ξ η η ξ
ξ η η ξ
xy x y
x y x y J
= −
∂
∂
∂
∂ − ∂
∂
∂
∂
evaluate A
ki
for element
7
Boundary Terms
(x
4
, y
4
(x 2 , y 2 )
(x 1 , y 1 )
(x
3
, y
3
- Have gradient (2 nd^ or 3rd^ kind) boundary
condition to include in solution
- Compute gradients from solution for
Dirichlet boundary condition
η^ = –
along line of constant ξ
= ±1 or constant η = ±
found along these lines
8
Boundary Integral
(x 4 , y 4 )
(x 2 , y 2 )
(x 1 , y 1 )
(x
3
, y
3
- Differential distance, ds = S ξ=± 1
dη/2 or
Sη=± 1 dξ/2 where S is length of side
η^ = –
Γ
end
start
k side
k
ds
n
u
ds
n
u ˆ
whenever one side of
element is on external
boundary
2 3 2
2 S (^) ξ= 1 = x 3 − x 2 + y − y
9
ξ = 1 Boundary as Example
(x 4 , y 4 )
(x 2 , y 2 )
(x
1
, y
1
(x 3 , y 3 )
- Differential distance, ds = S ξ=
dη/
η^ = –
along ξ = 1 boundary
ds
n
u
∫ k
Γ
2 3 2
2 1 3 2 S = x − x + y − y ξ=
2 3
−
1
1
1
3
2
1 2
ξ ξ
d S ds S
10
ξ = 1 Boundary Example II
(x
4
, y
4
(x 2 , y 2 )
(x
1
, y
1
(x 3 , y 3 )
η^ = –
integral
for φ 2 and
φ 3 at ξ = 1
ds
n
u
∫ k
Γ
2 1
ξ
=
1
1
1
1 1
2
=
− =
= Γ =
ξ
ξ
ξ ξ
S
n
d u
S
n
u
ds
n
u
11
ξ = 1 Boundary Equations
(x 4 , y 4 )
(x
2
, y
2
(x 1 , y 1 )
(x 3 , y 3 )
η^ = –
= –1 A 11 u 1 +^ A 12 u 2 + A 13 u 3 + A 14 u 4 =^0
A (^) 21 u 1 + A 22 u 2 + A 23 u 3 + A 24 u 4 = B
A 31 u 1 + A 32 u 2 + A 33 u 3 + A 34 u 4 = B
A 41 u 1 + A 42 u 2 + A 43 u 3 + A 44 u 4 = 0
1
1
=
=
ξ
ξ
S
n
u
B
term
- Equations with B not needed for
Dirichlet boundary conditions
- Used after solution to compute gradients 12
η = 1 Boundary Equations
(x 4 , y 4 )
(x
2
, y
2
(x 1 , y 1 )
(x 3 , y 3 )
η^ = –
= –1 A 11 u 1 +^ A 12 u 2 + A 13 u 3 + A 14 u 4 =^0
A 21 u 1 + A 22 u 2 + A 23 u 3 + A 24 u 4 = 0
A (^) 31 u 1 + A 32 u 2 + A 33 u 3 + A 34 u 4 = B
A 41 u 1 + A 42 u 2 + A 43 u 3 + A 44 u 4 = B
1
1
=
=
η
η
S
n
u
B
term
- Equations with B not needed for
Dirichlet boundary conditions
- Used after solution to compute gradients
19
Review April 27 Homework II
- Local coordinate system (h = 0.1 m)
- ξ 1 = -1, η 1 = -1 => x 1 = 0, y 1 = 0
- ξ 2 = 1, η 2 = -1 => x 2 = h, y 2 = 0 x 2 - x 1 = h
- ξ 4 = -1, η 4 = 1 => x 4 = h sin θ, y 4 = h cos θ
- ξ 3 = 1, η 3 = 1 => x 3 = x 4 + h, y 3 = h cos θ
- y 3 – y 2 = y 4 – y 1 = h cos θ
1
30 o
1
1
1
1 1 1 1
1
1 1
0
o
1
4 3
2
(x 4 , y 4 )
(x
2
, y
2
(x )
1
, y
1
(x 3 , y 3 )
20
Review April 27 Homework III
(x
4
, y
4
(x 2 y2)
(x
1
, y
1
(x 3 , y 3 )
sin ( sin ) 4
(sin ) 4
θ θ=
+ξ θ+
−ξ
+η
−η
+η
−η
η
ξ
ξ
h x h h
y
h x h h
4 1 3 2
2 1 3 4
2 1 3 4
x x x x x
y y y y y
x x x x x
+ξ − +
−ξ
+η − +
−η
+η − +
−η
η
ξ
ξ
21
Review April 27 Homework IV
(x
4
, y
4
(x 2 y2)
(x 1 , y 1 )
(x
3
, y
3
cos ( cos) 4
( cos ) 4
4 1 3 2
θ θ=
+ξ θ+
−ξ
+ξ − +
−ξ
η
h h h
y y y y y
2
cos 0 2
sin
2
cos
2
2 θ ⎟ = ⎠
⎞ ⎜ ⎝
⎛ θ −
θ = (^) ξη− ηξ=
hh h h J xy xy
( )
θ
θ
θ
θ
θ
=
θ
= θ
θ+ θ
θ
⎟ ⎠
⎞ ⎜ ⎝
⎛ θ ⎟+ ⎠
⎞ ⎜ ⎝
⎛ θ
=
ηξ ηξ
η η
cos
sin
4
cos
0 2
cos
2 2
sin
cos
1
4
cos
sin cos 4
4
cos
2
cos 2
sin
2
2
2 2
2
2
2 2 2 2
h
h h h
J
xx yy
h
h
h
h h
J
x y
22
Review April 27 Homework V
- Use one-point Gauss quadrature for
(symmetric) A
ki
from April 20-22 lecture
A f dd f byGauss quadrature
a J
J
xx y y
J
x y
J
xx y y
J
x y
f
ki
k i
k k
i k k
1
1
1
1
2
2 2
2 2
∫ ∫ − −
ξ ξ ηξ η ξ
η η ηξ η ξ
a = 0
23
Review April 27 Homework VI
θ
= θ
⎟+ ⎠
⎞ ⎜ ⎝
⎛
=
ξ+^ ξ
cos
1
4
cos
0 2 2
2
2
2 2
h
h
J
x y
⎥ ⎦
⎤ ⎢ ⎣
⎡
∂ξ
∂ϕ
θ
θ − ∂η
∂ϕ
∂η θ
∂ϕ ⎥+ ⎦
⎤ ⎢ ⎣
⎡
∂η
∂ϕ
θ
θ − ∂ξ
∂ϕ
∂ξ θ
∂ϕ ξ η= i k k i k k f cos
sin
cos
1
cos
sin
cos
1 (, )
⎭
⎬
⎫
⎩
⎨
⎧ ⎥ ⎦
⎤ ⎢ ⎣
⎡
∂ξ
∂ϕ − θ ∂η
∂ϕ
∂η
∂ϕ ⎥+ ⎦
⎤ ⎢ ⎣
⎡
∂η
∂ϕ − θ ∂ξ
∂ϕ
∂ξ
∂ϕ
θ
ξ η= i k k i k k f sin sin cos
1 (,)
⎭
⎬
⎫
⎩
⎨
⎧ ⎥ ⎦
⎤ ⎢ ⎣
⎡
∂ξ
∂ϕ − ∂η
∂ϕ
∂η
∂ϕ ⎥+ ⎦
⎤ ⎢ ⎣
⎡
∂η
∂ϕ − ∂ξ
∂ϕ
∂ξ
∂ϕ ξ η= i k k i k k f 2
1
2
1
3
2 (, )
For q = 30 o , sinθ = ½
and cosθ = 31/2^ /
24
Review April 27 Homework VII
integrals using Gauss
quadrature with one Gauss point
- See general form of shape function on next
chart for computational ease
- Assemble equations for three unknown
nodes in diagram
- Substitute boundary values and solve
resulting system of three equations for
the three unknowns
25
Review April 27 Homework VIII
shape functions as
i i i
+ a + b
b i a i i
ξ
η
ϕ
η
ξ
ϕ
i i i
i i i
b a
a b
these equations
to get integrals
for various Aki
26
Review April 27 Homework IX
⎭
⎬
⎫ ⎥ ⎦
⎤ ⎢ ⎣
⎡ + η −
⎩
⎨
⎧ ⎥ ⎦
⎤ ⎢ ⎣
⎡ + ξ −
4
( 1 )
2
1
4
( 1 )
4
( 1 )
4
( 1 )
2
1
4
( 1 )
4
( 1 )
3
2 (,)
i i k k k k
i i k k k k
b a b a a b
a b a b b a f
- Set x = h = 0 to get f(0,0) for Gauss
quadrature
⎭
⎬
⎫
⎩
⎨
⎧ ⎥ ⎦
⎤ ⎢ ⎣
⎡
⎤ ⎢ ⎣
⎡ = − 24
1
24 4 4
1
3 4 4
2 ( 0 , 0 )
ai ak bk bi bk ak f
163
2
3 16 32
2 ( 0 , 0 )
ai ak bibk aibk biak aiak bibk aibk biak f
⎞ ⎜ ⎝
⎛ + −
=
27
Review April 27 Homework X
= 4f(0,0) for zero-point Gauss
43
2 4 ( 0 , 0 ) ik ik ik ik ki
aa bb ab ba A f
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
=
0 6 0 6
2 0 2 0
0 6 0 6
2 0 2 0
23
1 A
12 44
31 33
22 24
11 13
u u
u u
u u
u u
A 11 = 2, A 13 = -2, A 22 = 6, A 24 = -6, A 31 = -2, A 33 = 2, A 42 = -6, A 44 = 6, A 12 = A 14 = A 21 = A 23 = A 32 = A 34 = A 41 = A 43 = 0
28
Review April 27 Homework XI
equations with
global indices for
node ij
i-1,j-1 (^) i,j-1 i+1,j-
i-1,j
i-1,j+
i,j+1 i+1,j+
i+1,j
(UR)
(LL)
(UL)
(LR)
i,j
[ ]
[ ] [ ]
[ ]
[ ] (^1) , 1 0
( ) , (^113)
( ) 23
( ) 14
1 , 1
( ) 1 , 24
( ) 43
( ) 12
,
( ) 44
( ) 33
( ) 22
( ) 1 , 11
( ) 34
( ) 21
1 , 1
( ) , 1 42
( ) 41
( ) 1 , 1 32
( ) 31
−
− − − + −
i j
UR ij
UR UL
i j
UR i j
UR LR
ij
UR UL LL LR i j
UL LL
i j
LR ij
LL LR i j
LL
A A u A u
A A u A u
A A u A A A A u
A u A A u A u
29
Review April 27 Homework XII
i-1,j-1 (^) i,j-1 i+1,j-
i-1,j
i-1,j+ i,j+1 i+1,j+
i+1,j
(UR)
(LL)
(UL)
(LR)
i,j
[ ] [ ] [ ]
[ 0 0 ] 6 [ 0 0 ] 2 0
2 0 0 6 0 0 2 6 2 6
1 , 1 , 1 , 1 1 , 1
1 , 1 , 1 1 , 1 1 , ,
− + + − + + + + + +
− − − + − −
i j i j ij i j
i j ij i j i j ij
u u u u
u u u u u
numerical values
(same for all
elements)
− 2 ui − 1 , j − 1 − 6 ui + 1 , j − 1 + 16 ui , j − 6 ui − 1 , j + 1 − 2 ui + 1 , j + 1 = 0
30
Higher order Shape Functions
- Equation on previous page is valid for
any shape functions in a quadrilateral
- Isoparameteric elements use the same
order shape functions for both the
geometry and the dependent variable
- Could use linear functions for geometry
higher order for dependent variable
- Higher order functions for geometry would
allow elements with curved sides
37
Triangular Coordinates
- Vertices have coordinates x (^) i , yi
- Natural (area) coordinates, λ i
start at
zero at side opposite vertex i, are
perpendicular to that side, and go to
one at vertex i
- Three such coordinates with λi = Ai /A
x 1 ,y 1 x 2 ,y 2
x3 ,y 3
λ 3
λ 3 = A 3 /A
λ 3 = 1
A 3
38
Triangular Coordinates II
- Triangle area = A = (base)(height)/2 =
(x 2 – x 1 ) (y 3 – y 1 )/2 (for this triangle)
has height which is λ
3
times total height, i.e. h 3 = λ 3 (y 3 – y 1 ) so
that A 3 = (x 2 – x 1 ) [λ 3 (y 3 – y 1 )]/
- Dividing A 3 by A gives λ 3 = A 3 /A
x 1 ,y 1 x 2 ,y 2
x3 ,y 3
λ 3
λ 3 = A 3 /A
λ 3 = 1
A 3
39
Triangular Coordinates III
- Have three separate area coordinate
variables, λ
1
2
, and λ
3
- Each goes from zero to one
λ 2
λ 3
λ 1
x 3 , y 3
x 1 , y 1 x 2 , y 2
- Each λi = Ai /A so that λ 1 + λ 2 + λ 3 = A 1 /A
+ A 2 /A + A 3 /A = A/A = 1
40
Triangular Coordinates IV
- Lines from point where all λi intersect to
each vertex defines three subareas
- Previous chart showed that each area, Ai
= (height) i (base) iλi /2 = λi A or λi = Ai /A
- Since A 1 + A 2 + A 3 = A, λ 1 + λ 2 + λ 3 = 1
λ 2
λ 3
λ 1
x 3 , y 3
x 2 , y 2 x 1 , y 1
A 2 A 1
A 3
41
Triangular Coordinates V
λ 2
λ 3
λ 1
x 3 , y 3
x 1 , y 1 x 2 , y 2
A 2 A 1
A 3
- At any point in the triangle the x, y and λi
coordinates are related as follows
x = x 1 λ 1 + x 2 λ 2 + x 3 λ 3 y = y 1 λ 1 + y 2 λ 2 + y 3 λ 3
- Can write matrix equation to relate x and y
to λ
i
plus constraint that λ
1
2
3
42
Coordinate Transformations
3
2
1
1 2 3
1 2 3
y y y
x x x
y
x
−
y
x
y y y
x x x
1
1 2 3
1 2 3
3
2
1
( )
Det A
M
b
ji
i j
ij
for bij, the
components
of B = A
is minor
determinant
43
Finding the Inverse
A
xy xy xy
xy xy xy
y y y
Det x x x 2
2 1 3 2 1 3
2 3 1 2 31
1 2 3
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
− − − −
− − − − −
− − − −
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
12 12 2 1 2 1
1 3 13 3 1 3 1
2 3 23 3 2 3 2
1
1 2 3
1 2 3 2
1
1 1 1
xy yx y y x x
xy yx y y x x
xy yx y y x x
A y y y
x x x
( )
Det A
M
b
ji
ij
ij
end of presentation
M 32
b 23
44
Getting λ i from x and y
- This equation gives all λ i
coordinates for
any Cartesian coordinate x, y
- Can express shape functions of any
order in terms of λi coordinates
y
x
xy yx y y x x
yx xy y y x x
xy yx y y x x
A
2 3 12 1 2 2 1
13 1 3 3 1 2 3
2 3 2 3 2 3 3 2
3
2
1
45
Triangle Shape Functions (N i ≡ φ i )
- Linear shape functions: N i
i
- Three nodes at triangle vertices
- Satisfy basic result that N i ( x j ) = δ ij
- Quadratic shape functions
- Nodes 1, 2, and 3 at vertices and nodes 4,
5, and 6 at midpoints of edge
( )
( )
( ) 3 3 3 6 3 1
2 2 2 5 2 3
1 1 1 4 12
N N
N N
N N
6
5
3 4
1
2
46
Shape Function Derivatives
- Finite element coefficients have
derivatives and integrals of shape
functions with respect to x and y
- Have to get derivatives and integrals
with respect to the λi
are independent
and λ
2
as independent
values of λ
i
- Have usual equations for transforms
47
Shape Function Derivatives II
- Transform derivative of any function, ψ
2
2
1
1
2
2
1
1
λ
λ ψ
λ
ψ λ ψ
λ
λ ψ
λ
ψ λ ψ
∂
x x x y y y
y
x
xy yx y y x x
yx xy y y x x
xy yx y y x x
A
2 3 1 2 1 2 2 1
13 1 3 3 1 2 3
2 3 23 2 3 3 2
3
2
1
y
A
x x
x
A
y y
A
xy yx
23 23 2 3 3 2 1
48
Shape Function Derivatives III
- Summary of partial derivatives required
for natural coordinate derivative
transformations
- Definitions of ai and bi given here
A
b
A
x x
A y
a
A
y y
x
A
b
A
x x
A y
a
A
y y
x
A
b
A
x x
A y
a
A
y y
x
3 1 2 3 3 2 1 3
2 3 1 2 2 1 3 2
1 2 3 1 1 3 2 1
λ λ
λ λ
λ λ
55
Border Terms
- Consider side 2-3, ds = L 23 dλ 3
- Others will be similar
1 , 2 , 3
ˆ ˆ ˆ
3
2
1
0
3 23
23 0 () 23
= ∂
∂
∂
∂
∂
∂
= Γ
Nd k n
u Nds L n
u ds n
u A u Nk k k
M
i
kii e
λ
- Along side 2-3, λ 1 = 0 we only have to
evaluate this integral for k = 2 and k = 3
23
23
1
0
2 3
23
23
1
0
3 3 23
23
( 23 ) 3 0
3
n
L u
n
u d L n
u Au B L
M
i
i i ∂
∑ ∫
λ λ λ
56
Border Terms II
- To get border integral for k = 2 with N 2 =
λ 2 , use λ 1 + λ 2 + λ 3 = 1 so dλ 3 = –dλ 1 –
dλ 2 = –dλ 2 since λ 1 is constant along 23
- On side 23, λ 2 goes from 1 to zero as we
integrate in a counterclockwise direction
23
23
0
1
2 3
23
23
0
1
2 2 23
23
( 23 ) 2 0
2
ˆ
2 2
ˆ ˆ
n
L u
n
u d L n
u Au B L
M
i
i i ∂
∂ ⎥= ⎦
⎤ ⎢ ⎣
⎡
∂
∂ − =− ∂
∂ = =
=
λ λ λ
- Similar results for other borders
- Terms appear in two of three element
equations
57
Border Terms III
- Element equations for border on side 23
( 23 ) 311 32 2 33 3 3
( 23 ) 211 22 2 233 2
11 1 122 133
Au A u A u B
A u A u A u B
Au A u Au
- Element equations for border on side 31
( 31 ) 311 32 2 33 3 3
211 22 2 23 3
( 31 ) 111 12 2 13 3 1
Au Au Au B
Au A u Au
Au Au Au B
23
23
( 23 ) 3
( 23 ) 2
ˆ
2 n
L u
B B
31
31
( 31 ) 3
( 31 ) 1
ˆ
2 n
L u
B B
58
Border Terms III
- Element equations for border on side 12
311 32 2 333
( 12 ) 211 22 2 233 2
( 12 ) 111 12 2 13 3 1
Au A u A u
Au A u Au B
Au Au Au B
= 0 for nodes not on external
boundary and R
i
= B
i
(jk)
otherwise
12
12
( 12 ) 2
( 12 ) 1
ˆ
2 n
L u
B B
311 32 2 33 3 3
211 22 2 23 3 2
111 12 2 133 1
A u A u Au R
A u A u A u R
Au A u Au R
59
Assembly
- Away from boundary typical node is part
of five or six triangles
- Must consider element equations for that
node from all triangles
- Each element has three equations
- To assemble equations for one node, pick the
one equation (from three) from each element
where node appears
- pick equation where the nodal value is multiplied by Akk (coefficient whose subscripts are the same)
60
Mesh
gular
mesh
from
MATLAB
PDE
toolbox
per node
61
Assembly for Six Triangles
- Central node g is part of
six elements labeled
from (α) to (ζ)
system is node 3 for all
triangles in local system
a
b
d c
e
f
(α)
(β)
(γ)
(δ)
(ε)
(ζ)
g
( ) 3
() 33
() 32
() 31
() 3
() 33
() 32
() 31
() 3
() 33
() 32
() 31
() 3
() 33
() 32
() 31
() 3
() 33
() 32
() 31
() 3
() 33
() 32
() 31
γ γ γ γ ς ς ς ς
β β β β ε ε ε ε
α α α α δ δ δ δ
A u A u A u R A u A u A u R
A u A u A u R A u A u A u R
A u A u A u R A u A u A u R
b c g e f g
a b g d e g
f a g c d g
62
Assembly for Six Triangles II
equations for node g
have all node g
coefficients as A 33 a
b
d c
e
f
(α)
(β)
(γ)
(δ)
(ε)
(ζ)
g
( ) ( ) ( )
( ) ( ) ( )
( )
( ) 3
() 3
() 3
() 3
() 3
() 3
() 33
() 33
() 33
() 33
() 33
() 33
() 31
() 32
() 31
() 32
() 31
() 32
() 31
() 32
() 32
() 31
() 31
() 32
α β γ δ ε ς
α β γ δ ε ς
δ ε ε ς ς α
α β γ β γ δ
R R R R R R
A A A A A A u
A A u A A u A A u
A A u A A u A A u
g
d e f
a b c
63
Triangle Quality
- For accuracy want all triangles to be as
close to equilateral triangles as possible
- Area of equilateral triangle with side b
2
2 2
b
bh bb b b A
A
- Quality of triangle with sides b 1 , b 2 , and b 3
b b b
A
Q
64
65
May 4 Homework
- Two problems
- Repeat finite element problem from last
week using gradient boundaries
- Triangular finite elements
1
30 o
1
1
1 1 1 1
0
1 1
Zero gradient on these elements
equations from
homework
solution
boundary node
0
66
May 4 Homework II
shape function form
i i i
+ a + b
i ai bi
for Aki integrals
4 cos
sin 3
ik ik
ikik ik ik
ki
ab ba
aabb aa bb
A
73
Jacobian Details
and show that numerator
gives previous expression for 2A in
terms of triangle vertices’ coordinates
( )( ) ( )( )
yx yx yx yx A
yx yx yx yx
AJ y y x x y y x x
3 3 32 13 12
33 31 2 3 2 1
3 2 3 1 3 1 3 2
2
1 2 1 2
λ λ dxdy dxdy Ad λ d λ
A
d d = Jdxdy = =
74
Integrating Triangle Area
- Use following limits for integration over
entire area of triangle
λ 2 λ 1
x 3 , y 3
x1 , y 1 x2 , y 2
Length = λ 1 = 1 – λ 2
∫ ∫
∫ ∫
−
Ω Ω
1
0
1
0
1 2 3 1 2
1 2 3 1 2 3 1 2
2
λ
A f d d
f dxdy f Ad d
λ 3 = 0
75
Confirm Integration Limits
- If f = 1 we should get area
[ ]
A ( ) d A A A
A dxdy Ad d A d d A d
= ⎥ ⎦
⎤ ⎢ ⎣
⎡ ⎥ = − ⎦
⎤ ⎢ ⎣
⎡ = − = −
= = = =
∫
∫ ∫ ∫ ∫ ∫
− − Ω Ω
2
1 2 1 2
2 1 2
2 2 2
1
0
2 2 2 2
1
0
2
1
0
1
0
2
1
0
1 (^121210)
2 2
λ λ λ λ
λ λ λλ λ λ
λ λ
- Confirms correct limits for integrating
complete area of a triangle using the
natural (area) coordinates, λ
i
- Also shows integral value = 1/
76
Linear Shape Functions
- Start with general two-dimensional
integral for Helmholtz equation example
NaN dxdy y
N
y
N
x
N
x
N
A (^) k i i k i k ki (^) ∫ Ω
2
- For linear shape functions Ni = λi
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
− −
− −
1
0
1
0
1
0
1
0
1 2
2 1 2
1
0
1
0
1
0
1
0
1 2
2 1 2
2 2
2 2
λ λ
λ λ
λ λ λλ λ λ
λ λ λλ λ λ
λ λ λ λ
d d a A d d A
b
A
b
A
a
A
a A
d d a A d d x x y y
A A
k i
i k i k
k i
i k i k ki
77
Linear Shape Functions II
- Use following general result (we have
shown it is correct for m
1
= m
2
= m
3
A
aa bb
A
aa bb
d d
A
aa bb
d d
A
b
A
b
A
a
A
a
A
ik ik ik ik ik ik
i k i k
1
0
1
0
1 2
1
0
1
0
1 2
2
2
∫ ∫
∫ ∫
−
−
λ
λ
( 2 )!
1 2 3
1 2 3 1 2 3
1 2 3
∫ Ω
m m m
m m m
d A
m m m
78
Linear Shape Functions III
- Use general form just stated to get
remainder of A
ki
integral for N
i
i
ik
k i k i
Aa
i k
Aa a A
i k
Aa a A
NaNdxdy a dxdy
∫ ∫ Ω Ω
2 2
2 2
2 2
