Transcript MECH300H Introduction to Finite Element Methods Lecture 7
MECH300H Introduction to Finite Element Methods Lecture 7
Finite Element Analysis of 2-D Problems
2-D Discretization
Common 2-D elements:
2-D Model Problem with Scalar Function - Heat Conduction
•
Governing Equation
x
T
(
x
,
x y
)
y
T
(
x
,
y y
)
Q
(
x
,
y
) 0 •
Boundary Conditions
Dirichlet BC:
in
W Natural BC: Mixed BC:
Weak Formulation of 2-D Model Problem
•
Weighted - Integral of 2-D Problem -----
W
w
x
x
y
y
Q x y dA
0 •
Weak Form from Integration-by-Parts -----
0 W
w
x
T
x
w
y
T
y
dxdy
W
x
w
T
x
dxdy
W
y
w
T
y
dxdy
Weak Formulation of 2-D Model Problem
•
Green-Gauss Theorem -----
W
x
w
T
x
W
y
w
T
y
dxdy
dxdy
G G
w
T
x
w
T
y
n ds x
n ds y
where n x
n
n and n y which is normal to the boundary x i
n y are the components of a unit vector, j
i
cos
j
sin
G
.
Weak Formulation of 2-D Model Problem
•
Weak Form of 2-D Model Problem -----
0 W
w
x
T
x
w
y
T
y
G
w
T
x
n x
T
y
n y
ds
dxdy
EBC: Specify T(x,y) on
G
T
x
n
T
x
i
T
y
j
outward flux on the boundary
G
T
y
n i x
n j y
G
at the segment ds.
FEM Implementation of 2-D Heat Conduction – Shape Functions
Step 1: Discretization – linear triangular element
T 1
T
T
1 1
T
2 2
T
3 3 T 3 T 2
i
i
Interpolation properties 1 at
ith
node 0 at other nodes 1 1
x y
1 1 1
x
1
x
2
x
3 Derivation of linear triangular shape functions: Let 1
c
0
c
1
x
c
2
y c
0
c
0
c
0
c x
1 1
c x
1 3
c y
2 1 2
c y
2 3 1 0 0
c
0 1
c c
1 2 1 1
x
1
x
2
x
3
y y
2 1 1
y
3 1
x
2
A e y
y
2
x
3 3
y
3
x
2 2 Same 2 1
x
2
A e y
y
3
x
1
y
1
x
3 3 3 1
x
2
A e y
2
y
1
x
2
y
2
x
1
y
1
y
2 1
y
3 1
FEM Implementation of 2-D Heat Conduction – Shape Functions
linear triangular element – area coordinates
T 1 T 3 A 2 A 1 A 3 1 1
x
2
A e y
x y
2 3
y
2
x
3
x y
3 2
y
3
x
2
A
1
A e
T 2 2 1
x
2
A e y
y
3
x
1
y
1
x
3
A
2
A e
1
3 1
x
2
A e y
x y
1
y x
2 2 1
x y
2 1
y
2
x
1
3
2
A
3
A e
Interpolation Function - Requirements
•
Interpolation condition
•
Take a unit value at node i, and is zero at all other nodes
•
Local support condition
• i
is zero at an edge that doesn’t contain node i.
•
Interelement compatibility condition
•
Satisfies continuity condition between adjacent elements over any element boundary that includes node i
•
Completeness condition
•
The interpolation is able to represent exactly any displacement field which is polynomial in x and y with the order of the interpolation function
Formulation of 2-D 4-Node Rectangular Element –
Let
u
Bi-linear Element
u
1 1
u
2 2
u
3 3
u
4 4
1
3
a
b
a
b
2
a
4
a
b
b
Note: The local node numbers should be arranged in a counter-clockwise sense. Otherwise, the area Of the element would be negative and the stiffness matrix can not be formed.
1
2
3
4
FEM Implementation of 2-D Heat Conduction – Element Equation
•
Weak Form of 2-D Model Problem -----
0 W
e
w
x
T
x
w
y
T
y
dxdy
G
e wq ds n
Assume approximation:
)
n
j
1
u j
j
and let w(x,y)=
i (x,y) as before, then
0 W
e
x i
x
j n
1
T j
j
y i
y
j n
1
T j
j
i
G
i
where
K ij j n
1
K T ij j
W
e
i Qdxdy
e
G
i q ds n
W
e
i
j x
i
j y
dxdy
)
FEM Implementation of 2-D Heat Conduction – Element Equation
j n
1
j
W
e
i Qdxdy
G
e
i
4
A e
l
23
l
23
l
23 2
l
31
l
12
l
23
l
31
l
31 2
l
31
l
12
l
23
l
31
l
12
l
12 2
l
12
Q q
Q Q
2 3
q q
2 3
Q i q i
W
e
i Qdxdy
G
e
i q ds n
Assembly of Stiffness Matrices
U 1
F i
u 1 ( 1 ) , U 2
W
u 2 ( 1 )
u 1 ( 2 ) , U 3
G
i q n
u 3 ( 1 )
ds
u ( 4 2 )
j n e
1
, U 4
K u ij j
u 2 ( 2 ) , U 5
u ( 3 2 )
Imposing Boundary Conditions
1 1 1 1 The meaning of
q i
: 1 3 3 2
q
1 (1) G 1
q n
(1) 1 (1)
ds
h
(1) 12
q n
(1) 1 (1)
ds
h
(1) 23
q n
(1) 1 (1)
ds
h
(1) 31
q n
(1) 1 (1)
ds
h
(1) 12
q n
(1) 1 (1)
ds
h
(1) 31
q n
(1) 1 (1)
ds
2 3 3
q
2 (1) G 1
q n
(1) 2 (1)
ds
h
(1) 12
q n
(1) 2 (1)
ds
h
(1) 23
q n
(1) 2 (1)
ds
h
(1) 31
q n
(1) 2 (1)
ds
h
(1) 12
q n
(1) 2 (1)
ds
h
(1) 23
q n
(1) 2 (1)
ds
1 2 2
q
3 (1) G 1
q n
(1) 3 (1)
ds
h
(1) 12
q n
(1) 3 (1)
ds
h
(1) 23
q n
(1) 3 (1)
ds
h
(1) 31
q n
(1) 3 (1)
ds
h
(1) 23
q n
(1) 3 (1)
ds
h
(1) 31
q n
(1) 3 (1)
ds
Consider
Imposing Boundary Conditions
q
2
q
2 (1)
q
1 (2)
q
(1) 2
h
(1) 12
q
(1)
n
2 (1)
ds
h
(1) 23
q
(1)
n
2 (1)
ds q
3
q
3 (1)
q
4 (2)
q
1 (2)
h
( 2) 12
q
(2)
n
1 (2)
ds
h
( 2) 41
q
(2)
n
1 (2)
ds q
3 (1)
h
(1) 23
q n
(1) 3 (1)
ds
h
(1) 31
q n
(1) 3 (1)
ds q
(2) 4
h
( 2) 34
q
(2)
n
4 (2)
ds
h
( 2) 41
q
(2)
n
4 (2)
ds
Equilibrium of flux:
q n
(1)
h
(1) 23
q n
(2)
h
( 2 ) 41 FEM implementation:
h
(1) 23
q n
(1) 2 (1)
ds q
2
h
( 2) 41
q n
(2) 1 (2)
ds
;
h
(1) 23
q n
(1) 3 (1)
ds
h
(1) 12
q
(1)
n
2 (1)
ds
h
( 2) 12
q n
(2) 1 (2)
ds q
3
h
( 2) 41
q n
(2) 4 (2)
ds
h
(1) 31
q
(1)
n
3 (1)
ds
h
( 2) 34
q n
(2) 4 (2)
ds
Example:
q n
0
Calculating the q Vector
T
293
K q n
1
2-D Steady-State Heat Conduction - Example
A D AB and BC: CD: convection
q n
0
h
50
W m
2
o C
DA:
T
180
o C T
25
o C
0.6 m y B 0.4 m C x