jn reddy - 1 lecture notes on nonlinear fem meen...
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INTRODUCTION AND OVERVIEW OF LINEAR FEM using a 2D model problem
J. N. Reddye-mail: [email protected]
Fall 2016
MEEN 673Nonlinear Finite Element Analysis
Chapter 3
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3
An Overview of The Finite Element Method
2D Problems involving a single unknown
• Model equation Discretization• Weak form development • Finite element model • Approximation functions• Interpolation functions of
higher-order elements• Post-computation of variables• Numerical examples• Transient analysis of 2-D problems
CONTENTS
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MODEL EQUATION
Model Differential Equation
11 22 inu u
a a fx x y y
2-D Problems: 4
coefficients that describe material behaviorsource term
ijaf
11 12 21 22ˆ ˆ, 0x y n
u u u uu u a a n a a n q
x y x y
Type of boundary conditions
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EXAMPLES OF MODEL EQUATION(1)− Deflection of a membrane
2-D Problems: 5
11 22 0u u
a a fx x y y
11 22
( , ) transverse deflection of a point in the membrane( , ) applied pressure
( , ), ( , ) tensions in the and directions, respectively
u x y
f x ya x y a x y x
y
=
==
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1-1
-0.5
0
0.5
1
Transverse deflection of a square membrane fixed on all its sides and subjected to uniform pressure.
x
y
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EXAMPLES OF MODEL EQUATION(2)− Torsion of a cylindrical member
2 2
2 2 0 inGx y
Ω
11 22
( , )
shear modulusangle of twi
warping func
t
tion
s
x ya a GG
0
, , ( , )
( , ) , ( , )
on
xz yz
x yy n x n
u zy v zx w x y
x y G y x y G xy
x
x
y
Γ
L
z
Mz
Mz
y
x
uvzθα =
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EXAMPLES OF MODEL EQUATION(3)− 2D Heat transfer
11 22, temperature; ,, thermal conductivities
in the and directions, respectively
internal heat generationheat flux normal to the boundary
x y
x y
n
u T a k a kk k
x y
fq
11 22
11 22
( , ) , ( , )
ˆ on
x y
x y n
u uq x y a q x y ax y
u ua n a n qx y
Γ
11 22 0u u
a a fx x y y
x
y
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11 22
, water head (velocity potential), permeabilities in the and directions,
respectively internal infiltrationflow normal to the boundaryn
ua a x y
fq
11 22
11 22
( , ) , ( , )
ˆ on
x y
x y n
v x y a v x y ax y
a n a n qx y
Γ
11 22 0a a f
x x y y
EXAMPLES OF MODEL EQUATION(4) − 2D Inviscid flow
x
y
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2-D Problems: 9
WEAK FORM DEVELOPMENT
Approximation ( , ) ( , )hu x y u x y
,e e
e e
x y
F Fdxdy F ds dxdy F dsn n
x y
Green-Gauss Theorem
Consider the identity:
1 11 1 1 1ori i
i i i i
w F F ww F F w w w F F
x x x x x x
11 220
e
h hi
u uw a a f dxdy
x x y y
Weak Form DevelopmentStep 1
1F2F
Step 2: Trade differentiations between and w hu
F
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2-D Problems: 10
WEAK FORM DEVELOPMENT continued
e
e
x
Fdxdy n F ds
x
Consider the identity: 11 1
ii i
F ww w F F
x x x
Step 2:
F
1e
e
x i
Fdxdy n w F ds
x
11e
e
hx i
uFdxdy n w a ds
x x
22e
e
hiy
uGdxdy w a s
yd
yn
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2-D Problems: 11
Step 1
WEAK FORM DEVELOPMENT (continued)
11 22
11 22
11 22
0e
e
e
h hi
i h i hi
h hi x y
u uw a a f dxdy
x x y y
w u w ua a w f dxdy
x x y y
u uw a n a n
x y
11 22
11 22 , flux normal to the boundary
e
e
i h i hi
i n
h hn x y
ds
w u w ua a w f dxdy
x x y y
w q ds
u uq a n a n
x y
Step 2
Step 3
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2-D Problems: 12
FINITE ELEMENT APPROXIMATION
x
y
n
q
n qn ˆq
x
y nixˆn
jyˆn
n i jx yˆ ˆˆ n n
ˆn i
ˆn j
ˆn i
n cos i sin jˆ ˆˆ
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2-D Problems: 13
Weak form
FINITE ELEMENT APPROXIMATION
11 220e
e
i h i hi
i n
w u w ua a w f dxdy
x x y y
w q ds
1 2 3 4
1
( , ) ( , ) ( terms)
( , )
h
n
j jj
u x y u x y c c x c y c xy n
u x y
Approximation
Finite element model [the ith equation is obtained by replacing the weight function w by ]( 1,2, , )i i n
11 220e
e
i h i hi
i n
u ua a f
x x y y
q ds
dxdy
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FINITE ELEMENT MODEL DEVELOPMENT(continued)
11 22
11 221 1
11
0
0
e
e
e
e e
i h i hi
i n
n nj ji i
j jj j
i i n
j
u ua a f dxdy
x x y y
q ds
u a u a dxdyx x y y
f dxdy q ds
u a
221 e
e
e
nj ji i
j
i i n
a dxdyx x y y
f dxdy q ds
2-D Problems: 14
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FINITE ELEMENT MODEL DEVELOPMENT(continued)
11 221
1 1
11 22
0
or
,
e
e
e
e
e
e
nj ji i
jj
i i n
n ne e e e e e e e e eij j i i ij j i
j j
j je i iij
ei i i
u a a dxdyx x y y
f dxdy q ds
K u f Q K u F
K a a dxdyx x y y
F f dxdy
K u F
nq ds
2-D Problems: 15
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2-D Problems: 16
APPROXIMATION FUNCTIONSLinear Triangular Element
1 2 3
1 1 1 2 1 3 1 1
2 2 1 2 2 3 2 2
3 3 1 2 3 3 3 3
( , )
( , )
( , )
( , )
eh
e eh
e eh
e eh
u x y c c x c y
u x y c c x c y u
u x y c c x c y u
u x y c c x c y u
u A c c A u1 11 1
12 2 2 2
3 33 3
1
1 or
1
e e
e e e e e e e
e e
u cx y
u x y c
x yu c
y
x
1
3
21 1( , )x y2 2( , )x y
3 3( , )x y
1 2
3
·
·
·
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2-D Problems: 17
APPROXIMATION FUNCTIONSLinear Triangular Element (continued)
1
1
3
2
ψ1
11
3
2
ψ21
1
3
2
ψ3
1
1 2 33
1 1 2 2 3 31
( , ) 1 1
( , ) ( , ) ( , ) ( , )
1( , )
2, , ( )
Area of the triangle, 2 Determinant of [ ]
e e eh
e e e e e e e ej j
j
ei i i ie
i j k j k i j k i j ke e
u x y c c x c y x y c x y A u
x y u x y u x y u u x y
x y x y
x y y x y y x x
A
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2-D Problems: 18
APPROXIMATION FUNCTIONSLinear Rectangular Element
1 2 3 4
11 1 1
22 1 2 2
3 3 1 2 3 4 3
4 4 1 3 4
( , )
( , )
( , )
( , )
( , )
eh
e eh
e eh
e eh
e eh
u x y c c x c y c x y
u x y c u
u x y c c a u
u x y c c a c b c ab u
u x y c c b u
1 2
3 4
( , ) 1 1 , ( , ) 1 ,
( , ) , ( , ) 1 ,
e e
e e
x y x yx y x y
a b a bx y x y
x y x ya b a b
y
x
1
4 3
2
b
a
y_
x_
( , )ej i i ijx y
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2-D Problems: 19
1
1
4
3
2
ψ1
11
4
3
2
ψ2
1
1
4
3
2
ψ3 1
1
43
2
ψ4
APPROXIMATION FUNCTIONSLinear Rectangular Element (continued)
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2-D Problems: 20
Interpolation Property of the Approximation Functions
3
1
1, , a fixed value( , ) ; ( , ) 1
0, , a fixed valuee ej i i ij j
j
i jx y x y
i j
x
1 231 2
34
y
x
223
u(x,y)
u1u3
Representation of the domainby 3-node triangles
u2
1
( , )ehu x y
x
1 232
4
yu(x,y)
Representation of the domainby 4-node rectangles
1u4u 2u
3u
23
1
4
4
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2-D Problems: 21
NUMERICAL EVALUATION OF COEFFICIENT MATRICES
Linear Right-Angled Triangular Element
2 2 2 2
2 2
2 2
[ ] 02
0
1
13
1
e e
ee e
a b b ak
K b bab
a a
ff
b
ax2
3
1
y
2
3
1
Right-angle triangle
2 e ab
11 22
1 1 0 1 0 1
[ ] 1 1 0 0 0 02 2
0 0 0 1 0 1
e e eb aK a a
a b
11 22When , we havee eea a k
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2-D Problems: 22
NUMERICAL EVALUATION OF COEFFICIENT MATRICESLinear Rectangular Element
y
x
1
4 3
2
b
a
y_
x_
11 22
2 2 1 1 2 1 1 2 1
2 2 1 1 1 2 2 1 1[ ] ,
1 1 2 2 1 2 2 1 16 6 41 1 2 2 2 1 1 2 1
e e e e ef abb aK a a f
a b
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1. Superparametric (m > n): The polynomial degree ofapproximation used for the geometry is of higher order than that used for the dependent variable.
2. Isoparametric (m = n): Equal degree of approximationis used for both geometry and dependent variables.
3. Subparametric (m < n): Higher-order approximation ofthe dependent variable is used.
Geometry:
Solution:
1 1
ˆ ˆ( ), ( )m m
e e e ej j j j
j j
x x , y y ,
1 1
( , ) ( ) ( ( ) ( ))n n
e e e ej j j j
j j
u x y u x,y u x , ,y ,
PARAMETRIC FORMULATIONS (2-D)
Thus, there are two meshes in finite element analysis.
Numer Integration: 23
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NUMERICAL EVALUATION OF INTEGRAL COEFFICIENTS
• Transformation of the integrals posed on arbitrary-shaped element to the master element domain so that evaluationof the integrals is made easy.
• The Gauss integration rule that evaluates an integral expression as a linear sum of the integrand evaluated at certain points (Gauss points) and weights (Gauss weights)is used.
x
y
Γ
Γ
Ω e
e
nα
Ω
∧
•
•
•
• eu2i
Degrees of freedom
12
34
eu1
1
( , ) ( , ) ( , )n
e e eh j j
j
u x y u x y u x y
1
( ( , ), ( , ))
( , )
eh
ne ej j
j
u x y
u
Numer Integration: 24
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ξ
η
ξ = 1
η = 1
x
y
eΩ
1Ω 2Ω 3ΩΩ
ξ
η
eΩ
ηξ ddJdydx =
)1(),1( ηη ,yy,xx ==
)1(),1( ,yy,xx ξξ ==)()(
y,xy,x
ηηξξ
==
)()(
ηξηξ,yy,xx
==
y
x
NUMERICAL INTEGRATION
1
1
ˆ( , ) ( )
ˆ( , ) ( )
me ej j
j
me ej j
j
x x ,
y y ,
Numer Integration: 25
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( ) ( ) General
( ) ( ) Gauss rule
b
a
NPTx
ij ij ij I Ix
I
NGP
ij ij ij I II
G F x dx F x W
K F d F W
1
1
11
Numerical Integration -26
,
NGPor NGPor NGP
pNGP
p
p
p
12
1 1
2 3 2
4 5 3
Nearest larger integer equal to (p+1)/2
( ) [ ]
( ) ( )
iF Fh
I F d F F
F F W
1
1 21 2
1
1 1
22 2
2 0
•
1
•
( )F 0
1
( )F
NUMERICAL INTEGRATIONJN Reddy - 25 Lecture Notes on NONLINEAR FEM
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11 22
ˆ
1 1ˆ
( , ) ( ( , ), ( , ))
ˆ( , ) ( , )
e
e
j je i iij
ij ij
NGP NGP
ij I J ij I JI J
K a a dxdyx x y y
F x y dxdy F x y J d d
F J d d W W F
WI – Weights(xI, yI) – Integration
points
TRANSFORMATION OF THE INTEGRAL
i i i i i i
i ii i i i
x y x yx y x x
x yx yy yx y
ψ ψ ψ ψ ψ ψξ ξ ξ ξ ξ ξ
ψ ψψ ψ ψ ψη ηη η η η
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⇒ = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ = + ∂ ∂∂ ∂∂ ∂ ∂ ∂ ∂ ∂
J
Numer Integration: 27
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NUMERICAL INTEGRATION
1 1
1 1
1 11 2
2 2
1 2
ˆ ˆ
Jˆ ˆ
ˆˆ ˆ
ˆˆ ˆ
m mi ii ii i
m mi ii ii i
m
m
m m
x y x y
x yx y
x yx y
x y
1 *
J J
i ii
i i i
ξ ξx
y η η
−
∂ ∂∂ ∂ ∂∂ = = ∂ ∂ ∂
∂ ∂ ∂
Jacobian matrix
Global derivatives in terms of the local derivatives
Numer Integration: 28
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ELEMENT CALCULATIONS
11 22
* * * *11 11 12 11 12
ˆ
* * * *22 21 22 21 22
( , )
( , )
e
j je i iij
j ji i
j ji i
K a a dxdyx x y y
a J J J J
a J J J J
1 1
ˆ 1 1
1 1
ˆ ˆ( , ) ( , )
ˆ ( , )
e eij ij
NGPNGPeij I J I J
I J
J d d
F d d F d d
F WW
Numer Integration: 29
,dxdy J dξ d Jη= = J
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ξ
η
1 2
34
ξ
η
51 2
346
7
89
ξ
η
1
3
2
ξ
η
21
3
4
56
ξ
η
1
2
34
ξ
η
1
3
2
ξ
η
1
3
24
56
ξ
η
1
2
34
5
6
78
9
Master element Actual element
Numer Integration: 30
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Computational Issues: - 31
η
ξ
ξ = − 13 ξ = 1
3
η = 13
η = − 13
η= 0
η
ξ
ξ = − 35 ξ = 3
5
η = 35
η = − 35
ξ= 0
η= 0.861...
ξ= − 0.861...
ξ
ηξ= 0.861...
η= 0.339...
η= − 0.339...η= − 0.861...
GAUSS QUADRATURE
Domain of the masterelement
Domain of the physical element
Gauss points Gauss weights
ˆ
, 1
ˆ,
ˆ ,
( ) ( )
( )
e e
ij ij
N
I J ij I JI J
F x y dxdy F ξ,η dξ dη
W W F ξ η
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2-D Problems: 32
ASSEMBLY OF ELEMENTS/EQUATIONS(1 DoF per node)
•
•
•
• 1 3
2
4
•
•
••
•
1
2
43
5
1
1
1
11
2
3
4
5
6
7
9
8
a relationship between certain property of global
node and global node .0, if node and do not
belong to the same elemen .t
IJK
I JI J
(1) (1) (1)11 11 13 14 15 13 14
(1) (2) (1) (2)22 22 11 25 23 14 26
(1) (2) (3) (4)55 33 44 22 11
(3) (4) (1) (1) (2)56 23 13 1 1 2 2 1
(1) (2) (3) (4)5 3 4 2 1
, , , 0,
, , 0,
,
, , ,
K K K K K K K
K K K K K K K
K K K K K
K K K F F F F F
F F F F F
1 2 3 4
1 2 5 3
2 4 7 5
3 5 6[ ]
5 7 6
7 8 9 6
B
Connectivity matrix
Element local node numbers
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2-D Problems: 33
POST-COMPUTATION OF VARIABLES
( ) ( )
1
( ) ( )( ) ( )
1 1
( , ) ( , ), ( , )
, , ( , )
ne e e eh j j
j
e en ne ej je e eh h
j jj j
u x y u x y x y
u uu u x y
x x y y
6
•
•
•
• 1 3
2
4
•
•
••
•
1
2
43
5
1
1
1
11
2
3
4
5
7
9
8
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2-D Problems: 34
Derivatives of the Solution
( )( )
1 1
( )( )
1 1
1, ( , )
2
1, ( , )
2
en neje e e eh
j j jej j
en neje e e eh
j j jej j
uu u x y
x x
uu u x y
y y
Linear triangular element
Linear Rectangular element( )
( )
1 1
( )( )
1 1
1( ), ( , )
2
1( ), ( , )
2
en neje e e e eh
j j j jej j
en neje e e e eh
j j j jej j
uu u y x y
x x
uu u x x y
y y
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1
1 2 3
6
9
1 1
1
1 2
3 4
x
y
a = 1
a = 1
u = 0
0=∂∂
−=∂∂
=yu
nuqn
(a)
4
0=∂∂
−xu
u = 0
7 8
5
52
6 10
1 3 4
15
25
1 11 x
ya = 1
u = 0
0=∂∂
−=∂∂
=yu
nuqn
(b)
16
1
20
11
21
u = 0
0=∂∂
−xu
a = 1
9
1 2 3 4
8
12
5
1613
13
8
18
Example 8.3.1 from the book2×2 mesh 4×4 mesh
11 1 12 2 14 4 15 5 1
1 1 1 1 1 1 11 2 4 511 12 14 13 1 1 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
For example, equation for node 1 is:
orK U K U K U K U F
K U K U K U K U f Q f
12 =∇− u
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Coordinate, x
Solu
tion,
u(x
,0)
• Ritz solution∆ FEM (mesh T2)
Analytical
FEM (mesh T1)
x
y
0=∂∂
xu
0=∂∂
yu
0=u
0=u
12 =∇− u
Coordinate, x
Solu
tion,
qx(x
,0)
• FEM (mesh T2)FEM (mesh T1)Analytical
x
y
0=∂∂
xu
0=∂∂
yu
0=u
0=u
12 =∇− u
•
•
•
Finite Element, Ritz, and Exact Solutions
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DISCUSSION (distributed boundary source and convection type boundary condition)
0
0
( ) ( )
( ) 1
h
i n i
n
Q q s s ds
sq s q
h
8 6
2 1,
3 2 3 2n nq h q h
Q Q
For linear elements
2-D Problems: 37
1
1
1
11 2
3 4
5
6
7 8
x
y
1.0
1.0
1 2
3 41
(0, ) 4 (1 )
0 1
u y y y
y
3 0nq u
( , 0) 4 (1 ), 0 1u x x x x
0 5q 0nq
s
h
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DISCUSSION(computation of convection type BC)
00 0
010
01 0 0
( ) ( ) ( ) ( )
( ) ( )
h h
i n i i
h n
j j ij
h hn
j i j ij
Q q s s ds u u s ds
u u s ds
u ds u ds
0
General form of the BC:
( ) 0nq u u
1
1
1
11 2
3 4
5
6
7 8
x
y
1.0
1.0
1 2
3 41
(0, ) 4 (1 )
0 1
u y y y
y
03 0 ( 3, 0)nq u u
( , 0) 4 (1 ), 0 1u x x x x
5nq
0nq
2-D Problems: 38
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Transient Analysis of 2-D Problems
( )2
1 2 11 222, ,
u u u uc c a a f x y t
t t x x y y
( )
2
1 2 11 222
2
1 2 11 222
0
, ,e
e
e
h h h hi
h h i h i hi i
i i
u u u uw c c a a
t t x x y yf x y t dxdy
u u w u w uw c w c a a
t t x x y y
w f dxdy qw ds
Weak formulation
Model Governing Differential Equation
2-D Problems: 39
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1
( ) ( ) ( ) ( ), , , , ,n
e e eh j j
j
u x y t u x y t u t x y
Approximation
Finite element model
1 2
11 22
Cu Mu Ku F
,e e
e
e e
e eij i j ij i j
j je i iij
ei i i
C c dxdy M c dxdy
K a a dxdyx x y y
F f dxdy q ds
+ + =
= =
∂ ∂ ∂ ∂= + ∂ ∂ ∂ ∂
= +
∫ ∫
∫
∫ ∫
SPATIAL DISCRETIZATION:Finite Element Model
2-D Problems: 40
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TIME APPROXIMATIONS (Parabolic)
K u F , K K C
F K C u F F
1 1 1 1 1 1
1 1 1 1 1
ˆ ˆ ˆ ,
ˆ (1 ) (1 )
s s s s s s
s s s s s s s
t
t t
u u u u
Cu Cu Cu Cu1 1 1
1 1 1
(1 )
(1 )
s s s s s
s s s s s
t
t
Alfa-family of approximation
Cu F K u1 1 1 1s s s s
C K u C K u
F F1 1 1
1 1
(1 )
(1 )
s s s s s s
s s s
t t
t
Cu F K us s s s
Cu Ku F, 0 t T
Semidiscrete FE model
Fully discretized model
2-D Problems: 41
NOTE: From here, the procedure is the same as in 1-D
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TIME APPROXIMATIONS (Hyperbolic)
C u M u K u Fe e e e e e e
u u u uu u u u u u
2121 ,
1 , , 1, (1 )
s s s s
s s s s s s
t t
t
Semidiscrete FE model
Newmark scheme (second-order equations)
K u F , K K M C
F F M u u u C u u u
1 1 1 1 1 3 1 5 1
1 1 1 3 4 5 1 5 6 7
ˆ ˆ ˆ
ˆ
s s s s s s s
s s s s s s s s s s
a a
a a a a a a
Fully discretized model
2-D Problems: 42
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An Example: Transient Heat Conduction Problem
1 ( , , )T T Tc k k f x y tt x x y y
52
6 10
1 3 4
15
25
x
y
16 20
11
21
12
34
78
910
2625 31
32
x
y
1.01.0
T = 0
)(insulated0=∂∂
−yT
T = 0
)(insulated
0=∂∂
−xT
0 0 0
0 1 1
x yT Tk n n x yx y
T y x
on and
on and0 1( , , )T x y
Governing equation
Boundary conditions
Initial condition
2-D Problems: 43
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Stability Characteristics
0.00 0.02 0.04 0.06 0.08 0.10 0.12Time, t
-0.15
-0.10
-0.05
0.00
0.05
0.10
0.15Te
mpe
ratu
re,
00518.0triangles linear of mesh44
cri =∆×t
01050 .,. =∆= tα005000 .,. =∆= tα01000 .,. =∆= tα
T(0,
0,t)
2-D Problems: 44
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0.00 0.20 0.40 0.60 0.80 1.00Distance, x
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Tem
pera
ture
,T(x
,0,t f
ixed
)
500050 .,. ==∆ αtt = 1.0 (steady state)
t = 0.005
t = 0.1
t = 0.2
t = 0.5
0.0 0.4 0.8 1.2 1.6 2.0Time, t
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Tem
pera
ture
, 500050 .,. ==∆ αt
T(0,
0,t)
A NUMERICAL EXAMPLE (Parabolic)
•
2-D Problems: 45
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y
2ft4ft
x
x
y
52
6 10
u = 0
0=∂∂
−yu
1 3 4
15
2516
0=∂∂
−xu
2011
21u = 0
y
x1 2 45 8
121613
9
3
14 15
TRANSIENT RESPONSE OF A MEMBRANE
0.0 0.5 1.0 1.5 2.0 2.5Time, t
-0.50
-0.25
0.00
0.25
0.50
Defle
ction
,
u(2,
1, t
)
1058.0domain original the of quadrant a in
elements triangular linear of mesh44
cri =∆
×
t
• 5.0,5.0,125.0 ===∆ γαt
3/1,5.0,125.0 ===∆ γαt3/1,5.0,1.0 ===∆ γαt
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5Time, t
-8
-6
-4
-2
0
2
4
6
8
Defle
ction
,
♦
2-D Problems: 46
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2-D Problems: 47
SUMMARY 1. Starting point is a model differential equation in u2. Construct an integral statement – weak form,
which has three steps. Integration by parts that (a) relaxes (“weakens” differentiability on the variable u, and (b) brings in secondary variable into the integral form.
3. Substitute suitable approximation for u and obtain the finite element model (i.e., a set of algebraic relations between nodal values of u and Q).
4. Numerical evaluation of coefficients 5. Assemble equations, impose boundary conditions, and
solve the assembled equations. 6. Post-computation of variables follows same procedure as
in 1-D FEM.7. Reviewed time-dependent problems.
andij iK f
JN Reddy - 46 Lecture Notes on NONLINEAR FEM