Post on 28-Jul-2020
Finite Element Modeling & Analysis
Human-centered CAD Lab.
1 2007-10-24
Finite Element Modeling and Analysis
Structural problems Stress, strain, distribution
Heat transfer problems Temperature distribution
Electrostatic potential problems
Fluid mechanics problems
Vibration analysis problems
2
Finite Element Modeling and Analysis – cont’
When geometry of problem domain is complicated, analytic solution can not be derived Numerical approximate solution
Even for simple geometry, analytic solution cannot be obtained if the object is composed of composite material
3
Problems which cannot be solved analytically
4
Approximation of each object by an assemblage of finite element
5
Analysis Procedure
Input material properties and boundary condition before elements are generated B.C
Known displacement
Known external force
Known temperature
Pre-processor Allow the interactive input of B.C’s
Generate meshes automatically
6
Choice of Element
Select element type among those supported by the FEM package in use
Select element type mostly suitable for the given problem ← “know-how”
Determining element size is also a “know-how”
Use small elements where rapid change in stress or strain is expected
7
Analysis Procedure – cont’
System equations are composed by FEM package
System equation Relation between B.C. and unknown (displacement,
temperature, …) at every node
→ By Solving system equation, unknowns at each node
is obtained
Unknown values (displacement, temperature, stress,strain, …) at arbitrary location is obtained byinterpolating the values at nodes
Post-processor displays color coding or contour plot
8
Formulation of System Equation
9
Formulation of System Equation – cont’
: External surface tractions
: Body forces
: Concentrated external forces
10
SfBfiF
Formulation of System Equation – cont’
Only displacement at any X, Y, Z
needs to be solved
→ Strain and stress are derived from
11
Z)]Y, W(X, Z),Y, V(X, Z),Y, U(X,[UT
ZXYZXYZZYYXXT ε
ZXYZXYZZYYXXT τ
U
Formulation of System Equation – cont’
Principle of virtual displacement total external virtual work
= total internal virtual work
‘ – ’ indicates the quantities caused by virtual displacement
12
i
i
iS
S
SB
V
T
V
T TT
dSdVdv FUfUfU ε
Formulation of System Equation – cont’
Let displacement of all the nodes
where N : Number of nodes
n = 3N : Degree of freedom
13
U
NNN222111T w v u w v u w v u U
U U U U n321
Formulation of System Equation – cont’
Displacement at arbitrary location in an element
→ Is approximated by proper shape function that
interpolates the displacements at nodes of the
element
→ Thus is the unknown to be solved
14
TU
Formulation of System Equation – cont’
Displacement at arbitrary x, y, z in m-element
Strain is obtained by differentiating displacement
Stress – Strain relation
↑
Element initial stress
15
UHu
)z,y,x()z,y,x( )m()m(
UB
)z,y,x()z,y,x(ε )m()m(
I(m)(m))m()m( τ ετ C
Formulation of System Equation – cont’
→ Virtual nodal displacement
16
i
ii
mS
mmSmS
mV
mmBm
mV
mmm
T
m
T
m
T
m
T
dS
dVdV
FUfu
fu
)(
)()(
)()()(
)()()()()()(
τε
FB
fH
fHUUBCBU
)()()(
)()()(
)()()()()()()(
)(
)(
)()(
m
mV
mIm
m
mS
mSmS
m
mV
mBmTmm
mV
mmT
dV
dS
dVdV
m
T
m
T
m
T
m
T
τ
Formulation of System Equation – cont’
17
FB
fH
fHUBCB
)()()(
)()()(
)()()()()()()(
)(
)(
)()(
m
mV
mIm
m
mS
mSmS
m
mV
mBmmm
mV
mm
dV
dS
dVdV
m
T
m
T
m
T
m
T
τ
RUK
Formulation of System Equation – cont’
Role of FEM package is to evaluate the integrals for K, R
→ Solve for displacements at all the nodes
→ Evaluate displacement, stress, strain at any
location in each element using shape function
18
Example of Formulation
Derive the system equations of the plate using a two-element model
Thickness is 1 cm
Load is applied very slowly so that inertial effect is ignored
19
yP
Example of Formulation – cont’
Deriving the displacement transformation matrix of any element
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yxu 321
yxv 321
131211 131211 yxvyxu
232212 232212 yxvyxu
333213 333213 yxvyxu
The displacements of each node
Example of Formulation – cont’
The constants can be determined in terms of
and
21
ii ,
ii vu , ii yx ,
a
uauaua
2332211
1
a
ububub
2332211
2
a
ucucuc
2332211
3
a
vavava
2332211
1
a
vbvbvb
2332211
2
a
vcvcvc
2332211
3
23132123321 xxcyybyxyxa
31213231132 xxcyybyxyxa
12321312213 xxcyybyxyxa
3212 aaaa
Example of Formulation – cont’
Substituting the values of of element 1 into the previous equations
22
321321 ,,,,, yyyxxx
yvvxvvvv
yuuxuuuu
)4
1
4
1()
10
1
10
1(
)4
1
4
1()
10
1
10
1(
31211
31211
Example of Formulation – cont’
23
8
7
6
5
4
3
2
1
)1(
004
10
10
10)
4
1
10
11(0
0004
10
10
10)
4
1
10
11(
U
U
U
U
U
U
U
U
yxyx
yxyx
v
u
8
7
6
5
4
3
2
1
)2(
)410
1(0)10
1(0)4
1(000
0)410
1(0)10
1(0)4
1(00
U
U
U
U
U
U
U
U
yxxy
yxxy
v
u
Example of Formulation – cont’
Determining the strains or the matrix using the following two-dimensional strain-displacement relations
24
)(mB
y
u
x
v
y
vx
u
xy
y
x
Example of Formulation – cont’
25
UB )1(
8
7
6
5
4
3
2
1
)1(
)1(
0004
1
10
10
10
1
4
1
004
1000
4
10
0000010
10
10
1
U
U
U
U
U
U
U
U
xy
y
x
ε
UB )2(
8
7
6
5
4
3
2
1
)2(
)2(
10
1
4
1
10
100
4
100
4
1000
4
1000
010
10
10
10000
U
U
U
U
U
U
U
U
xy
y
x
ε
Example of Formulation – cont’
The stress-strain relations for a homogeneous, isotropic plane-stress element are given by
26
xy
y
x
xy
y
xm E
)1(2
100
01
01
1 2)(τ
Example of Formulation – cont’
We assumed that there is no initial stress on the structure
The stiffness matrix of each element is derived as follows by substituting the previous results
Since two elements are made of the same material, the same expression for in Equation can be used for both the elements
27
)(mτ)( mC
Example of Formulation – cont’
28
)1()1()1()1()1()1(
dVT
VBCBK )1()1()1()1(
)1(dA
T
ABCB dxx
T
)10
44()1()1(
)1(10
0 BCB
dxx
E
T
)10
44(
0004
1
10
10
10
1
4
1
004
1000
4
10
0000010
10
10
1
2
100
01
01
0004
1
10
10
10
1
4
1
004
1000
4
10
0000010
10
10
1
1
10
02
Example of Formulation – cont’
29
00000000
00000000
0016
100
4016
1
40
00032
1
80
10
80
1
32
1
00080
1
200
10
200
1
80
1
0040
00100
1
40100
1
0016
1
80
1
200
1
40200
1
16
1
80
1
004032
1
80
1
100
1
80
1
32
1
100
1
1
202
)1(
E
K
Example of Formulation – cont’
30
)2()2()2()2()2()2(
dVT
VBCBK )2()2()2()2(
)2(dA
T
ABCB dxx
T
10
4)2()2()2(10
0BCB
200
1
16
1
80
1
200
1
4016
1
80
100
80
1
32
1
100
1
80
1
100
1
4032
100
200
1
80
1
200
100
80
100
40100
10
100
1
40000
16
1
400
4016
1000
80
1
32
1
80
100
32
100
00000000
00000000
1
202
E
Example of Formulation – cont’
31
2001
161
801
2001
40161
80100
801
321
1001
801
1001
4032100
2001
801
2001
16100
801
161
40
4010010
321
1001
8010
801
321
161
400
801
2001
1610
2001
801
801
321
80100
321
1001
401001
00161
801
2001
402001
161
801
004032
180
1100
180
132
1100
1
120
2)2(
E
K
Example of Formulation – cont’
The load vector is equal to because there are only concentrated forces on the nodes
32
CRR
y
y
x
y
x
P
F
F
F
F
0
0
0
3
3
1
1
R
Example of Formulation – cont’
Solving for the unknown nodal point displacements from the following equation
33
y
y
x
y
x
P
F
F
F
F
U
U
U
U
U
U
U
U
kkkkk
kkkkk
kkkkk
kkkkk
0
0
0
3
3
1
1
8
7
6
5
4
3
2
1
8887838281
7877737271
2827232221
1817131211
Example of Formulation – cont’
Applying the boundary conditions as below
34
y
y
x
y
x
P
F
F
F
F
U
U
U
U
kkkkk
kkkkk
kkkkk
kkkkk
0
0
0
0
0
0
0
3
3
1
1
8
7
4
3
8887838281
7877737271
2827232221
1817131211
Example of Formulation – cont’
Partitioned into two parts as follows
35
yPU
U
U
U
kkkk
kkkk
kkkk
kkkk
0
0
0
8
7
4
3
88878483
78777473
48474443
38373433
y
x
y
x
F
F
F
F
U
U
U
U
kkkk
kkkk
kkkk
kkkk
3
3
1
1
8
7
4
3
68676463
58575453
28272423
18171413