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FOUNDMENTALS FOR FINITE ELEMENTMETHOD
CHAPTER1:
The FThe Finite Element Methodinite Element MethodA Practical CourseA Practical Course
CONTENTSCONTENTS STRONG AND WEAK FORMS OF GOVERNING EQUATIONS HAMILTON’S PRINCIPLE FEM PROCEDURE
– Domain discretization– Displacement interpolation– Formation of FE equation in local coordinate system– Coordinate transformation– Assembly of FE equations– Imposition of displacement constraints– Solving the FE equations
STATIC ANALYSIS EIGENVALUE ANALYSIS TRANSIENT ANALYSIS (reading materials) REMARKS
STRONGSTRONG AND AND WEAKWEAK FORMS OF FORMS OF GOVERNING EQUATIONSGOVERNING EQUATIONS
System equations: strong form (PDE), difficult to solve.
Weak (integral) form: requires weaker continuity on the dependent variables (e.g., u, v, w).
Weak form is often preferred for obtaining an approximated solution.
Formulation based on a weak form leads to a set of algebraic system equations – FEM.
HAMILTON’S PRINCIPLEHAMILTON’S PRINCIPLE
“Of all the admissible time histories of displacement the most accurate solution makes the Lagrangian functional a minimum.”
An admissible displacement must satisfy:– The compatibility conditions– The essential or the kinematic boundary conditions
– The conditions at initial (t1) and final time (t2)
HAMILTON’S PRINCIPLEHAMILTON’S PRINCIPLE
Mathematically
02
1 dtL
t
t
where L=T+Wf
VUUT T
V
d2
1
VcVΠ T
V
T
V
dd εε2
1σε
2
1
fsT
Sb
T
Vf SfUVfUW
f
dd
(Kinetic energy)
(Potential energy)
(Work done by external forces)
Lagrangian functional
FEM PROCEDUREFEM PROCEDURE
Step 1: Domain discretization Step 2: Displacement interpolation Step 3: Formation of FE equation in local
coordinates Step 4: Coordinate transformation Step 5: Assembly of FE equations Step 6: Imposition of displacement constraints Step 7: Solving the FE equations
Step 1: Domain Step 1: Domain discretizationdiscretization
The solid body is divided into Ne elements with proper connectivity – compatibility.
All the elements form the entire domain of the problem without any gap or overlapping – compatibility.
There can be different types of element with different number of nodes.
The density of the mesh depends upon the accuracy requirement of the analysis.
The mesh is usually not uniform, and a finer mesh is often used in the area where the displacement gradient is larger.
Triangular elements Nodes
Step 2: Displacement Step 2: Displacement interpolationinterpolation
Bases on local coordinate system,
the displacement within element is
interpolated using nodal displacements.
eii
n
i
zyxzyxzyxd
dNdNU ),,( ),,(),,(1
1
2
displacement compenent 1
displacement compenent 2
displacement compenent f
i
n f
d
d
d n
d
1
2
displacements at node 1
displacements at node 2
displacements at node d
e
n dn
d
dd
d
nf: Degree of freedoms at a node
x, u
y, v
1 (x1, y1) (u1, v1)
2 (x2, y2) (u2, v2)
3 (x3, y3) (u3, v3)
A fsx
fsy
nd: number of nodes in an element
Step 2: Displacement interpolationStep 2: Displacement interpolation
N is a matrix of shape functions
1 2( , , ) ( , , ) ( , , ) ( , , )
for node 1 for node 2 for node
dn
d
x y z x y z x y z x y z
n
N N N N
fin
i
i
i
N
N
N
000
000
000
000
2
1
Nwhere
Shape function for each displacement component at a node
Displacement interpolationDisplacement interpolation
Constructing shape functions– Consider constructing shape function for
a single displacement component– Approximate in the form
1
( ) ( ) ( )dn
hi i
i
Tu p
x x p x α
1 2 3 ={ , , , ......, }d
Tn α
pT(x)={1, x, x2, x3, x4,..., xp} (1D)
Basis function
Pascal triangle of monomialsPascal triangle of monomials: : 2D2D
xy x2
x3
x4
x5
y2
y3
y4
y5
x2y
x3y
x4y x3y2
xy2
xy3
xy4 x2y3
x2y2
Constant terms: 1
x y
1
Quadratic terms: 3
Cubic terms: 4
Quartic terms: 5
Quintic terms: 6
Linear terms: 2 3 terms
6 terms
10 terms
15 terms
21 terms
2 2( ) ( , ) 1, , , , , ,..., ,T T p px y x y xy x y x y p x p
Pascal pyramid of monomialsPascal pyramid of monomials : : 3D3D
x
x2
x3
x4
y
y2
y3
y4
xy
z
xz yz
x2y xy2
x2z zy2
z2
xz2 yz2
xyz
z3
x3y
x3z
x2y2
x2z2 x2yz
xy3
zy3
z2y2
xy2z xyz2
xz3
z4 z3y
1 Constant term: 1
Linear terms: 3
Quadratic terms: 6
Cubic terms: 10
Quartic terms: 15
4 terms
10 terms
20 terms
35 terms
2 2 2( ) ( , , ) 1, , , , , , , , , ,..., , ,T T p p px y z x y z xy yz zx x y z x y z p x p
Displacement interpolationDisplacement interpolation
– Enforce approximation to be equal to the nodal displacements at the nodes
di = pT(xi) i = 1, 2, 3, …,nd
or
de=P
where
1
2=
d
e
n
d
d
d
d
T1
T2
T
( )
( )
( )dn
p x
p xP
p x
,
Moment matrix
Displacement interpolationDisplacement interpolation
– The coefficients in can be found by
e 1α P d
– Therefore, uh(x) = N( x) de
1 2
1 1 1 11 2
( ) ( ) ( )
1 2
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
n
T T T Tn
N N N
nN N N
x x x
N x p x P p x P p x P p x P
x x x
Displacement interpolationDisplacement interpolation
Sufficient requirements for FEM shape functions
1 , 1,2, ,
0 , , 1, 2, ,d
i j ijd
i j j nN
i j i j n
x
1. (Delta function
property)
1
( ) 1n
ii
N
x2. (Partition of unity property – rigid body movement)
1
( )dn
i ii
N x x x
3. (Linear field reproduction property)
Step 3: Formation of FE equations in local Step 3: Formation of FE equations in local coordinatescoordinates
Since U= Nde
Therefore, = LU = L N de= B de
Strain matrix
or where
(Stiffness matrix)
eT
Ve
Tee
TTe
Ve
T
Ve
VcVcVcΠ ddBBddBdBdd )(2
1
2
1εε
2
1
VcT
Ve
e dBBk 1
2Te e e d k d
Step 3: Formation of FE equations in local Step 3: Formation of FE equations in local coordinatescoordinates
Since U= Nde eU Nd
or eeTeT dmd
2
1 where
(Mass matrix)
1 1 1d d ( d )
2 2 2e e e
T T T T Te e e e
V V V
T V V V U U d N Nd d N N d
de
Te
V
Vm N N
Step 3: Formation of FE equations in local Step 3: Formation of FE equations in local coordinatescoordinates
eTes
Teb
TefW FdFdFd
sbe FFf (Force vector)
d d ( d ) ( d )e e e e
T T T T T T T Tf e b e s e b e s
V S V S
W V S V S d N f d N f d N f d N f
de
Tb b
V
VF N f de
Ts s
S
SF N f
Step 3: Formation of FE equations in local Step 3: Formation of FE equations in local coordinatescoordinates
)(d
d)
d
d( T
e
TeT
e ttd
dd
ttt ee
t
t
Teee
t
t
Te
t
teeTeee
t
t
Te ddd
2
1
2
1
2
1
2
1
dmddmddmddmd
0d)(2
1
teeee
Te
t
tFkddmd
0d)2
1
2
1(
2
1
te
Teee
Teee
Te
t
tFddkddmd
eeeee fdmdk
FE Equation
(Hamilton’s principle)
2
1
( )d 0
t T T Te e e e e e e et
t d m d d k d d f
Step 4: Coordinate Step 4: Coordinate transformationtransformation
eeee fdmkd
x
y
x'y'
y'
x'
Local coordinate systems
Global coordinate systems
ee TDd
eeeee FDMDK
TkTK eT
e TmTM eT
e eT
e fTF , ,
where
(Local)
(Global)
Step 5: Assembly of FE equationsStep 5: Assembly of FE equations
Direct assembly method– Adding up contributions made by elements
sharing the node
FDMKD
FKD (Static)
eeeee FDMDK
Step 6: Impose displacement constraintsStep 6: Impose displacement constraints
No constraints rigid body movement (meaningless for static analysis)
Remove rows and columns corresponding to the degrees of freedom being constrained
K is semi-positive definite
Step 7: Solve the FE equationsStep 7: Solve the FE equations
Solve the FE equation,
for the displacement at the nodes, D
The strain and stress can be retrieved by using = LU and = c with the interpolation, U=Nd
FDMKD
STATIC ANALYSISSTATIC ANALYSIS
Solve KD=F for D
– Gauss elimination– LU decomposition– Etc.
EIGENVALUE ANALYSISEIGENVALUE ANALYSIS
0 DMKD (Homogeneous equation, F = 0)
Assume )exp( tiD
0][ 2 MK
Let 2 0][ MK
0]det[ MKMK
[ K i M ] i = 0 (Eigenvector)
(Roots of equation are the eigenvalues)
EIGENVALUE ANALYSISEIGENVALUE ANALYSIS
Methods of solving eigenvalue equation– Jacobi’s method– Given’s method and Householder’s method– The bisection method (Sturm sequences)– Inverse iteration– QR method– Subspace iteration– Lanczos’ method
TRANSIENT ANALYSISTRANSIENT ANALYSIS
Structure systems are very often subjected to transient excitation.
A transient excitation is a highly dynamic time dependent force exerted on the structure, such as earthquake, impact, and shocks.
The discrete governing equation system usually requires a different solver from that of eigenvalue analysis.
The widely used method is the so-called direct integration method.
TRANSIENT ANALYSISTRANSIENT ANALYSIS(reading material)(reading material)
The direct integration method is basically using the finite difference method for time stepping.
There are mainly two types of direct integration method; one is implicit and the other is explicit.
Implicit method (e.g. Newmark’s method) is more efficient for relatively slow phenomena
Explicit method (e.g. central differencing method) is more efficient for very fast phenomena, such as impact and explosion.
REMARKSREMARKS In FEM, the displacement field U is expressed by
displacements at nodes using shape functions N defined over elements.
The strain matrix B is the key in developing the stiffness matrix.
To develop FE equations for different types of structure components, all that is needed to do is define the shape function and then establish the strain matrix B.
The rest of the procedure is very much the same for all types of elements.
Newmark’s method (Implicit)Newmark’s method (Implicit)
Assume that
2 1
2t t t t t t tt t
D D D D D
1t t t t t tt D D D D
KD CD MD FSubstitute into
2 1
2
1
t t t t t
t t t t t t t t
t t
t
K D D D D
C D D D MD F
Typically
= 0.5
= 0.25
Newmark’s method (Implicit)Newmark’s method (Implicit)residual
cm t t t t K D F
where
2
cm t t K K C M
2residual 11
2t t t t t t t t tt t t
F F K D D D C D D
Therefore, 1cm
residualt t t t
D K F
Newmark’s method (Implicit)Newmark’s method (Implicit)
Start with D0 and 0D
Obtain 0D KD CD MD Fusing
1cm
residualt t t t
D K FObtain tD using
Obtain Dt and tD using
2 1
2t t t t t t tt t
D D D D D
1t t t t t tt D D D D
March forward in time
Central difference method (explicit)Central difference method (explicit)int residual MD F CD KD F F F
residual 1D M F (Lumped mass – no need to solve matrix equation)
2t t t t tt D D D
2t t t t tt D D D
2
12t t t t t t
t
D D D D
2
2t t t t t
tt
D D D D
Central Central difference difference
method method (explicit)(explicit)
D,
t
x
x
x x
x
t0t-t -t/2 t/2
Find average velocity at time t = -t/2 using
Find using the average acceleration at time t = 0.
Find Dt using the average velocity at time t =t/2
Obtain D-t using
D0 and are prescribed and
can be obtained from
Use to obtain assuming . Obtain using
Time marching in half the time step
0D
0D
residual 1D M F
2
2t t t t t
tt
D D D D
/ 2t D
/ 2 / 2t t t t tt D D D
/ 2tD
/ 2 / 2t t t t tt D D D
/ 2 / 2t t t t tt D D D
/ 2 / 2t t t t tt D D D
tD / 2 0t D D
tD residual 1D M F
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