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An Introduction to the Finite Element Analysis
Joint Advanced Student School
St.Petersburg 2005
Agenda
PART I
Introduction and Basic Concepts
1.0 Computational Methods1.1 Idealization
1.2 Discretization
1.3 Solution
2.0 The Finite Elements Method2.1 FEM Notation
2.2 Element Types
3.0 Mechanichal Approach3.1 The Problem Setup
3.2 Strain Energy
3.3 External Energy
3.4 The Potential Energy Functional
Joint Advanced Student School
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Agenda
PART II
Mathematical Formulation
4.0 The Mathematics Behind the Method
4.1 Weighted Residual Methods
4.2 Approxiamting Functions
4.3 The Residual
4.4 Galerkin’s Method
4.5 The Weak Form
4.6 Solution Space
4.7 Linear System of Equations
4.8 Connection to the physical system
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Agenda
PART III
Finite Element Discretization
5.0 Finite Element Discretization
4.1 The Trial Basis
4.2 Matrix Form of the Problem
4.3 Element Stiffness Matrix
4.4 Element Mass Matrix
4.5 External Work Integral
4.6 Assembling
4.7 Linear System of Equations
6.0 References
7.0 Question and Answers
PART I
Introduction and Basic Concepts
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1.0 Computational Methods
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1.1 Idealization
• Mathematical Models• “A model is a symbolic device built to
simulate and predict aspects of behavior of a system.”
• Abstraction of physical reality
• Implicit vs. Explicit Modelling• Implicit modelling consists of using existent
pieces of abstraction and fitting them into the particular situation (e.g. Using general purpose FEM programs)
• Explicit modelling consists of building the model from scratch
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1.2 Dicretization
1. Finite Difference Discretization
• The solution is discretized
• Stability Problems
• Loss of physical meaning
2. Finite Element Discretization
• The problem is discretized
• Physical meaning is conserved on elements
• Interpretation and Control is easier
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1.3 Solution
1. Linear System Solution Algorithms
• Gaussian Elimination
• Fast Fourier Transform
• Relaxation Techniques
2. Error Estimation and Convergence
Analysis
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2.0 Finite Element Method
• Two interpretations
1. Physical Interpretation:The continous physical model is divided into finite pieces called elements and laws of nature are applied on the generic element. The results are then recombined to represent the continuum.
2. Mathematical Interpretation:The differetional equation reppresenting the system is converted into a variational form, which is approximated by the linear combination of a finite set of trial functions.
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2.1 FEM Notation
Elements are defined by the following
properties:
1. Dimensionality
2. Nodal Points
3. Geometry
4. Degrees of Freedom
5. Nodal Forces
(Non homogeneous RHS of the DE)
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2.2 Element Types
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3.0 Mechanical Approach
• Simple mechanical problem
• Introduction of basic mechanical concepts
• Introduction of governing equations
• Mechanical concepts used in mathematical
derivation
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3.1 The Problem Setup
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• Hooke’s Law:
where
• Strain Energy Density:
3.2 Strain Energy
dx
dux )(
)()( xEx
)()(2
1xx
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• Integrating over the Volume of the Bar:
• All quantities may depend on x.
3.2 Strain Energy (cont’d)
dxEAuuU
dxuEAudxpdVU
L
LL
V
0
00
''2
1
')'(2
1
2
1
2
1
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3.3 External Energy
• Due to applied external loads
1. The distributed load q(x)
2. The point end load P. This can be
included in q.
• External Energy:
L
dxquW0
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3.4 The Total Potential
Energy Functional
• The unknown strain Function u is found by
minimizing the TPE functional described below:
)]([)]([)]([ xuWxuUxu
or
WU
PART II
Mathematical Formulation
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4.0 Historical Background
• Hrennikof and McHenry formulated a 2D structural problem as an assembly of bars and beams
• Courant used a variational formulation to approximate PDE’s by linear interpolation over triangular elements
• Turner wrote a seminal paper on how to solve one and two dimensional problems using structural elements or triangular and rectangular elements of continuum.
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4.1 Weighted Residual Methods
The class of differential equations containing also the one
dimensional bar described above can be described as
follows :
.0)1()0(
)1(10),()())((][
uu
xxquxzdx
duxp
dx
duL
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It follows that:
Multiplying this by a weight function v and integrating over
the whole domain we obtain:
For the inner product to exist v must be “square integrable”
Therefore:
Equation (2) is called variational form
0][ quL
4.1 Weighted Residual Methods
)2(0)][,()][(1
0 quLvdxvquL
)1,0(2Lv
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4.2 Approximating Function
• We can replace u and v in the formula with their approximation function i.e.
• The functions fj and yj are of our choice and are meant to be suitable to the particular problem. For example the choice of sine and cosine functions satisfy boundary conditions hence it could be a good choice.
N
j
jj
N
j
jj
xdxVxv
xcxUxu
1
1
)()()(
)()()(
y
f
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4.2 Approximating Function
• U is called trial function and V is called test function
• As the differential operator L[u] is second order
• Therefore we can see U as element of a finite-diemnsional
subspace of the infinite-dimensional function space C2(0,1)
• The same way
)1,0()1,0( 22 CUCu
)1,0()1,0( 2CSU N
)1,0()1,0(ˆ 2LSV N
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4.3 The Residual
• Replacing v and u with respectively V and U (2) becomes
• r(x) is called the residual (as the name of the method suggests)
• The vanishing inner product shows that the residual is orthogonal to
all functions V in the test space.
qULxr
SVrV N
][)(
ˆ,0),(
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• Substituting into
and exchanging summations and integrals we obtain
• As the inner product equation is satisfied for all choices of V in SN the above equation has to be valid for all choices of dj
which implies that
4.3 The Residual
N
j
jj xdxV1
)()( y 0)][,( qULV
NjdqULd j
N
j
jj ,...,2,1,0)][,(1
y
NjqULj ,...,2,10)][,( y
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• One obvious choice of would be taking it equal to
• This Choice leads to the Galerkin’s Method
• This form of the problem is called the strong form of the
problem. Because the so chosen test space has more
continuity than necessary.
• Therefore it is worthwile for this and other reasons to convert
the problem into a more symmetrical form
• This can be acheived by integrating by parts the initial strong
form of the problem.
4.4 Galerkin’s Method
jyjf
NjqULj ,...,2,10)][,( f
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• Let us remember the initial form of the problem
• Integrating by parts
4.5 The Weak Form
dxqzupuvquLv
1
0
])''([)][,(
.0)1()0(
10),()())((][
uu
xxquxzdx
duxp
dx
duL
0')''(])''([1
0
1
0
1
0
vpudxvqvzupuvdxqzupuv
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4.5 The Weak Form
• The problem can be rewritten as
where
• The integration by parts eliminated the second derivatives
from the problem making it possible less continouity than the
previous form. This is why this form is called weak form of the
problem.
• A(v,u) is called Strain Energy.
0),(),( qvuvA
dxvzupuvuvA )''(),(
1
0
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4.6 Solution Space
• Now that derivative of v comes into the picture v needs to have more
continoutiy than those in L2. As we want to keep symmetry its
appropriate to choose functions that produce bounded values of
• As p and z are necessarily smooth functions the following restriction is
sufficient
• Functions obeying this rule belong to the so called Sobolev Space
and they are denoted by H1. We require v and u to satisfy boundary
conditions so we denote the resulting space as
dxzuupuuA ))'((),(
1
0
22
1
0
1
0
22)'(
H
dxuu
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• The solution now takes the form
• Substituting the approximate solutions obtained earlier in the
more general WRM we obtain
• More explicitly substituting U and V (remember we chose
them to have the same base) and swapping summations and
integrals we obtain
4.7 Linear System of Equations
1
0),(),( HvqvuvA
N
N
SVqVUVA
HSVU
0
1
00
),(),(
,
N
k
jkjk NjqAc1
,...,2,1,),(),( fff
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4.8 Connection to the Physical System
Mechanical Formulation Mathematical Formulation
dxEAuuUL
0
''2
1
L
dxquW0
0 WU
dxvzupuvuvA )''(),(
1
0
0),(),( qvuvA
),( qv
PART III
Finite Element Discretization
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• Let us take the initial value problem with constant coefficients
• As a first step let us divide the domain in N subintervals with
the following mesh
• Each subinterval is called finite element.
5.0 Finite Element Discretization
Njxx jj :1),,( 1
1...0 10 Nxxx
.0)1()0(
0,
10),(''
uu
zp
xxqzupu
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• Next we select as a basis the so called “hat function”.
5.1 The Trial Basis
otherwise
xxxxx
xx
xxxxx
xx
x jj
jj
j
jj
jj
j
j
0
)( 1
1
1
1
1
1
f
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• With the basis in the previous slide we construct our
approximate solution U(x)
• It is interesting to note that the coefficients correspond to the
values of U at the interior nodes
5.1 The Trial Basis
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• The problem at this point can be easily solved using the
previously derived Galerkin’s Method
• A little more work is needed to convert this problem into
matrix notation
5.1 The Trial Basis
N
k
jkjk NjqAc1
,...,2,1,),(),( fff
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• Restricting U over the typical finite element we can write
• Which in turn can be written as
in the same way
5.2 Matrix Form of the Problem
],[)]()([)(
)(][)( 1
1
1
1
1 jj
j
j
jj
j
j
jj xxxc
cxx
x
xccxU
ff
f
f
],[)()()( 111 jjjjjj xxxxcxcxU ff
],[)]()([)(
)(][)( 1
1
1
1
1 jj
j
j
jj
j
j
jj xxxd
dxx
x
xddxV
ff
f
f
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• Taking the derivative
• Derivative of V is analogus
5.2 Matrix Form of the Problem
1
1
1
1 ],[]/1/1[/1
/1][)('
jjj
jj
j
j
jj
j
j
jj
xxh
xxxc
chh
h
hccxU
],[]/1/1[/1
/1][)(' 1
1
1 jj
j
j
jj
j
j
jj xxxd
dhh
h
hddxV
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• The variational formula can be elementwise defined as
follows:
5.2 Matrix Form of the Problem
j
j
j
j
j
j
x
x
x
x
M
j
x
x
S
j
M
j
S
jj
N
j
jj
dxVqqV
dxzVUUVA
dxUpVUVA
UVAUVAUVA
qVUVA
1
1
1
),(
),(
''),(
),(),(),(
0]),(),([1
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• Substituting U,V,U’ and V’ into these formulae we obtain
4.3 The Element Stiffnes Matrix
11
11
][11
11][),(
]/1/1[/1
/1][),(
1
1
1
21
1
1
1
1
j
j
j
j
jjj
j
jx
xj
jj
S
j
j
jx
xjj
j
j
jj
S
j
h
pK
c
cKdd
c
cdx
h
pddUVA
c
cdxhh
h
hpddUVA
j
j
j
j
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4.4 The Element Mass Matrix
• The same way
21
12
6
][),(
][][),(
1
1
1
1
1
11
j
j
j
j
jjj
S
j
j
jx
xjj
j
j
jj
M
j
zhM
c
cMddUVA
c
cdxzddUVA
j
j
fff
f
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• The external work integral cannot be evaluated for every
function q(x)
• We can consider a linear interpolant of q(x) for simplicity.
• Substituting and evaluating the integral
4.5 External Work
Integral
j
j
x
xdxVqqV
1
),(
],[),()()( 111 jjjjjj xxxxqxqxq ff
jj
jjj
j
jjj
j
j
jj
x
xj
j
jjj
qqhl
ldddxq
qddqV
j
j
2
2
6
],[],[],[),(
1
1
1
1
1
1
11
fff
f
Element load vector
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• Now the task is to assemble the elements into the whole
system in fact we have to sum each integral over all the
elements
• For doing so we can extend the dimension of each element
matrix to N and then put the 2x2 matrix at the appropriate
position inside it
• With all matrices and vectors having the same dimension the
summation looks like
4.6 Assembling
N
j
S
jA1
KcdT
21
121
.........
121
121
12
h
pK
1
2
1
1
2
1
......
NN d
d
d
d
c
c
c
c
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• Doing the same for the Mass Matrix and for the Load Vector
4.6 Assembling
N
j
M
jA1
McdT
41
141
.........
141
141
14
6
zhM
N
j
jqV1
),( ldT
NNN qqq
qqq
qqq
h
12
321
210
4
...
4
4
6l
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• Substituting this Matrix form of the expressions in
we obtain the following set of linear equations
• This has to be satisfied for all choices of d therefore
4.7 Linear System of Equations
0lcMKdT ])[(
N
j
jj qVUVA1
0]),(),([
0lcMK )(
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References
• Carlos Felippahttp://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.pdf
• Joseph E Flaherty,Amos Eaton Professorhttp://www.cs.rpi.edu/~flaherje/FEM/fem1.ps
• Gilbert Strang, George J. Fix
An Analysis of the Finite Element Method
Prentice-Hall,1973