Transcript An Introduction to the Finite Element Analysis

An Introduction to the
Finite Element Analysis
Presented by
Niko Manopulo
Agenda
PART I
Introduction and Basic Concepts
1.0
Computational Methods
1.1
1.2
1.3
2.0
The Finite Elements Method
2.1
2.2
3.0
Idealization
Discretization
Solution
FEM Notation
Element Types
Mechanichal Approach
3.1
3.2
3.3
3.4
The Problem Setup
Strain Energy
External Energy
The Potential Energy Functional
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Agenda
PART II
Mathematical Formulation
4.0
The Mathematics Behind the Method
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Weighted Residual Methods
Approxiamting Functions
The Residual
Galerkin’s Method
The Weak Form
Solution Space
Linear System of Equations
Connection to the physical system
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Agenda
PART III
Finite Element Discretization
5.0
Finite Element Discretization
4.1
4.2
4.3
4.4
4.5
4.6
4.7
6.0
7.0
The Trial Basis
Matrix Form of the Problem
Element Stiffness Matrix
Element Mass Matrix
External Work Integral
Assembling
Linear System of Equations
References
Question and Answers
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PART I
Introduction and Basic Concepts
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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3.2 Strain Energy
• Hooke’s Law:
 ( x)  E ( x)
where
du
 ( x) 
dx
• Strain Energy Density:
1
   ( x) ( x)
2
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3.2 Strain Energy (cont’d)
• Integrating over the Volume of the Bar:
1
1 L
1 L
U    dV   p dx   ( EAu' )u ' dx
2 V
2 0
2 0
1 L
U   u ' EAu' dx
2 0
• All quantities may depend on x.
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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
W   qu dx
0
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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:
  U W
or
[u ( x)]  U [u ( x)]  W [u ( x)]
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PART II
Mathematical Formulation
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 :
L[u ]  
d
du
( p( x) )  z ( x)u  q( x),
dx
dx
u (0)  u (1)  0.
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0  x 1
(1)
4.1 Weighted Residual Methods
It follows that:
L[u]  q  0
Multiplying this by a weight function v and integrating over
the whole domain we obtain:
1
 (L[u]  q)v dx  (v, L[u]  q)  0
0
(2)
For the inner product to exist v must be “square integrable”
Therefore:
v  L2 (0,1)
Equation (2) is called variational form
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4.2 Approximating Function
• We can replace u and v in the formula with their approximation
function i.e.
N
u ( x )  U ( x )   c jf j ( x )
j 1
N
v( x)  V ( x)   d jy j ( x)
j 1
• 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.
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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
u  C 2 (0,1)  U  C 2 (0,1)
• Therefore we can see U as element of a finite-diemnsional
subspace of the infinite-dimensional function space C2(0,1)
U  S N (0 ,1)  C 2 (0 ,1)
• The same way
V  Sˆ N (0 ,1)  L2 (0 ,1)
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4.3 The Residual
•
Replacing v and u with respectively V and U (2) becomes
(V , r )  0,
r ( x)  L[U ]  q
•
•
V  Sˆ N
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.
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4.3 The Residual
• Substituting V ( x) 
N
d y
j 1
j
j
( x) into
(V , L[U ]  q)  0
and exchanging summations and integrals we obtain
N
 d (y
j 1
j
j
, L[U ]  q)  0
d j ,
j  1, 2 , ..., N
• 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
(y j , L[U ]  q)  0
j  1, 2 , ..., N
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4.4 Galerkin’s Method
• One obvious choice of y j would be taking it equal to f j
• This Choice leads to the Galerkin’s Method
(f j , L[U ]  q)  0
j  1, 2 , ..., N
• 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.
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4.5 The Weak Form
• Let us remember the initial form of the problem
L[u ]  
d
du
( p( x) )  z ( x)u  q( x),
dx
dx
0  x 1
u (0)  u (1)  0.
1
(v, L[u ]  q)   v [( pu' )' zu  q] dx
0
• Integrating by parts
1
1
 v [( pu' )' zu  q] dx   (v' pu'vzu  vq) dx  vpu'
0
0
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1
0
0
4.5 The Weak Form
• The problem can be rewritten as
A(v , u)  (v , q)  0
where
1
A(v , u )   (v' pu'vzu) dx
0
• 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.
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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
1
A(u , u )   ( p(u ' ) 2  zu 2 ) dx
•
0
As p and z are necessarily smooth functions the following restriction is
sufficient
1
2
2
(
u
'
)

u
dx

0
•
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
1
conditions so we denote the resulting space as H 0
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4.7 Linear System of Equations
• The solution now takes the form
A(v , u)  (v , q)
v  H01
• Substituting the approximate solutions obtained earlier in the
more general WRM we obtain
U ,V  S0N  H 01
A(V ,U )  (V , q)
V  S0N
• More explicitly substituting U and V (remember we chose
them to have the same base) and swapping summations and
integrals we obtain
N
 c A(f
k 1
k
j
, fk )  (f j , q) ,
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j  1, 2 , ..., N
4.8 Connection to the Physical System
Mechanical Formulation
Mathematical Formulation
1
1 L
U   u ' EAu ' dx
2 0
L
A(v , u )   (v' pu'vzu) dx
0
W   qu dx
(v , q )
  U  W  0
A(v , u)  (v , q)  0
0
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PART III
Finite Element Discretization
5.0 Finite Element Discretization
• Let us take the initial value problem with constant coefficients
 pu' ' zu  q( x),
0  x 1
p, z  0
u (0)  u (1)  0.
• As a first step let us divide the domain in N subintervals with
the following mesh
0  x0  x1  ...  xN  1
• Each subinterval ( x j 1 , x j ), j  1: N is called finite element.
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5.1 The Trial Basis
• Next we select as a basis the so called “hat function”.
 x  x j 1

 x j  x j 1
 x x
f j ( x)   j 1
 x j 1  x j


 0
x j 1  x  x j
x j  x  x j 1
otherwise
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5.1 The Trial Basis
• 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
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5.1 The Trial Basis
• The problem at this point can be easily solved using the
previously derived Galerkin’s Method
N
 c A(f
k 1
k
j
, fk )  (f j , q) ,
j  1, 2 , ..., N
• A little more work is needed to convert this problem into
matrix notation
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5.2 Matrix Form of the Problem
• Restricting U over the typical finite element we can write
U ( x)  c j 1f j 1 ( x)  c jf j ( x)
x [ x j 1 , x j ]
• Which in turn can be written as
U ( x)  [c j 1
f j 1 ( x)
c j 1 
c j ]
  [f j 1 ( x) f j ( x)] c 
f
(
x
)
 j

 j 
x  [ x j 1 , x j ]
in the same way
V ( x)  [d j 1
f j 1 ( x)
d j 1 
d j ]
  [f j 1 ( x) f j ( x)] d 
f
(
x
)
 j

 j 
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x  [ x j 1 , x j ]
5.2 Matrix Form of the Problem
• Taking the derivative
U ' ( x)  [c j 1
 1 / h j 
c j 1 
c j ]
  [1 / h j 1 / h j ] c 
1
/
h
j 

 j 
h j  x j  x j 1
x  [ x j 1 , x j ]
• Derivative of V is analogus
V ' ( x)  [d j 1
 1 / h j 
d j 1 
d j ]
 [1 / h j 1 / h j ]


1
/
h
d
j 

 j 
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x  [ x j 1 , x j ]
5.2 Matrix Form of the Problem
• The variational formula can be elementwise defined as
follows:
N
 [ A (V ,U )  (V , q)
j 1
j
j
]0
A j (V , U )  ASj (V ,U )  A Mj (V ,U )
A (V , U )  
S
j
xj
x j 1
A (V , U )  
M
j
xj
x j 1
pV 'U ' dx
zVU dx
xj
(V , q )   Vq dx
x j 1
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4.3 The Element Stiffnes Matrix
• Substituting U,V,U’ and V’ into these formulae we obtain
A (V , U )  [d j 1
S
j
A (V , U )  [d j 1
S
j
 xj
d j ] 
 x j1

 xj p
d j ] 
 x j1 h 2
j

 c j 1 
 1 / h j 
p
[1 / h j 1 / h j ] dx   

 cj
 1/ hj 
 
c j 1 
 1  1  c j 1 
 1 1  dx   c   [d j 1 d j ]K j  c 

  j 
 j 
p  1 1 
Kj  
h j  1  1
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4.4 The Element Mass Matrix
• The same way
 x j f j 1 
 c j 1 
A (V , U )  [d j 1 d j ]  z  [f j 1 f j ] dx   
 x j1 f j
 cj
 

 
c j 1 
S
A j (V , U )  [d j 1 d j ]M j  
 cj 
M
j
zh j 2 1
Mj 
6 1 2
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4.5 External Work
Integral
• The external work integral cannot be evaluated for every
function q(x)
xj
(V , q)   Vq dx
x j 1
• We can consider a linear interpolant of q(x) for simplicity.
q( x)  q j 1f j 1 ( x)  q jf j ( x),
x [ x j 1, x j ]
• Substituting and evaluating the integral
f j 1 
q j 1 
(V , q) j   [d j 1 , d j ] [f j 1 , f j ]
dx  [d j 1 , d j ] l j

x j 1
 fj 
 qj 
h j 2q j 1  q j 
lj  
Element load vector

6 q j 1  2q j 
xj
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4.6 Assembling
• 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
N

j 1
 c1 
c 
S
T
A j  d Kc c   2 
 ... 


c
 N 1 
 d1 
d 
d  2 
 ... 


d
 N 1 
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 2 1

 1 2  1





1 2 1
p
K 

... ... ...
h


 1 2  1


 1 2 

4.6 Assembling
• Doing the same for the Mass Matrix and for the Load Vector
N
M
T
A

d
Mc
 j
j 1
N
 (V , q)
j 1
j
 dTl
4 1

1 4 1




zh  1 4 1
M 

... ... ... 
6 

1 4 1


1 4

 q0  4q1  q2 


h  q1  4q2  q3 
l

...
6


q N  2  4q N 1  q N 
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4.7 Linear System of Equations
• Substituting this Matrix form of the expressions in
N
[ A (V ,U )  (V , q)
j 1
j
j
]0
we obtain the following set of linear equations
dT[(K  M)c  l]  0
• This has to be satisfied for all choices of d therefore
(K  M)c  l  0
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References
• Carlos Felippa
http://caswww.colorado.edu/courses.d/IFEM.d/IFE
M.Ch06.d/IFEM.Ch06.pdf
• Joseph E Flaherty,Amos Eaton Professor
http://www.cs.rpi.edu/~flaherje/FEM/fem1.ps
• Gilbert Strang, George J. Fix
An Analysis of the Finite Element Method
Prentice-Hall,1973
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