Lecture 1 test

Swiss Federal Institute of Technology

The Finite Element Method
for the Analysis of Linear Systems

Prof. Dr. Michael Havbro Faber
ETH Zurich, Switzerland

Method of Finite Elements I

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Contents of Today's Lecture

•

Motivation, overview and organization of the course

•

Introduction to the use of finite element
- Physical problem, mathematical modeling and finite element
solutions
- Finite elements as a tool for computer supported design and
assessment

•

Basic mathematical tools


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Page 3


•

Motivation
In this course we are focusing on the assessment of the response
of engineering structures



Page 4


•

Motivation
In this course we are focusing on the assessment of the response
of engineering structures



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•

Motivation
What we would like to establish is the response of a structure
subject to “loading”.
The Method of Finite Elements provides a framework for the
analysis of such responses – however for very general problems.
The Method of Finite Elements provides a very general approach
to the approximate solutions of differential equations.
In the present course we consider a special class of problems,
namely:
Linear quasi-static systems, no material or geometrical or
boundary condition non-linearities and also no inertia effect!




•

Organisation
The lectures will be given by:
M. H. Faber
Exercises will be organized/attended by:
J. Qin
By appointment, HIL E13.1


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Page 7


•

Organisation
PowerPoint files with the presentations will be uploaded on our
homepage one day in advance of the lectures
http://www.ibk.ethz.ch/fa/education/ss_FE
The lecture as such will follow the book:
"Finite Element Procedures" by K.J. Bathe, Prentice Hall, 1996




•

Overview


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•

Overview


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- we are only working on the
basis of mathematic models!
- choice of mathematical
model is crucial!
- mathematical models must
be reliable and effective

Improve mathematical model

Physical problem,
mathematical modeling and
finite element solutions
Change physical problem

•

Physical problem

Mathematical model governed by
differential equations and assumptions on
-geometry
-kinematics
-material laws
-loading
-boundary conditions
-etc.
Finite element solution
Choice of
-finite elements
-mesh density
-solution parameters
Representation of
-loading
-boundary conditions
-etc.
Assessment of accuracy of finite element
Solution of mathematical model

Refinement of analysis
Design improvements

Interpretation of results


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•

Reliability of a mathematical model
The chosen mathematical model is reliable if the required response
is known to be predicted within a selected level of accuracy
measured on the response of a very comprehensive mathematical
model

•

Effectiveness of a mathematical model
The most effective mathematical model for the analysis is surely
that one which yields the required response to a sufficient
accuracy and at least costs




•

Example
Complex physical problem modeled by a simple mathematical
model

M = WL
= 27,500 Ncm
1 W ( L + rN )3 W ( L + rN )
δ at load W =
+
5
3
EI
AG
6
= 0.053cm



•

Example
Detailed reference model – 2D plane stress model – for FEM
⎫
∂τ xx ∂τ xy
analysis
+
= 0⎪
∂x
∂τ yx

∂y
∂τ yy

⎪
⎬ in domain of bracket
+
= 0⎪
⎪
∂x
∂y
⎭

τ nn = 0, τ nt = 0 on surfaces except at point B
and at imposed zero displacements
Stress-strain relation:
⎡
⎤
⎢1 ν
⎡τ xx ⎤
0 ⎥ ⎡ε xx ⎤
⎥⎢ ⎥
E ⎢
⎢ ⎥
τ yy ⎥ =
ν 1
0 ⎥ ⎢ε yy ⎥
⎢
1 −ν 2 ⎢
⎢
⎢τ xy ⎥
1 −ν ⎥ ⎢γ xy ⎥
⎣ ⎦
0 0
⎢
⎥⎣ ⎦
2 ⎦
⎣
∂u
∂v
∂u ∂v
+
Strain-displacement relation: ε xx = ; ε yy = ; γ xy =
∂x
∂y
∂y ∂x



•

Example
Comparison between simple and more refined model results
M = WL
= 27,500 Ncm

1 W ( L + rN )3 W ( L + rN )
δ at load W =
+
5
3
EI
AG
6
= 0.053cm

δ

at load W

M


x =0

= 0.064cm

= 27,500 Ncm

Reliability and efficiency
may be quantified!


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•

Observations
Choice of mathematical model must correspond to desired
response measures
The most effective mathematical model delivers reliable answers
with the least amount of efforts
Any solution (also FEM) of a mathematical model is limited to
information contained in the model – bad input – bad output
Assessment of accuracy is based on comparisons with results
from very comprehensive models – however, in practice often
based on experience



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•

Observations
Sometimes the chosen mathematical model results in problems
such as singularities in stress distributions
The reason for this is that simplifications have been made in the
mathematical modeling of the physical problem
Depending on the response which is really desired from the
analysis this may be fine – however, typically refinements of the
mathematical model will solve the problem




•

Finite elements as a tool for computer supported design and
assessment
FEM forms a basic tool
framework in research and
applications covering many
different areas
- Fluid dynamics
- Structural engineering
- Aeronautics
- Electrical engineering
- etc.


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Page 18


•

Finite elements as a tool for computer supported design and
assessment
The practical application necessitates that solutions obtained by
FEM are reliable and efficient
however
also it is necessary that the use of FEM is robust – this implies
that minor changes in any input to a FEM analysis should not
change the response quantity significantly
Robustness has to be understood as directly related to the
desired type of result – response




•

Vectors and matrices
Ax = b

⎡ a11
⎢
⎢
A = ⎢ ai1
⎢
⎢
⎢ am1
⎣

AT is the transpose of A
a1i
aii

⎡ x1 ⎤
⎡ b1 ⎤
⎢x ⎥
⎢b ⎥
2⎥
x=⎢ , b=⎢ 2⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
xn ⎦
⎣
⎣bm ⎦

a1n ⎤
⎥
⎥
⎥
⎥
⎥
amn ⎥
⎦

if A = AT there is
m = n (square matrix)
and aij = a ji (symmetrical matrix)

⎡1
⎢0
⎢
I = ⎢0
⎢
⎢
⎢0
⎣

0 0
1 0
0 1
0 0

0⎤
0⎥
⎥
0 ⎥ is a unit matrix
⎥
⎥
1⎥
⎦



•

Banded matrices
symmetric banded matrices
aij = 0 for j > i + mA , 2mA + 1 is the bandwidth
⎡3
⎢2
⎢
A = ⎢1
⎢
⎢0
⎢0
⎣

0⎤
0⎥
⎥
1⎥
⎥
4⎥
0 1 4 3⎥
⎦

2
3
4
1


1
4
5
6

0
1
6
7

mA = 2



•

Banded matrices and skylines
mA + 1

⎡3
⎢2
⎢
A = ⎢0
⎢
⎢0
⎢0
⎣


2 0 0 0⎤
3 0 1 0⎥
⎥
0 5 6 1⎥
⎥
1 6 7 4⎥
0 1 4 3⎥
⎦



•

Matrix equality

A ( m × p ) = B ( n × q ) if and only if
m = n,
p = q,
and aij = bij


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•

Matrix addition
A ( m × p ) , B ( n × q ) can be added if and only if
m = n, p = q, and
if C = A + B, then
cij = aij + bij


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•

Matrix multiplication with a scalar

A matrix A multiplied by a scalar c by multiplying all elements of A with c
B = cA
bij = caij


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•

Multiplication of matrices

Two matrices A ( p × m ) and B ( n × q ) can be multiplied only if m = n
C = BA
m

cij = ∑ air brj , C ( p × q )
r =1


Page 25



•

Multiplication of matrices

The commutative law does not hold, i.e. AB = CB does not imply that A = C
AB ≠ BA, unless A and B commute

however does hold for special cases
(e.g. for B = I)

The distributive law hold, i.e.

Special rule for the transpose of matrix products
E = ( A + B ) C = AC + BC

( AB )

T

The associative law hold, i.e.
G = (AB)C = A(BC) = ABC

= BT AT



•

The inverse of a matrix
The inverse of a matrix A is denoted A −1
if the inverse matrix exist then there is:
AA −1 = A −1A = I
The matrix A is said to be non-singular
The inverse of a matrix product:

( AB ) = B-1A -1
-1


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•

Sub matrices

A matrix A may be sub divided as:
⎡ a11
A = ⎢ a21
⎢
⎢ a31
⎣

a12
a22
a32

⎡ a11
A=⎢
⎢ a21
⎣

a12 ⎤
⎥
a22 ⎥
⎦


a13 ⎤
a23 ⎥
⎥
a33 ⎥
⎦

Page 28



•

Trace of a matrix

The trace of a matrix A ( n × n ) is defined through:
n

tr ( A) = ∑ aii
i =1


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•

The determinant of a matrix
The determinant of a matrix is defined through the recurrence formula
n

det( A ) = ∑ (−1)1+ j a1 j det(A1 j )
j =1

where A1 j is the ( n − 1) × ( n − 1) matrix obtained by eliminating
the 1st row and the j th column from the matrix A and where there is
if A = [ a11 ] , det A = a11


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•

The determinant of a matrix
It is convenient to decompose a matrix A by the so-called
Cholesky decomposition
⎡ 1 0 0⎤
T
L = ⎢l21 1 0 ⎥
A = LDL
⎢
⎥
⎢l31 l32 1 ⎥
⎣
⎦
where L is a lower triangular matrix with all diagonal elements
equal to 1 and D is a diagonal matrix with components dii then
the determinant of the matrix A can be written as
n

det A = ∏ d ii
i =1


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•

Tensors

Let the Cartesian coordinate frame be defined by
the unit base vectors ei

x3

A vector u in this frame is given by
e3

3

u = ∑ ui ei
e2
e1

x1

x2

i =1

simply we write
u = ui ei
i is called a dummy index or a free index




•

Tensors
An entity is called a tensor of first order
if it has 3 components ξi in the unprimed frame
and 3 components ξi' in the primed frame,
and if these components are related by the characteristic law

ξi' = pik ξi
where pik = cos ( ei' , e k )
In the matrix form, it can be written as
ξ ' = Pξ

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•

Tensors

An entity is called a second-order tensor
if it has 9 components tij in the unprimed frame
'
and 9 components tij in the primed frame,

and if these components are related by the characteristic law
'
tij = pik p jl tkl


Page 34

Lecture 1 test

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