Lecture 1 test

Swiss Federal Institute of Technology Page 1

Method of Finite Elements I

The Finite Element Method for the Analysis of Linear Systems

Prof. Dr. Michael Havbro FaberSwiss Federal Institute of Technology

ETH Zurich, Switzerland



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



Motivation, overview and organization of the course

• Motivation

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




• Motivation

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




• 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 approachto 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




• 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




• Overview



Introduction to the use of finite element

• Physical problem, mathematical modeling and finite element solutions

- we are only working on the basis of mathematic models!

- choice of mathematical model is crucial!

- mathematical models must be reliable and effective

Physical problem

Mathematical model governed by differential equations and assumptions on-geometry-kinematics-material laws-loading-boundary conditions-etc.

Finite element solutionChoice of-finite elements-mesh density-solution parametersRepresentation of -loading-boundary conditions-etc.

Assessment of accuracy of finite elementSolution of mathematical model

Interpretation of resultsRefinement of analysis

Design improvements

Impr

ove

mat

hem

atic

al m

odel

Cha

nge

phys

ical

pro

blem



• 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





27,500M WL

Ncm==

3

( ) ( )1

536

0.053

N Nat load W

W L r W L rEI AG

cm

δ + += +

=

• Example

Complex physical problem modeled by a simple mathematical model



• Example

Detailed reference model – 2D plane stress model – for FEM analysis


0in domain of bracket

0

xyxx

yx yy

x y

x y

ττ

τ τ

∂ ⎫∂+ = ⎪∂ ∂ ⎪

⎬∂ ∂ ⎪+ = ⎪∂ ∂ ⎭

0, 0 on surfaces except at point Band at imposed zero displacementsnn ntτ τ= =

2

Stress-strain relation:

1 01 0

110 0

2

xx xx

yy yy

xy xy

Eτ ν ετ ν ε

ντ ν γ

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

Strain-displacement relation: ; ;xx yy xyu v u vx y y x

ε ε γ∂ ∂ ∂ ∂= = = +∂ ∂ ∂ ∂




27,500M WL

Ncm==

3

( ) ( )1

536

0.053

N Nat load W

W L r W L rEI AG

cm

δ + += +

=

• Example

Comparison between simple and more refined model results

0.064at load W cmδ =

0 27,500xM Ncm= =

Reliability and efficiency may be quantified!



Introduction to the use of finite element• 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



Introduction to the use of finite element• 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 themathematical 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



Introduction to the use of finite element• 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.



Introduction to the use of finite element• 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 impliesthat 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 thedesired type of result – response



Basic mathematical tools• Vectors and matrices

Ax = b

11 1 1

1

1

i n

i ii

m mn

a a a

a a

a a

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A =

1 1

2 2,

n m

x bx b

x b

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x = b =

1 0 0 00 1 0 0

is a unit matrix0 0 1 0

0 0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

I =

is the transpose of if there is

(square matrix)and (symmetrical matrix)

T

T

ij ji

m na a

==

=

A AA A



Basic mathematical tools• Banded matrices

symmetric banded matrices

3 2 1 0 02 3 4 1 01 4 5 6 10 1 6 7 40 0 1 4 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A =2Am =

0 for , 2 1 is the bandwidthij A Aa j i m m= > + +



Basic mathematical tools• Banded matrices and skylines

3 2 0 0 02 3 0 1 00 0 5 6 10 1 6 7 40 0 1 4 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A =

1Am +



Basic mathematical tools• Matrix equality

( ) ( ) if and only if, ,

and ij ij

m p n qm np qa b

× ×

==

=

A = B



Basic mathematical tools• Matrix addition

( ) ( ), can be added if and only if, , and

if , then

ij ij ij

m p n qm n p q

c a b

× ×

= == +

= +

A B

C A B



Basic mathematical tools• Matrix multiplication with a scalar

A matrix multiplied by a scalar by multiplying all elements of with

ij ij

c c

cb ca==

A A

B A



Basic mathematical tools• Multiplication of matrices

( ) ( )

( )1

Two matrices and can be multiplied only if

, m

ij ir rjr

p m n q m n

c a b p q=

× × =

= ×∑

A B

C = BA

C



Basic mathematical tools• Multiplication of matrices

( )

The commutative law does not hold, i.e.

, unless and commute

The distributive law hold, i.e.

The associative law hold, i.e.

≠AB BA A B

E = A + B C = AC + BC

G = (AB)C = A(BC) = ABC

( )

does not imply that

however does hold for special cases (e.g. for )

Special rule for the transpose of matrix products

T T T

AB = CB A = C

B = I

AB = B A



Basic mathematical tools• The inverse of a matrix

( )

1

1 1

-1 -1 -1

The inverse of a matrix is denoted

if the inverse matrix exist then there is:

The matrix is said to be non-singular

The inverse of a matrix product:

−

− −

A A

AA = A A = I

A

AB = B A



Basic mathematical tools• Sub matrices

11 12 13

21 22 23

31 32 33

11 12

21 22

A matrix may be sub divided as:

a a aa a aa a a

a a

a a

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

A

A =

A =



Basic mathematical tools• Trace of a matrix

( )

1

The trace of a matrix is defined through:

( )n

iii

n n

tr a=

×

∑

A

A =



Basic mathematical tools• The determinant of a matrix

( ) ( )

11 1

1

1

The determinant of a matrix is defined through the recurrence formula

det( ) ( 1) det( )

where is the 1 1 matrix obtained by eliminating

the 1 row and the column from the matrix

nj

j jj

j

st th

a

n n

j

+

=

−

− × −

∑A = A

A

[ ]11 11

and where there is

if ,deta a= =

A

A A



Basic mathematical tools• The determinant of a matrix

It is convenient to decompose a matrix by the so-called Cholesky decomposition

where is a lower triangular matrix with all diagonal elementsequal to 1 and is a diagonal matrix with componen

T

A

A = LDL

LD

1

ts then the determinant of the matrix can be written as

det

ii

n

iii

d

d=

=∏

A

A

21

31 32

1 0 01 0

1ll l

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

L =



Basic mathematical tools• Tensors

3

1

Let the Cartesian coordinate frame be defined bythe unit base vectors

A vector in this frame is given by

simply we write

is called a dummy index or a free index

i

i ii

i i

u

u

i

=

=

=

∑

e

u

u e

u e

1e

2e

3e

1x

2x

3x




'

'

An entity is called a tensor of first orderif it has 3 components in the unprimed frame

and 3 components in the primed frame,and if these components are related by the characteristic law

i

i

i ikp

ξ

ξ

ξ ξ=

( )'

'

where cos ,

In the matrix form, it can be written as

i

ik i kp =

=

e e

ξ Pξ




'

'

An entity is called a second-order tensorif it has 9 components in the unprimed frame

and 9 components in the primed frame,

and if these components are related by the characteristic law

ij

ij

ij ik

t

t

t p= jl klp t