The finite element method

Swiss Federal Institute of Technology

The Finite Element Method for the Analysis of Linear Systems

Prof. Dr. Michael Havbro Faber Swiss Federal Institute of Technology ETH Zurich, Switzerland

Method of Finite Elements I

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Contents of Today's Lecture

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Motivation, overview and organization of the course

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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

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Basic mathematical tools


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



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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!




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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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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




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Overview


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Overview


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Introduction to the use of finite element

- 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

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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 Method of Finite Elements I

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

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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


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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 Method of Finite Elements I

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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 ν = 1 0 ⎥ ⎢ε yy ⎥ ⎢ 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 Method of Finite Elements I

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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




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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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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


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Vectors and matrices Ax = b

a1n ⎤ ⎡ a11 " a1i ⎢ # % ⎥ ⎢ ⎥ aii A = ⎢ ai1 # ⎥ ⎢ ⎥ # % ⎢ ⎥ ⎢⎣ am1 amn ⎥⎦ " ⎡ x1 ⎤ ⎡ b1 ⎤ ⎢x ⎥ ⎢b ⎥ 2 x = ⎢ ⎥, b = ⎢ 2 ⎥ ⎢#⎥ ⎢#⎥ ⎢ ⎥ ⎢ ⎥ x ⎣ n⎦ ⎣bm ⎦ Method of Finite Elements I

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

⎡1 ⎢0 ⎢ I = ⎢0 ⎢ ⎢# ⎢⎣0

0 0 " 0⎤ 1 0 " 0 ⎥⎥ 0 1 " 0 ⎥ is a unit matrix ⎥ # # % #⎥ 0 0 " 1 ⎥⎦

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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

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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 ⎥⎦



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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


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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 Method of Finite Elements I

= BT AT



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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 ⎥⎦

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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⎤ ⎢l ⎥ T 1 0 L = A = LDL ⎢ 21 ⎥ ⎢⎣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




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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 Method of Finite Elements I

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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


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The finite element method

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