Introduction to Finite Element Method

227 Pages·2005·1.33 MB·English

by  Tai Hun Kwon

Introduction to
Finite Element Method

Tai Hun Kwon

DEPARTMENT OF MECHANICAL ENGINEERING
POHANG UNIVERSITY OF SCIENCE & TECHNOLOGY

© 2005 by T. H. Kwon
1. INTRODUCTION

* Numerical Methods

1. Finite Difference Method (FDM)
(cid:122) Pointwise approximation to differential equation (DE)
(cid:122) Array of grid points

2. Finite Element Method (FEM)
(cid:122) Global approximation or integral approximation to DE
(cid:122) Assembly of finite elements (subdomains, subregions)

3. Boundary Element Method (BEM)
(cid:122) Deal with integral equation rather than differential equation
(cid:122) Discretization over boundary only

4. Finite Volume Method (control volume method)
5. Spectral Method (spectral element method)

* FEM is

good for i) various problems (engineering, physical problems)
ii) arbitrary geometry (complicated, irregular)

difficult in implementation

© 2005 by T. H. Kwon 1
* Application Examples of FEM

(cid:122) Structural Analysis (steady, time-dependent dynamics, eigenvalue)

Beam Plate and Shell

(cid:122) Thermal System Analysis

T1 T2

Conduction

(cid:122) Flow Analysis

Flow + Convection
Heat transfer

(cid:122) Thermomechanical Process Analysis

Forging Rolling Injection Molding
© 2005 by T. H. Kwon 2
* Notation Convention

[A] : m x n matrix
m x n
⎣A⎦ : 1 x n row matrix
1 x n
{ }
A : n x 1 column matrix
n x 1
[ ] [ ] [ ]
C = A B : matrix multiplication
m x n m x l l x n
l
C = ∑A B = A B (Einstein summation convention)
ij ik kj ik kj
k=1
[ ]{ } { }
A x = f : matrix equation
{x}=[A]−1{f} : inverse matrix
[A][A]−1 =[A]−1[A]=[I] : identity matrix
A T = A : transposed matrix
ji ij
[ ]{ } { }
A x =λ x : eigenvalue, eigenvector
[[ ] [ ]]{ }
A −λ I x =0
[[ ] [ ]]
det A −λ I =0 : characteristic equation
−λ3 +Iλ2 −IIλ+III = 0
where I = A
ii
1( ) [ ]
II = A A − A A : invariants of A
2 ii jj ij ji
[ ]
III =det A
© 2005 by T. H. Kwon 3
⎧0 : i ≠ j
δ = ⎨ : Kronecker delta
ij ⎩1 : i = j
⎧1 : even permutation
⎪
ε = ⎨-1 : odd permutation : permutation symbol
ijk
⎪
0 : any other indices are repeated
⎩
[ ]
det A = A =ε A A A =ε A A A
ijk i1 j2 k3 pqr 1p 2q 3r
{ }
⎣A⎦ B = scalar
1 x n n x 1
{ } [ ]
A ⎣B⎦ = C = matrix
m x 1 1 x n m x n

© 2005 by T. H. Kwon 4
* General Procedure of FEM
1. Identify the system (governing) equation. (usually DE)
L(φ) =0. (1.1)
2. Introduce an integral form equation. (Weak form equation) FEM Formulation
(cid:122) Direct Approach
(cid:122) Variational Approach : weak form
(cid:122) Method of Weighted Residual Approach : weak form
∫ψL(φ)dΩ =0 ⇒ ∫L′(ψ)L′′(φ)dΩ =0 (1.2)
Ω Ω
3. Discretize the domain of interest into elements. Element Types
y

x

4. Introduce an approximation of the field variable over an element. Interpolation
φ
y 3
φ(x)

φ
x 1
φ
2
φ(x) = N (x)φ + N (x)φ + N (x)φ (1.3)
1 1 2 2 3 3
φ : Nodal values of the field variable
i
N : Interpolation functions, Shape functions
i
© 2005 by T. H. Kwon 5
5. Evaluate the integral form over each element. Numerical Integral
[K]e{φ}e ={f}e (1.4)
6. Assemble the global matrix equation. Assembly Procedure
[K]{φ}={F} (1.5)
7. Solve the matrix equation to get the unknowns Solution Techniques
{φ}=[K]−1{F} (1.6)
8. Calculate the values of interest from the approximate solution.
∂φ ∂φ
e.g. , , etc.
∂x ∂x

© 2005 by T. H. Kwon 6
2. FINITE ELEMENT FORMULATIONS

2.1 Direct approach for discrete systems

Direct approach has the following features:
(cid:122) It applies physical concept (e.g. force equilibrium, energy conservation, mass
conservation, etc.) directly to discretized elements.
It is easy in its physical interpretation.
(cid:122) It does not need elaborate sophisticated mathematical manipulation or concept.
(cid:122) Its applicability is limited to certain problems for which equilibrium or
conservation law can be easily stated in terms of physical quantities one wants
to obtain. In most cases, discretized elements are self-obvious in the physical
sense.

There are several examples of direct approaches as illustrated in the following.
Using the first example, important features of FEM will be discussed in detail.

Example 1: Force Balance (Linear Spring System)
[Bathe P.79, Ex.3.1]

The problem of a linear spring system is depicted in the following figure.

F F F F F F
1 2 3 4 5 6

K1 K2 K3 K4 K5
x

δ δ δ δ δ δ
1 2 3 4 5 6

In this particular problem, let us assume that δ , F , , F are specified. Solve
L
1 2 6
the displacements at the nodes, and the reaction force at the node number 1.
One can typically follow several steps as an FEM procedure described below:

Node 1 Node 2
One element: f f
1 2

δ δ
1 2
© 2005 by T. H. Kwon 7
Equilibrium:
f = k(δ −δ )
1 1 2
f = k(δ −δ ),
2 2 1
(Note that f = −f for force equilibrium.)
1 2
Element Matrix Equation for one element:
⎡ k −k⎤⎧δ ⎫ ⎧f ⎫
⎢ ⎥⎨ 1⎬= ⎨ 1⎬ ⇔ [K]e{φ}e ={f}e
⎣−k k ⎦⎩δ ⎭ ⎩f ⎭
2 2
Global matrix equation by Assembly:
[K]{δ}={F}
where [K] is called ‘Stiffness Matrix’ and
{F} is called ‘Resultant Nodal Force Matrix’.
The physical meaning of assembly procedure can be found in the force balance as
described below:

f2(1) f1(2)

K K
f1(1) 1 2 f2(2)

(−k δ +k δ )+(k δ −k δ )= f (1) + f (2) = F
1 1 1 2 2 2 2 3 2 1 2
1442443 1442443
f (1) f(2)
2 1

from element 1 from element 2

External nodal force applied at node 2

The typical assembly can be done as shown below:

© 2005 by T. H. Kwon 8
⎡ k −k ⎤⎧δ ⎫ ⎧F ⎫
1 1 1 1
⎢ ⎥⎪ ⎪ ⎪ ⎪
−k k +k −k 0 δ F
⎢ 1 1 2 2 ⎥⎪ 2⎪ ⎪ 2⎪
⎢ −k k +k −k ⎥⎪⎪δ ⎪⎪ ⎪⎪F ⎪⎪
⎢ 2 2 3 3 ⎥⎨ 3⎬= ⎨ 3⎬ (2.1)
−k k +k −k δ F
⎢ 3 3 4 4 ⎥⎪ 4⎪ ⎪ 4⎪
⎢ 0 −k k +k −k ⎥⎪δ ⎪ ⎪F ⎪
⎢ 4 4 5 5 ⎥⎪ 5⎪ ⎪ 5⎪
⎢⎣ −k k ⎥⎦⎪⎩δ ⎪⎭ ⎪⎩F ⎪⎭
5 5 6 6

banded and symmetric matrix

Now, let us pay attention to the boundary condition in this particular case.

Displacements: δ = 0 : specified δ ,δ ,δ ,δ ,δ : unknown
1 2 3 4 5 6

Geometric condition, Essential boundary condition

Forces: F ,F ,F ,F ,F : specified F : unkown reaction force
2 3 4 5 6 1

Force condition, Natural boundary condition

One should recognize that for each node only one of the displacement and force can
be specified as a boundary condition. Nature does not allow to specify both the
displacement and force simultaneously at any node. If none of the two is known, then the
problem is not well posed, in other words, one does not have a problem to solve.
It should also be noted that if there is no geometry constraint at all, there is no
unique solution. One can get a solution only up to a constant. (In other words, the
linear spring system can be moved in x-axis without further deformation.) In this case,
the stiffness matrix becomes a singular matrix. You will easily understand that
equation (2.1) is singular since the summation of six rows becomes null. Think about
the physical meaning of the fact that the summation of six rows becomes null. It just
indicates that the force applied on the system is in balance! In this regard, it is obvious
that at least one geometry constraint should be assigned in order to get a unique
solution. Later, we will discuss the methods of introducing boundary conditions into
equation (2.1).
© 2005 by T. H. Kwon 9