Introduction to Finite Element Methodstrana.snu.ac.kr/lecture/fem_2018/Note18_FEM.pdf · 1. Introduction 2. Approximation of Functions & Variational Calculus 3. Differential Equations

Dept. of Civil and Environmental Eng., SNU

Structural Analysis Lab.

Prof. Hae Sung Lee, http://strana.snu.ac.kr

Introduction to

Finite Element Method

Fall Semester, 2018

Hae Sung Lee

Dept. of Civil and Environmental Engineering

Seoul National University




Contents

1. Introduction

2. Approximation of Functions & Variational Calculus

3. Differential Equations in One Dimension

4. Multidimensional Problems-Elasticity

5. Discretization

6. Two Dimensional Elasticity Problems

7. Various Types of Elements

8. Numerical Integration

9. Convergence Criteria in Isoparameteric Element

10. Miscellaneous Topics

11. Problems with Higher Continuity Requirement - Beams

12. Mixed Formulation




Chapter 2

Approximation of Functions

and Variational Calculus




Fundamental Considerations

What is the best solution for a given problem ?

Needless to say, it is the exact solution…

What is the exact solution?

The solution that satisfies the governing equations as well as boundary conditions if any.

How many do the exact solutions exist?

Of course, one… In case several or infinite numbers of the exact solutions exist for a giv-

en equation, we call the problem as an ill-posed problem, and have great difficulties in

determining a solution of the given problem…

What if it is impossible to determine the exact solution for various reasons?

We need approximate solutions.

What is an approximate solution?

Fundamental Questions

- What is the best approximation?

- How can we represent the best approximation?

A Good (or Robust or Well Formulated) Approximation Should

- Yield the best approximation to the exact solution for a given degree of approxima-

tion.

- Converge to the exact solution as higher degree of approximation is employed.

What is the Definition of the Best Approximation?

- May be defined as the closest solution to the exact solution.

- But, how close is “the closest”?

- “Close or Far” implies the distance between two spatial points.

- We should define some sort of ruler to measure the distance between two elements in

a function set…




Normes of Functions: A measure of a function set

A function set is said to be a normed space if to every f there associated a

nonnegative real number f , called the norm of f, in a such way that

- 0f if and only if 0f

- ff || for any real number .

- gfgf

Every normed space may be regarded as a metric space, in which the distance between

any two elements in the space is measured by the defined norm. Various types of norm

can be defined for a function space. Among them the following norms are important.

- L1 norm: V

LdVff

1

- L2 norm: 2/12 )(2

V

LdVff

- H1 norm: 2/12 ))((1 V

HdVffff

Discretization

- Representation of a continuously distributed quantities with some numbers.

)()(1

XX

n

i

ii gaf )(Xf where gi are the basis functions of a function set,

- Set : Collection of some objectives with the same characteristics

- The basis functions should be linearly independent to each other.

0)(1

Xn

i

ii ga if and only if all 0ia

- Taylor series, Fourier series, etc…

-

Approximations

hm

i

ii

hn

i

ii gafgaf )()()()(11

XXXX where nm

Summation Notation: Repeated indices denote summation ii

m

i

ii baba 1

.




General Ideas for the Best Approximation

Let’s find out a approximate function that is closest to the given function by use of a

norm defined in the function space. If this is the case, the characteristics of an approxima-

tion method depend on those of the norm used in the approximation.

Least Square Error(LSE) Minimization

Error: hffe

Minimize V

h

L

h

LdVffffe 2

22)(

2

1

2

1

2

1

22

mkFaKfdVgadVgg

fdVgdVagg

dVgffdVa

fff

a

ki

m

i

ki

V

ki

m

i V

ik

V

k

V

m

i

iik

V

k

h

V k

hh

k

,,1for 0

)()(

11

1

Final System Equation : FKa

If the basis functions are orthogonal, K becomes diagonal.




Variation of a function

- The variation of a function means a possible change in the function for the fixed x.

Variational Calculus

- if ii gaf then

the variation of f is defined as ii gaf or by definition i

i

aa

ff

.

- ff

Fa

a

f

f

Fa

a

FFfF i

i

i

i

: )(

- hfhf )(

- hffhfh )(

- , )()(dx

fda

a

f

dx

da

dx

df

adx

dfi

i

i

i

-

fdxdxa

a

fafdx

afdx i

i

i

i

f

f

Variation of a function




Minimization by Variational Calculus

Min

l

hh dxfff0

2)(2

1)(

k

k

k

l

k

hh

l

hh

l

h

l

hh

aa

adxa

fff

dxfffdxffdxfff

0

00

2

0

2

)(

)()(2

1))(

2

1()(

Min k

V

hh adVfff possible allfor 0)(2

1)( 2

Euler Equation

Min ka

dxxffFfk

l

allfor 0),,()(0

l

kk

l

k

l

kk

l

kk

l

k

l

kk

dxgf

F

dx

dg

f

Fg

f

F

dxgf

Fg

f

Fdx

a

f

f

F

a

f

f

Fdx

a

FdxxffF

aa

00

0000

)(

)()(),,(

In case the basis functions vanish at the boundary, then

0 allfor 0)(0

f

F

dx

d

f

Fkdxg

f

F

dx

d

f

F

a

l

k

k

llll

dxff

F

dx

df

f

Ff

f

Fdxf

f

Ff

f

FdxxffFf

0000

)()(),,()(

In case the variation vanishes at the boundaries, then

k

k

k

l

k

l

aa

adxgf

F

dx

d

f

Ffdx

f

F

dx

d

f

Ff

00

)()()( .

Therefore,

Min 0




Example 1

Min 2

1

2)(1)(

x

x

dxyy subject to 11)( yxy , 22 )( yxy

0))(1())(1(0 2/122

yydx

dy

ydx

d

f

F

dx

d

f

F

0))(1())(1

)(1())(1(

))(1)(2

1())(1())(1(

2/32

2

22/12

2/322/122/12

yyy

yyy

yyyyyyyydx

d

baxyy 0 . By applying BC, 12

2112

12

120xx

yxyxx

xx

yyyy

Example 2

Min

l

dxufuu0

2 ))(2

1()( subject to 0)0( u , 0)( lu

00))(2

1( 2

fuufu

dx

dfu

udx

df

u

F

dx

d

u

F

Homework 1

1. Approximate a cosine function xl

y

2

cos by polynomials based the minimization of

the least square errors. Use polynomials up to the 20th order. You may use a numerical inte-

gration algorithm such as Simpson’s rule, the trapezoidal rule or the rectangular rule. You

also need a numerical solver to solve linear simultaneous equations in the “Linpack”. For

the accuracy of your calculation, please use “double precision” in your program. You

should present proper discussions on your results together with some graphs that show your

approximate functions and the given function.

2. Derive Euler equation for the following minimization function, and proper boundary condi-

tions.

Min

l

dxxfffFf0

),,,()(

3. Derive the governing equation and the boundary conditions for the following problem.

Min

l

dxwqdx

wdw

0

2

2

2

))(2

1()(




Chapter 3

Elliptic Differential Equations

in One Dimension




3.1. Problems with homogenous displacement boundary conditions

Problem Definition

0)()0( ,0 02

2

luulxfdx

ud

Approximation – Discretization

m

i

ii

h gau1

where 0)()0( luu hh

Residuals

Verbal Definition : Something left over, or resulting from subtraction...

Equation Residual : lxfdx

udR

h

E 0 02

2

Function Residual : lxuuR h

F 0 0

Error Estimator :

l hh

l

EF

R dxfdx

uduudxRR

0

2

2

0

))((2

1

2

1

Least Square Error

Error SquareLeast ))((2

1

))((2

1))((

2

1

))((2

1

))((2

1

0

00

0

2

2

2

2

0

2

2

LS

l hh

l hhl

hh

l hh

l hhR

dxdx

du

dx

du

dx

du

dx

du

dxdx

du

dx

du

dx

du

dx

du

dx

du

dx

duuu

dxdx

ud

dx

uduu

dxfdx

uduu

Energy Functional – Total Potential Energy

RR

l

h

l hhl

l hhl

hh

l

h

l

h

lhl

h

l

hh

hhl h

hR

Cfdxudxdx

du

dx

duufdx

dxdx

du

dx

du

dx

duudxu

dx

udu

dx

du

dx

duudxfuuf

dxfudx

uduuf

dx

ududxf

dx

uduu

)2

1(

2

1

})({2

1

)(2

1))((

2

1

000

000

2

2

000

0

2

2

2

2

0

2

2

f




Minimization Problems

RRLSR Min Min Min w.r.t. hhu

RRMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy

- 1st Order Necessary Condition for Minimization Problem

FKa

mkFaK

fdxgadxdx

dg

dx

dg

fdxgada

ddx

dx

dga

dx

dga

da

d

fdxda

dudx

dx

du

dx

du

da

d

a

m

i

kiki

l

k

m

i

i

l

ik

l m

i

ii

k

l m

i

ii

m

i

ii

k

l

k

hl hh

kk

RR

,1for 0

)())((

)(

1

01 0

0 10 11

00

0 RR : Variational principle or Principle of Virtual Work

0)()( 0)(

)(

)2

1(

1 1

1 01 0

0000

00

FKaaT

m

k

m

i

kikik

m

k

l

k

m

i

i

l

ikk

l

h

l hhl

h

l hh

l

h

l hhRR

FaKa

fdxgadxdx

dg

dx

dga

fdxudxdx

du

dx

udfdxudx

dx

du

dx

du

fdxudxdx

du

dx

du

Solution Space

- })( , 0)()0(|{0

2 l

dxdx

duluuuu

- h : The minimization problems yield the exact solution.

- h : The minimization problems yield an approximate solution.

Properties of K

- Symmetry : ji

l

ijl

ji

ij Kdxdx

dg

dx

dgdx

dx

dg

dx

dgK

00

- Positive Definiteness :

m

i

T

j

m

j

ijij

m

j

jl

im

i

i

j

m

j

jl

i

m

i

i

l h

aKaadxdx

dg

dx

dga

dxadx

dga

dx

dgdx

dx

du

1 11 01

10 1

2

0

0

)(

Kaa




Absolute Minimum Property of Total Potential Energy

eh uuu

)for only holdssign equality (The )(2

1

2

1

2

1

)(2

1

2

1

2

1

2

1

2

1

2

1

)()()(

2

1

2

1

0

2

00 0

0

2

2

0 00

00

2

2

00 00

000 00

0000

Cudxdx

du

dxdx

du

dx

duufdxdx

dx

du

dx

du

dxfdx

ududx

dx

du

dx

duufdxdx

dx

du

dx

du

fdxudxdx

udu

dx

duudx

dx

du

dx

duufdxdx

dx

du

dx

du

fdxudxdx

du

dx

dudx

dx

du

dx

duufdxdx

dx

du

dx

du

fdxuudxdx

uud

dx

uudfdxudx

dx

du

dx

du

eE

l eE

l eel l

l

e

l l eel

l

e

l

e

l

e

l l eel

l

e

l el l eel

l

e

l eel

h

l hhh

Weighted Residual Method

mkdxfdx

uddxR

l h

k

l

Ekk ,,1for 0)(0

2

2

0

- if kk g : Galerkin Method (elliptic System or self-adjoint system)

FKa

,,1for 0

)(

01 0

00 100

0000

2

2

mkfdxgadxdx

dg

dx

dg

fdxgdxadx

dg

dx

dgfdxgdx

dx

du

dx

dg

fdxgdxdx

du

dx

dg

dx

dugdxf

dx

ud

l

ki

m

i

l

ik

l

k

l m

i

i

ik

l

k

l h

k

l

k

l h

k

lh

k

l h

kk

- Identical result to the Rayleigh-Ritz Method or Variational Principle

- if kk g : Petrov-Galerkin Method (hyperbolic system or non-self adjoint system)

Weighted Residual Method vs. Variational Principle

ii

m

i

ik aamk

0 ,,1for 01

h

l l

hhhl h

h

l h

i

m

i

i

m

i

l h

iii

m

i

i

ufdxudxdx

du

dx

uddxf

dx

udu

dxfdx

udgadxf

dx

udgaa

possible allfor 0)()(

)()(

0 00

2

2

0

2

2

11 0

2

2

1




Example 1 : 12

2

dx

ud, 0)1()0( uu

- Exact solution : xxu2

1

2

1 2

- First trial : 2

321 xaxaau h

Applying BCs : 01 a , 32 aa )( 2

3 xxau h , )21(3 xadx

du h

3

1)21(

1

0

2

11 dxxK , 6

1)(1

1

0

2

1 dxxxF

2

1

6

1

3

133 aa . Therefore, uu h

- Second Trial : 3

4

2

321 xaxaxaauh

Applying BCs : 01 a , 432 aaa

21

)()( 3

4

2

3

gg

h xxaxxau

)31(),21( 221 xdx

dgx

dx

dg

3

1)21(

1

0

2

11 dxxK , 5

4)31(

1

0

22

22 dxxK

2

1)31)(21(

1

0

2

2112 dxxxKK

6

1)(1

1

0

2

1 dxxxF , 4

1)(1

1

0

3

2 dxxxF

System Equation : 0,2

1

4

16

1

5

4

2

12

1

3

1

43

4

3

aa

a

a

Example 2 : xdx

ud sin2

2

2

, 0)1()0( uu

- Exact Solution : xu sin

- First trial : 2

321 xaxaau h

Applying BCs : )( 2

2 xxauh , )21(2 xadx

du h

3

1)21(

1

0

2

11 dxxK ,

4

)(sin

1

0

22

1 dxxxxF

124

3

122 aa . Therefore, )(

12 2xxu h

955.03

)5.0(

hu Error = 4.5 %




- Second trial : 3

3

2

210 xaxaxaau h

Applying BCs : 00 a , 321 aaa

21

)()( 3

3

2

2

gg

h xxaxxau

)31(),21( 221 xdx

dgx

dx

dg

3

1)21(

1

0

2

11 dxxK , 5

4)31(

1

0

22

22 dxxK

2

1)31)(21(

1

0

2

2112 dxxxKK

4)(sin

1

0

22

1 dxxxxF ,

6

)(sin

1

0

32

2 dxxxxF

System Equation : 0,12

6

4

5

4

2

12

1

3

1

32

3

2

aa

a

a

????

- In general )(2

m

i

i

i

h xxau

Function Error=

2/1

1

0

2

1

0

2

sin

)(sin

xdx

dxux h

Derivative Error=

2/1

1

0

22

1

0

2

cos

))(cos(

xdx

dxux h

To evaluate numerator in the error expressions, the midpoint rule with 100 subinter-

vals is employed.

- Raw Output

***** 2th-order Polynomial ***** a 2 = -0.3819719E+01 Errors for 2th-order polynomial

Function error = 0.2009211E+01 % Derivative error = 0.6010036E+01 %




***** 3th-order Polynomial ***** a 2 = -0.3819719E+01 a 3 = 0.0000000E+00 Errors for 3th-order polynomial Function error = 0.2009211E+01 % Derivative error = 0.6010036E+01 % ***** 4th-order Polynomial ***** a 2 = 0.4193832E+00 a 3 = -0.7065170E+01 a 4 = 0.3532585E+01 Errors for 4th-order polynomial

Function error = 0.4048880E-01 % Derivative error = 0.1956108E+00 % ***** 5th-order Polynomial ***** a 2 = 0.4193832E+00 a 3 = -0.7065170E+01 a 4 = 0.3532585E+01 a 5 = 0.7833734E-12 Errors for 5th-order polynomial Function error = 0.4048880E-01 % Derivative error = 0.1956108E+00 %

***** 6th-order Polynomial ***** a 2 = -0.1405727E-01 a 3 = -0.5042448E+01 a 4 = -0.5128593E+00 a 5 = 0.3640900E+01 a 6 = -0.1213633E+01 Errors for 6th-order polynomial Function error = 0.4444114E-03 % Derivative error = 0.2944068E-02 % ***** 7th-order Polynomial ***** a 2 = -0.1405727E-01

a 3 = -0.5042448E+01 a 4 = -0.5128593E+00 a 5 = 0.3640900E+01 a 6 = -0.1213633E+01 a 7 = 0.5029577E-09 Errors for 7th-order polynomial Function error = 0.4444114E-03 % Derivative error = 0.2944068E-02 % ***** 8th-order Polynomial ***** a 2 = 0.2231430E-03 a 3 = -0.5170971E+01 a 4 = 0.2265606E-01 a 5 = 0.2462766E+01

a 6 = 0.2001274E+00 a 7 = -0.8751851E+00 a 8 = 0.2187963E+00




Errors for 8th-order polynomial Function error = 0.3045376E-05 % Derivative error = 0.2548945E-04 % ***** 9th-order Polynomial ***** a 2 = 0.2231387E-03 a 3 = -0.5170971E+01 a 4 = 0.2265578E-01 a 5 = 0.2462767E+01 a 6 = 0.2001259E+00 a 7 = -0.8751837E+00 a 8 = 0.2187955E+00 a 9 = 0.1698185E-06 Errors for 9th-order polynomial

Function error = 0.3045376E-05 % Derivative error = 0.2548945E-04 % ***** 10th-order Polynomial ***** a 2 = -0.2313690E-05 a 3 = -0.5167665E+01 a 4 = -0.4905381E-03 a 5 = 0.2553037E+01 a 6 = -0.1050486E-01 a 7 = -0.5742830E+00 a 8 = -0.3911909E-01 a 9 = 0.1217931E+00 a10 = -0.2435856E-01 Errors for 10th-order polynomial

Function error = 0.1289526E-07 % Derivative error = 0.1450501E-06 % ***** 11th-order Polynomial ***** a 2 = -0.4281311E-05 a 3 = -0.5167629E+01 a 4 = -0.7974500E-03 a 5 = 0.2554541E+01 a 6 = -0.1501611E-01 a 7 = -0.5656904E+00 a 8 = -0.4955267E-01 a 9 = 0.1296181E+00 a10 = -0.2766239E-01 a11 = 0.6006860E-03

Errors for 11th-order polynomial Function error = 0.2263193E-07 % Derivative error = 0.2253144E-06 %




- Result plots

Variation of errors

Plot of approximate function for different orders of polynomial

10-8

10-6

10-4

10-2

100

2 3 4 5 6 7 8 9 10 11

Function

Derivative

Err

or

(%)

Order of polynomial

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Exact

2nd order

4th order

F(X

)

X-Coordinate




Plot of derivative approximate function for different orders of polynomial

Homework 2

1. Derive RR for the following ODE for a beam subject to axial force P (positive for com-

pression) as well as a traverse load q. Assume homogeneous displacement BCs.

02

2

4

4

qdx

wdP

dx

wdEI

2. On what condition does the principle of minimum potential energy hold for Prob. 1 ?

Discuss the physical and the mathematical meanings of the condition.

3. Approximate solutions for a fixed-fixed end beam with a uniform load by polynomials

with one unknown and two unknowns. No axial force is applied. Compare your solutions

including displacement, rotation, moment and shear force to exact solution and discuss.

-4

-3

-2

-1

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Exact

2nd order

4th order

F'(

X)

X-Coordinate




3.2. Problems with Traction Boundary Conditions

Problem Definition

- Differential Equation

lxfdx

ud 0 0

2

2

- Boundary Conditions

at or 0 0 Tdx

duux and at or 0 T

dx

duulx

Error Minimization: Error Estimator

))((2

1

))((2

1))((

2

1))((

2

1

))((2

1))((

2

1

0

000

00

2

2

LS

l hh

lh

h

l hhl

hh

lh

h

l hhR

dxdx

du

dx

du

dx

du

dx

du

dx

du

dx

duuudx

dx

du

dx

du

dx

du

dx

du

dx

du

dx

duuu

dx

du

dx

duuudxf

dx

uduu


RR

lh

l

h

l hhl

l

lh

hhh

l

o

hhl

hh

l

h

l

h

lhl

h

lh

h

l

hh

hh

lh

h

l hhR

C

Tufdxudxdx

du

dx

duTuufdx

dx

duu

dx

duu

dx

duu

dx

duu

dxdx

du

dx

du

dx

duudxu

dx

udu

dx

du

dx

duudxfuuf

dx

du

dx

duuudxfu

dx

uduuf

dx

udu

dx

du

dx

duuudxf

dx

uduu

000

00

0

00

2

2

000

00

2

2

2

2

00

2

2

2

1)(

2

1

)(2

1

}{2

1)(

2

1

))((2

1)(

2

1

))((2

1))((

2

1


RRLSR Min Min Min w.r.t. hhu





- 1st Order Necessary Condition of Minimization Problem

miFaKTgfdxgadxdx

dg

dx

dg

Tgada

dfdxga

da

ddx

dx

dga

dx

dga

da

d

Tda

dufdx

da

dudx

dx

du

dx

du

da

d

a

m

i

kiki

l

k

l

k

m

i

i

l

ik

lm

i

ii

k

l m

i

ii

k

l m

i

i

i

m

i

i

i

k

l

k

hl

k

hl hh

kk

RR

,1for 0

)())((

)(

10

01 0

010 10 11

000

FKa

0 RR : Variational Principle or Principle of Virtual Work

0)()( 0)(

)(

)2

1(

1 1

1 00

1 0

00

00

00

FKaaT

m

k

m

i

kikik

m

k

ll

kk

m

i

i

l

ik

k

ll

hh

l hhl

h

l

h

l hhRR

FaKa

Tgfdxgadxdx

dg

dx

dga

Tufdxudxdx

du

dx

udTufdxudx

dx

du

dx

du

Absolute Minimum Property of Total Potential Energy by the Exact Solution

eh uuu

)for only holdssign equality (The )(2

1

2

1

2

1

)()(2

1

2

1

2

1

2

1

2

1

2

1

)()()()(

2

1

2

1

0

2

00

0 0

00

2

2

0 00

0

000

2

2

00 00

0

0000 0

00

000

000

Cudxdx

du

dxdx

du

dx

duTuufdxdx

dx

du

dx

du

dx

duTudxf

dx

ududx

dx

du

dx

duTuufdxdx

dx

du

dx

du

Tufdxudxdx

udu

dx

duudx

dx

du

dx

duTuufdxdx

dx

du

dx

du

Tufdxudxdx

du

dx

dudx

dx

du

dx

duTuufdxdx

dx

du

dx

du

Tuufdxuudxdx

uud

dx

uud

Tufdxudxdx

du

dx

du

eE

l eE

l eel

l l

l

e

l

e

l l eel

l

le

l

e

l

e

l

e

l l eel

l

le

l

e

l el l eel

l

le

l

e

l ee

lh

l

h

l hhh




Weighted Residual Method

FKa

,,1for 0

)(

)()()(

00

1 0

000

0000

00

2

2

00

mkTgfdxgadxdx

dg

dx

dg

Tgfdxgdxdx

du

dx

dg

dx

du

dx

dugfdxgdx

dx

du

dx

dg

dx

dug

dx

du

dx

dugdxf

dx

udg

dx

du

dx

dugdxRg

ll

kki

m

i

l

ik

l

k

l

k

l hk

lh

k

l

k

l hk

lh

k

lh

k

l h

k

lh

k

l

Ekk

Weighted Residual vs. Variational Principle

0

)(

)(

possible allfor 0 ,,1for 0

000

10

00

10

00

1

RRl

h

l

h

l hh

m

k

l

kk

l

kk

l hkk

m

k

l

k

l

k

l h

kk

ii

m

i

ik

Tufdxudxdx

du

dx

ud

Tgafdxgadxdx

du

dx

gad

Tgfdxgdxdx

du

dx

dga

aamk




3.3. Integrability Condition - Regularity (continuity) Requirement

- Integration of functions with discontinuity dxxuf

l

l

))(( ??

Original function u Function with transition zone u

dxxufdxxuf

l

l

l

l

))((lim))((0

where

x

lxu

uux

uuxlu

u for

for 22

for

, )( uu , )( uu

- Can we integrate dxu

l

l

on what condition ?

l

l

l

l

dxudxudxudxu

)()22

( uudxuu

xuu

dxu

))((lim)(lim00

uudxudxudxudxudxuudx

l

l

l

l

l

l

The last integral vanishes as far as u and u are finite, and the integral becomes

l

l

l

l

udxudxudx0

0

u-

u+

u-

u+

2




- Can we integrate dxu

l

l

2 on what condition ?

l

l

l

l

dxudxudxudxu 2222

2)(

6)()

22( 2222

uuuudxuu

xuu

dxu

)2

)(6

)((lim

)(lim

2222

0

222

0

2

uuuudxudxu

dxudxudxudxu

l

l

l

l

l

l

The last integral vanishes as far as u and u are finite, and the integral becomes

l

l

l

l

dxudxudxu0

2

0

22

- Can we integrate dxdx

dul

l

2)( ??

dxdx

uduudx

dx

ud

dxdx

uddx

uudx

dx

ud

dxdx

uddx

dx

uddx

dx

uddx

dx

ud

l

l

l

l

l

l

l

l

22

2

222

2222

)(2

)()(

)()2

()(

)()()()(

2

)(lim)()()(lim)(

2

0

2

0

2

0

2

0

2 uudx

dx

dudx

dx

dudx

dx

uddx

dx

dul

l

l

l

l

l

Therefore, the given definite integral has a finite value if and only if u is continuous.

)(lim)(lim00

uu

From the physical point of view, the aforementioned continuity condition represents the

compatibility condition, which states that the displacement field in a continuum should

be uniquely determined, ie, defined by single valued functions.




- Can we integrate dxdx

duu

l

l

on what condition ?

l

l

l

l

dxdx

ududx

dx

ududx

dx

ududx

dx

udu

2

)()()

2)(

22(

uuuudx

uuuux

uudx

dx

udu or

2

)()())()((

2

1 222

uuuuuudxu

dx

uddxu

dx

ududx

dx

udu

)(][

2

)()(

)2

)()((lim

)(lim

0

0

0

0

0

0

uudxdx

duudx

dx

duu

uuuudx

dx

duudx

dx

duu

uuuudx

dx

ududx

dx

udu

dxdx

ududx

dx

ududx

dx

ududx

dx

duu

l

l

l

l

l

l

l

l

l

l

Therefore, the given definite integral has a unique & finite value even if u is discontinu-

ous.

Homework 3

1. Derive R , LS , RR for the following ODE with the homogeneous displacement BCs.

04

4

qdx

wd

2. Prove the absolute minimum property of RR derived in Prob. 1 by the exact solution.

3. Approximate solutions for a cantilever beam subject to a uniform load (q) over the span

and a concentrate load applied at the free end by polynomials with one unknown, two un-

knowns and three unknowns. Compare your solutions including displacement, rotation,

moment and shear force to exact solution and discuss.

4. Identify the integrability condition for the following integrals, and evaluate the integrals for

the identified condition.

dxdx

wdl

2

0

2

2

)( and dxdx

wd

dx

dwl

0

2

2




3.4. The Other Side of the Principle of Virtual Work

Physical Viewpoint

If a deformable body is in equilibrium under a Q-force system and remains in equilibrium

while it is subjected a small virtual deformation, the external virtual work done by exter-

nal Q forces acting on the body is equal to the internal virtual work of deformation done

by the internal Q-stresses.

V

ijij

S

ii dVdSQu where )(2

1

i

j

j

i

ijx

u

x

u

Mathematical Viewpoint - Continuous Problem

If 0)( uA should hold for a given system, then the following statement should hold.

Here, u , and the order of A should be the same as u.

uuAu 0)( dvv

( : A proper Function space)

- Example : Beam problem

The governing equation for a beam problem : qdx

wdEI

4

4

The expression for virtual work :

vvv

wqdxdxdx

wdEI

dx

wddxq

dx

wdEIw

2

2

2

2

4

4

0)(

If w is the displacement induced by the unit load applied load at xj , the expression

for the principle of virtual work becomes as follows.

)()( j

v

j

vv

Q

xwdxxxwwqdxdxEI

MM

where M is the moment induced by the unit load applied at xj and )( jxx is a de-

lac delta function applied at xj.




Mathematical Viewpoint - Discrete Problem

uuuAu 0 )( )( ii Au ( : A proper Function space)

- Example : Truss problem

Equilibrium Equations at joints

0 , 0)(

1

)(

1

iim

j

i

j

iim

j

i

j YVXH for ni ,,1

Virtual Work Expression

0))( )((1

)(

1

)(

1

n

i

iiim

j

i

j

iiim

j

i

j vYVuXH

0))sin(- )cos((1

)(

1

)(

1

n

i

iiim

j

i

j

i

j

iiim

j

i

j

i

j vYFuXF

n

i

iiiin

i

im

j

i

j

i

j

iim

j

i

j

i

j

i vYuXFvFu11

)(

1

)(

1

) ()sin cos(

iY

iX

iF1

i

jF

i

imF )(

iY

iX

ivv sin)( 12

iuu cos)( 12

i i

)( 12 uu

)( 12 vv




n

i

iiiin

i

iiiinmb

ii

iii

i

iinmb

i

inmb

i

ii

nmb

i

i

ii

i

iii

nmb

i

ii

i

iii

i

i

vYuXvYuXEA

lF

EA

lFlF

vvuuF

vvFuuF

11111

1

1212

1

1212

) () ()()(

)sin)( cos)((

))(sin )(cos(

If force system consists of an unit load applied at k-th joint in arbitrary direction,

then

nmb

ii

iii

kkkk

EA

lFvYuX

1 )(coscos uuX

3.5. Finite Element Discretization

Domain Discretization

hlh

e l

h

e l

hh

lhn

e l

hn

e l

hh

lh

l

h

l hh

uTudxufdxdx

du

dx

ud

Tudxufdxdx

du

dx

ud

Tudxufdxdx

du

dx

ud

ee

ee

admissible allfor 00

011

0

00

Interpolation of Displacement Field in an Element

= +

L

eu R

eu 1 1 L

eu

R

eu

el

L

eN R

eN

ex 1ex

1U 1nU 1l 2l nl 3l . . .

node 01 x 2x 3x lxn 1

2U 3U

4U




e

e

ee

eR

e

e

e

ee

eL

e

eR

e

L

eR

e

L

e

R

e

R

e

L

e

L

e

h

e

eR

e

L

eR

e

L

e

R

e

R

e

L

e

L

e

h

e

l

xx

xx

xxN

l

xx

xx

xxN

u

uNNuNuNxu

u

uNNuNuNxu

= , =

),()(

),()(

1

1

1

1

uN

uN

Discretized Form of Variational Statement

eR

e

L

e

R

e

L

eR

e

R

eL

e

L

e

h

e

eR

e

L

e

R

e

L

eR

e

R

eL

e

L

e

h

e

u

u

dx

dN

dx

dNu

dx

dNu

dx

dN

dx

ud

u

u

dx

dN

dx

dNu

dx

dNu

dx

dN

dx

du

uB

uB

),(

),(

0

))0()0()()(()()( 1

0

TuFuuKu

TuNuuBBu

b

e

e

T

ee

e

e

T

e

b

e l

TT

ee

e l

TT

e

LR

n

e l

Th

e

e l

h

eT

h

e

lh

e l

h

e l

hh

ee

ee

ee

fdxdx

TulTlufdxudxdx

du

dx

ud

Tudxufdxdx

du

dx

ud

Compatibility Condition – Continuity Requirement

1

11 ,

eL

e

R

e

eR

e

L

e UuuUuu

UCuUCu

eee

n

e

e

R

e

L

e

e

U

U

U

U

u

u ,

0100

0010

1

1

1

Global System Equation – Global Stiffness Equation

.

UFKUU

TCFCUUCKCU

TCUFCUUCKCU

admissible allfor 0)(

)()(

T

T

b

e

e

T

e

T

e

e

e

T

e

T

T

b

T

e

e

T

e

T

e

e

e

T

e

T

FKU




Rayleigh-Ritz Type Interpretation of FEM

ii

h Ugu

h

l

h

l hh

ufdxudxdx

du

dx

ud

admissible allfor 00

e

e

x

x

TT

e

n

i

x

x

i

i

i

ii

i

i

i

x

x

n

n

n

nn

n

x

x

n

n

n

nn

n

x

x

n

n

n

nn

n

x

x

i

i

i

ii

i

x

x

i

i

i

ii

i

x

x

i

i

i

ii

i

x

x

i

i

i

ii

i

x

x

i

i

i

ii

i

x

x

i

i

i

ii

i

x

x

x

x

x

x

n

i

n

j

l

j

ji

i

l hh

i

i

i

i

n

n

n

n

n

n

i

i

i

i

i

i

i

i

i

i

i

i

dx

dxUdx

dgU

dx

dg

dx

dgU

dx

dgU

dxUdx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgUdxU

dx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgUdxU

dx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgUdxU

dx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgUdxU

dx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgUdxU

dx

dgU

dx

dg

dx

dgU

dxUdx

dgU

dx

dg

dx

dgU

dxUdx

dg

dx

dgUdx

dx

du

dx

ud

UBBU

1

1

1

1

1

2

1

1

1

1

1

1

2

3

2

2

1

2

1

1

1

11

1

1

11

1

1

1

1

1

2

2

1

11

11

11

1

1

1

1

1

1

11

11

1

2

21

1

3

3

222

222

112

2

22

111

1

1

1

1

1 00

))((

)(

)( )(

)( )(

)( )(

)()(

)( )(

)(

gi gi+1

iU 1iU

gi-1

1iU

g1 gn+1

1U 1nU




Finite Element Procedure

1. Governing equations in the domain, boundary conditions on the boundary.

2. Derive weak form of the G.E. and B.C. by the variational principle or equivalent.

3. Descretize the given domain and boundary with finite elements.

S , 2121 mn SSSVVVV

4. Assume the displacement field by shape functions and nodal values within an element.

eeeUNu

5. Calculate the element stiffness matrix and assemble it according to the compatibility.

dVel

e

DBBKT

, e

eKK

6. Calculate the equivalent nodal force and assemble it according to the compatibility.

lT

l

Te dVfe

0TNNF ,

e

eFF

7. Apply the displacement boundary conditions and solve the stiffness equation.

8. Calculate strain, stress and reaction force.

Example with Three Elements and Four Nodes

- Shape Function Matrix

e

e

ee

eR

e

e

e

ee

eL

e

eeR

e

L

eR

e

L

e

R

e

R

e

L

e

L

e

h

e

l

xx

xx

xxN

l

xx

xx

xxN

u

uNNuNuNxu

= , =

),()(

1

1

1

1

uN

]1 , 1[1

),( e

R

e

L

ee

ldx

dN

dx

dNB

- Element Stiffness Matrix

11

111]1 , 1[

1

1

11][

eeV e

el

dxll

e

K

1U 1l 2l 3l

01 x 2x 3x lx 4

2U 3U 4U




- Compatibility Matrix

0010

00011

4

3

2

1

1

1

1 UCu

U

U

U

U

u

uR

L

, 0100

00102

4

3

2

1

2

2

2 UCu

U

U

U

U

u

uR

L

1000

01003

4

3

2

1

3

3

3 UCu

U

U

U

U

u

uR

L

- Global Stiffness Matrix

33

3322

2211

11

321

3

21

333222111

1100

11110

01111

0011

1100

1100

0000

0000

1

0000

0110

0110

0000

1

0000

0000

0011

0011

1

1000

0100

11

11

10

01

00

00

1

0100

0010

11

11

00

10

01

00

1

0010

0001

11

11

00

00

10

01

1

ll

llll

llll

ll

lll

l

ll

TTT

e

ee

T

e CKCCKCCKCCKCK




- Force Term

1

1

21

1 1

1 e

x

x e

e

e

e

ldx

xx

xx

l

e

e

F

2

22

22

2

1

1

10

01

00

00

21

1

00

10

01

00

2

1

1

00

00

10

01

2

3

32

21

1

321

332211

l

ll

ll

l

lll

TTT

e

e

T

e FCFCFCFCF

- System Equation

2

22

22

2

1100

11110

01111

0011

3

32

21

1

4

3

2

1

33

3322

2211

11

l

ll

ll

l

U

U

U

U

ll

llll

llll

ll

- Case 1 : 0)1(

, 0)0( 3

1321

dx

duulll

5.0

1

1

9

1

110

121

012

2

22

22

2

1100

11110

01111

0011

4

3

2

3

32

21

1

4

3

2

1

33

3322

2211

11

U

U

U

l

ll

ll

l

U

U

U

U

ll

llll

llll

ll

18

9 ,

18

8 ,

18

5432 UUU

Displacement Strain (or Stress)




- Case 2 : 0)1( , 0)0( 3

1321 uulll

1

1

9

1

21

12

2

22

22

2

1100

11110

01111

0011

3

2

3

32

21

1

4

3

2

1

33

3322

2211

11

U

U

l

ll

ll

l

U

U

U

U

ll

llll

llll

ll

9

1 ,

9

132 UU

3.6. Finite Difference Discretization

Differential Equation

0 0 2

2

2

i

i fuDfdx

ud where D2 is a 2nd-order finite difference operator.

Displacement Strain (or Stress)

1ix ix 1ix

iU

1iU

1iU




Finite Difference Operator (Central Difference)

- Suppose u is approximated by a 2nd-order parabola ,ie, cbxaxu 2

)1

)11

(1

(1

1

1

1

112

1

1

2

11

i

i

i

ii

i

iii

iii

i

iii

Ul

Ull

Ulll

a

cblalU

cU

cblalU

)1

)11

(1

(2

2 1

1

1

11

2

2

2

i

i

i

ii

i

iii

i

xx

Ul

Ull

Ulll

auDdx

ud

i

Finite Difference Equations for interior nodes

i

ii

i

i

i

ii

i

i

fll

Ul

Ull

Ul 2

1)

11(

1 1

1

1

1

1

Finite Difference Equations for Boundary Nodes with Displacement BCs

In case that a displacement BC is specified at a boundary nodes, the finite difference

equations need to be set up for only interior nodes. The BC can be applied to the finite

difference equation for the node adjacent to the boundary nodes.

- Example – Case 2

3

32

4

3

3

32

2

2

221

3

2

2

21

1

1

2

1)

11(

1

2

1)

11(

1

fll

Ul

Ull

Ul

fll

Ul

Ull

Ul

Finite Difference Equations for Boundary Nodes with Traction BCs

In case that a traction BC is specified at a boundary nodes, a special treatment for bounda-

ry condition such as ghost node is introduced.

il 1 il

iU

1iU

1iU

0

1nx nx 1nx 2nx




- The finite difference equation at the node n+1 :

1

1

2

1

1

1 2

1)

11(

1

n

nn

n

n

n

nn

n

n

fll

Ul

Ull

Ul

- Approximation of the traction BC by the finite difference operator.

nn

nn

nn

lx

UUll

UU

dx

du

2

1

2 0

- Substitution of the FD traction BC into FD equations for the boundary node.

1

1

1

11 2)

11()

11(

n

nn

n

nn

n

nn

fll

Ull

Ull

Since the location of the ghost node is arbitrary, nn ll 1 can be assumed without

loss of generality. The final equation for the boundary node becomes

112

11 n

n

n

n

n

n

fl

Ul

Ul

- Example – Case 1

3

32

4

3

3

32

2

2

221

3

2

2

21

1

1

2

1)

11(

1

2

1)

11(

1

fll

Ul

Ull

Ul

fll

Ul

Ull

Ul

4

3

4

3

3

3 2

11f

lU

lU

l

Homework 4

1. Using the continuity requirements for beam problems (4th-order ODE) considered in

homework 2, propose suitable interpolation (shape) functions for a beam element, and de-

rive the element stiffness matrix of the beam element.

2. Derive a FDM equation for a cantilever beam subject to a concentrated load at the free

end. Propose FDM equations for all boundary conditions for beam problems using

proper ghost nodes. Discretize the beam with 5 nodes including two boundary nodes.

Do not solve the system equations.




Chapter 4

Multidimensional Problems

– Elasticity Problems–

b

a




4.1. Problem Definition

Governing Equations and Boundary Conditions

Equilibrium Equation : 0 b in V

Constitutive Law : :D in V

Strain-Displacement Relationship : ))((2

1 Tuu

in V

Displacement Boundary condition : 0 uu on uS

Traction Boundary Condition : 0TT on tS

Cauchy’s Relation on the Boundary : nT on S

Strain Energy (Refer to pp. 244 - 246 of Theory of Elasticity by Timoshenko)

dVx

u

dVx

u

x

u

x

u

x

udV

V

ij

j

i

jiij

V

ij

i

j

j

i

i

j

j

i

ij

V

ijij

2

1

)(2

1

2

1

)(2

1

2

1int

Total Potential Energy

dSTudVbudVS

ii

V

ii

V

ijij 2

1

V, E ,

TT

0 uu Su

St




4.2. Error Minimization

Error Estimator : S

h

i

h

iii

h

jij

V

h

ii

R dSTTuudVbuu ))((2

1))((

2

1,

Least Square Error

LSh

lklkijkl

V

h

jiji

h

klklijkl

V

h

jiji

h

ijij

V

h

jiji

S

h

i

h

ii

h

i

S

h

ii

h

ijij

V

h

jiji

S

h

i

h

iijij

h

ij

S

h

iiij

h

ij

V

h

jiji

S

h

i

h

iijij

h

jij

V

h

ii

R

dVuuDuu

dVDuu

dVuu

dSTTuudSTTuudVuu

dSTTuudSnuudVuu

dSTTuudVuu

)()(2

1

)()(2

1

))((2

1

))((2

1))((

2

1))((

2

1

))((2

1))((

2

1))((

2

1

))((2

1))((

2

1

,,,,

,,

,,

,,

,,

,,


RR

S

i

h

i

V

i

h

i

h

ij

V

h

ij

V

ij

j

i

j

S

ij

h

i

V j

ijh

i

h

ij

V

h

ij

V

ij

j

i

V

ij

j

h

ih

ij

V j

h

i

V

ij

j

i

h

ij

V

ij

j

h

i

V

ij

j

h

i

V

ij

j

i

h

ij

V

ij

j

h

i

V l

kijkl

j

h

i

V

ij

j

i

h

ij

V

ij

j

h

i

l

h

k

ijkl

V j

i

V

ij

j

i

h

ij

V

ij

j

h

ih

ij

V j

i

V

ij

j

i

h

ij

V

ij

j

h

i

j

iLS

C

dSTudVbudVdVx

u

dSnudVx

udVdVx

u

dVx

udV

x

udV

x

u

dVx

udV

x

udV

x

u

dVx

udV

x

uD

x

udV

x

u

dVx

udV

x

uD

x

udV

x

u

dVx

udV

x

udV

x

u

dVx

u

x

u

t

2

1

2

1

2

1

2

1

2

1

2

1

)(2

1

2

1

2

1

)(2

1

2

1

2

1

)(2

1

2

1

2

1

)(2

1

2

1

2

1

))((2

1





RRLSR Min Min Min w.r.t. h

i

h

iu


Find hh u such that minimize RR

where } , on 0|{ V l

kijkl

j

iu dV

dx

duD

dx

duSuuu

- h : The exact solution.

- h : An approximate solution.

0 RR : Variational Principle or Principle of Virtual Work

0

)(2

1

)(2

1

)2

1(

dSTudVbudV

dSTudVbudVD

dSTudVbudVDD

dSTudVbudV

dSTudVbudV

t

t

t

t

t

S

i

h

i

V

i

h

i

V

h

ij

h

ij

S

i

h

i

V

i

h

i

V

h

klijkl

h

ij

S

i

h

i

V

i

h

i

V

h

klijkl

h

ij

h

klijkl

h

ij

S

i

h

i

V

i

h

i

V

h

ij

h

ij

h

ij

h

ij

S

i

h

i

V

i

h

i

h

ij

V

h

ij

RR

Absolute Minimum Property of the Total Potential Energy

e

ii

h

i uuu

)(

2

1

2

1

)()(2

1

2

1

2

1

2

1

)()()()(

2

1

2

1

dSTudVbudVx

uD

x

u

dVx

uD

x

udSTudVbudV

x

uD

x

u

dSTuudVbuudVx

uD

x

u

dVx

uD

x

udV

x

uD

x

udV

x

uD

x

u

dSTuudVbuudVx

uuD

x

uu

dSTudVbudV

t

t

t

t

t

S

i

e

i

V

i

e

i

l

k

V

ijkl

j

e

i

l

e

k

V

ijkl

j

e

i

S

ii

V

ii

l

k

V

ijkl

j

i

S

i

e

ii

V

i

e

ii

l

e

k

V

ijkl

j

e

i

l

e

k

V

ijkl

j

i

l

k

V

ijkl

j

e

i

l

k

V

ijkl

j

i

S

i

e

ii

V

i

e

ii

l

e

kk

V

ijkl

j

e

ii

S

i

h

i

V

i

h

i

h

ij

V

h

ij

h




dVx

uD

x

u

dVbudVx

uD

x

u

dVbudVudVx

uD

x

u

dSTudVbudSTudVudVx

uD

x

u

dSTudVbudSnudVudVx

uD

x

u

dSTudVbudVx

udV

x

uD

x

u

l

e

k

V

ijkl

j

e

iE

V

ijij

e

i

l

e

k

V

ijkl

j

e

iE

V

i

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

i

S

i

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

ijij

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

i

V

ij

j

e

i

l

e

k

V

ijkl

j

e

iEh

tt

t

t

2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

,

,

,

,

Since Dijkl is positive definite,

u

xD

u

x

i

e

j

ijklk

e

l

0 0

j

e

i

x

uand

u

xD

u

xdVi

e

j

ijkl

V

k

e

l

0 if and only if 0

j

e

i

x

u. Therefore

.)?? =for only holdssign equality The ( i

h

i

Eh uu

4.3. Principle of Virtual Work

If the following inequality is valid for all real number , the principle of virtual work holds.

ii

RR

ii

RR vuvu )()(

) admissible allfor ( 0)0(

)(

)()(

2

1

2

1

)()()()(

2

1

)()(

2

ii

S

ii

V

ii

l

k

V

ijkl

j

i

l

i

V

ijkl

j

i

S

ii

V

ii

l

k

V

ijkl

j

i

S

iii

V

iii

l

i

V

ijkl

j

i

l

k

V

ijkl

j

i

l

k

V

ijkl

j

i

S

iii

V

iii

l

kk

V

ijkl

j

ii

ii

RR

vvdSTvdVbvdVx

uD

x

vg

dVx

vD

x

vdSTvdVbvdV

x

uD

x

vg

dSTvudVbvu

dVx

vD

x

vdV

x

uD

x

vdV

x

uD

x

u

dSTvudVbvudVx

vuD

x

vu

vug

t

t

t

t




If the principle of virtual work holds, then the principle of minimum potential energy holds

because the boxed equation of the total potential energy vanishes identically. The approximate

version of the principle of virtual work is

h

i

S

i

h

i

V

i

h

i

V

h

ij

j

h

i vdSTvdVbvdVx

v

t

admissible allfor 0

Equivalence to the PDE

- Exact form

ii

S

iii

V

jiji vdSTTvdVbv

t

0 )()( , iiijij TTb , 0 ,

- Approximate form

hh

i

S

i

h

i

h

i

V

i

h

jij

h

i vdSTTvdVbv

t

0 )()( . iiijij TTb , 0 ,

Since i

S

i

h

ii

V

i

h

jiji vdSTTvdVbv

t

0)()( . .

Equivalence to the Weighted Residual Method

- Discretization : kik

h

ikik

h

i gaugav ,

kdSTTgdVbg

adSTTgadVbga

t

t

S

i

h

ik

V

i

h

jijk

ik

S

i

h

ikik

V

i

h

jijkik

allfor )()(

possiblefor )()(

.

.

Uniqueness of solution

If two solutions satisfy the principle of virtual work, then

i

S

ii

V

ii

V

ij

j

i

i

S

ii

V

ii

V

ij

j

i

vdSTvdVbvdVx

v

vdSTvdVbvdVx

v

t

t

admissible allfor 0

admissible allfor 0

2

1

By subtracting two equations,

0 admissible allfor 0)( 2121

ijiji

V

ijij

j

i vdVx

v

00)(2121

l

k

l

k

l

k

l

kijkl

x

u

x

u

x

u

x

uD




Chapter 5

Discretization

5.1. Rayleigh-Ritz Type Discretization


5.3. Finite Element Programming (Linear static case)




5.1. Rayleigh-Ritz Type Discretization

Approximation

n

p

pipninii

h

i gcgcgcgcu1

2211

j

p

ip

n

p j

p

ip

j

h

i

x

gc

x

gc

x

u

1

Principle of Minimum Potential Energy

dSTgcdVbgcdVx

gcD

x

gc

tS

ipip

V

ipip

l

q

kq

V

ijkl

j

p

ip

h

2

1Min or 0

mr

h

c for all m, r

mrfcK

dSTgdVbgdVcx

gD

x

g

dSTgdVbgdVx

gcD

x

g

dSTgdVbgdVx

gD

x

gcdV

x

gcD

x

g

dSTgdVbg

dVx

gD

x

gcdV

x

gcD

x

g

c

rmkprmkp

S

mr

V

mrkp

V l

p

mjkl

j

r

S

mr

V

mr

l

p

kp

V

mjkl

j

r

S

mr

V

mr

l

r

V

ijml

j

p

ip

l

q

kq

V

mjkl

j

r

S

irmi

V

irmi

l

q

V

rqmkijkl

j

p

ip

l

q

kq

V

ijkl

j

p

rpmi

mr

h

t

t

t

t

and allfor 0

2

1

2

1

2

1

2

1

Principle of Virtual Work

0 V

i

h

i

V

h

i

h

i

V

h

ij

h

ij dVudVbudV

ip

S

ip

V

ipkq

V l

q

ijkl

j

p

ip

S

ip

V

ipkq

V l

q

ijkl

j

p

ip

S

ipip

V

ipip

V l

q

kqijkl

j

p

ip

h

cdSTgdVbgdVcx

gD

x

gc

dSTgdVbgdVcx

gD

x

gc

dSTgcdVbgcdVx

gcD

x

gc

t

t

t

allfor 0)(

)(

and allfor 0 ipfcK pikqpikq




Matrix Form – Virtual Work Expression

dVdV

dV

dV

dV

hT

V

hhhhhhhhhhh

V

hh

hhhhhhhhhh

V

hh

hhhhhhhhhhhhhhhh

V

hh

V

h

ij

h

ij

)(

)222(

)(

232313131212333322221111

232313131212333322221111

323223233131131321211212333322221111

dSdVdV

tS

h

V

hh

V

h

Tubu

Displacement

Ncu

n

n

n

n

n

n

h

h

h

h

c

c

c

c

c

c

c

c

c

g

g

g

g

g

g

g

g

g

u

u

u

3

2

1

32

22

12

31

21

11

2

2

2

1

1

1

3

2

1

00

00

00

00

00

00

00

00

00

Virtual Strain

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

23

13

12

33

22

11

)(

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

ii

ii

ii

ii

ii

ii

ii

ii

ii

hh

hh

hh

h

h

h

h

h

h

h

h

h

h




Virtual Strain - Matrix Form

cB

n

n

n

nn

nn

nn

n

n

n

h

h

h

h

h

h

h

c

c

c

c

c

c

c

c

c

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

3

2

1

32

22

12

31

21

11

23

13

12

3

2

1

2

2

3

2

1

2

3

2

1

2

2

2

3

2

2

2

1

2

2

1

3

1

1

1

3

1

1

1

2

1

3

1

2

1

1

1

23

13

12

33

22

11

0

0

0

00

00

00

0

0

0

00

00

00

0

0

0

00

00

00

)(

Stress-strain (displacement) Relation

DBcD

h

h

h

h

h

h

h

h

h

h

h

h

h

h

v

v

v

v

v

vv

v

v

vv

v

v

vv

v

v

v

vv

E

23

13

12

33

22

11

23

13

12

33

22

11

)1(2

2100000

0)1(2

210000

00)1(2

21000

000111

0001

11

00011

1

)21)(1(

)1()(

Final System Equation

cDBBc dVdVdVV

TT

V

hTh

V

h

ij

h

ij

dSdSdSTu

dVdVdVbu

ttt S

TT

S

Th

S

i

h

i

V

TT

V

Th

V

i

h

i

TNcTu

bNcbu

fKccTNbNcDBBcT or admissible allfor 0)( dSdVdV

tS

T

V

T

V

T




Example

Displacement Field

By the elementary beam solution, the displacement field of the structure is assumed as

)2

6

(

)2

(

23

2

lxx

bv

ylxx

au

2

100

01

01

1 ,

22

00

0)(

,

260

0)2

(

222

23

2

E

lxx

lxx

ylx

lxx

ylxx

DBN

Stiffness Matrix

)2(15

2

2

1)2(

15

2

2

1

)2(15

2

2

1)2(

15

2

2

1

332

1

)2

(2

1)

2(

2

1

)2

(2

1)

2(

2

1)(

1

22

00

0)(

2

100

01

01

200

20)(

1

55

5533

2

0

1

0 22

22

22

22

22

2

0

1

022

2

2

2

hlhl

hlhlhl

E

dzdydx

lxx

lxx

lxx

lxx

ylxE

dzdydx

lxx

lxx

ylx

lxx

lxx

ylxE

l h

h

l h

h

K

x, u

y, v

l

2h

h

PTy

2




Load Term

P

ldxdy

hl

yl

h

h

0

32

P0

30

02

3

3

2

1

0

f

System Equation

P

l

b

a

hlhl

hlhlhl

E 0

3)2(

15

2

2

1)2(

15

2

2

1

)2(15

2

2

1)2(

15

2

2

1

332

1

3

55

5533

2

PEIl

h

vbP

EIP

Eha

22

2

3

2 1))(

1

1

3

51( ,

11

2

3

Displacement and Stress

)2

6

(

1))(

1

1

3

51(

)2

(

1

2322

22

lxx

PEIl

h

vv

ylxx

PEI

u

Pxlx

Il

h

yI

M

yI

My

I

lxP

xy

yy

xx

)2

(

1)(

6

5

)(

22




Home Work 5

1. The displacement field of a thin plate is expressed as follows.

),( , ),,( , ),,( yxwwzy

wzyxvz

x

wzyxu

1) Drive the strain component using the given displacement field.

2) Assume 033 , and the plate is under plane stress condition, drive stress components.

3) Drive expressions for the total potential energy and the virtual work in case the plate is

subject to a traverse load a on the upper surface. (hint: perform analytic integration in the

direction of the thickness)

4) Drive the governing equation and all possible boundary conditions on the four edges.

2. Analyze the structure shown in the following figure under the plane stress condition by

Rayleigh-Ritz method

x

z

y

t

l

2h

q





Domain Discretization

, 2121 mnVVVV

e S

Th

e V

Th

e V

hΤh

Th

V

Th

V

hΤh

dSdVdV

ddVdV

et

ee

t

bubu

Tubu

The displacement field in an element

ee

e

n

n

n

n

n

n

e

e

e

e

u

u

u

u

u

u

u

u

u

N

N

N

N

N

N

N

N

N

u

u

u

UNu

3

2

1

32

22

12

31

21

11

2

2

2

1

1

1

3

2

1

00

00

00

00

00

00

00

00

00

where Ni j ij( )X .




The virtual strain field in an element

e

e

n

n

n

nn

nn

nn

n

n

n

ii

ii

ii

ii

ii

ii

ii

ii

ii

ee

ee

ee

e

e

e

e

e

e

e

e

e

e

u

u

u

u

u

u

u

u

u

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

N

x

Nu

x

Nu

x

Nu

x

Nu

x

Nu

x

Nu

x

Nu

x

Nu

x

Nu

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

UB

3

2

1

32

22

12

31

21

11

23

13

12

3

2

1

2

2

3

2

1

2

3

2

1

2

2

2

3

2

2

2

1

2

2

1

3

1

1

1

3

1

1

1

2

1

3

1

2

1

1

1

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

23

13

12

33

22

11

0

0

0

00

00

00

0

0

0

00

00

00

0

0

0

00

00

00

The stress field in an element

ee

e

e

e

e

e

e

e

e

e

e

e

e

e E

DBUD

23

13

12

33

22

11

23

13

12

33

22

11

)1(2

2100000

0)1(2

210000

00)1(2

21000

000111

0001

11

00011

1

)21)(1(

)1(




Stiffness Equation

0

0

e

eTee

e

eTe

e S

TTe

e V

TTe

e

e

V

TTe dSdVdVet

ee

fUUKU

TNUbNUUDBBU

Compatibility Conditions

0fKU

UfTUUTKTU

UTUUTU

admissible allfor 0

,

e

eTeT

e

eeTeT

eeee

Finite Element Procedure

1. Governing equations in the domain, boundary conditions on the boundary.

2. Derive weak form of the G.E. and B.C. by the variational principle or equivalent.

3. Descretize the given domain and boundary with finite elements.

S , 2121 nn SSSVVVV

4. Assume the displacement field by shape functions and nodal values within an element.

eeeUNu

5. Calculate the element stiffness matrix and assemble it according to the computability.

dVeV

Te

DBBK , e

eKK

6. Calculate the equivalent nodal force and assemble it according to the computability.

dSdVet

e S

T

V

Te

TNbNf , e

eff

7. Apply the displacement boundary conditions and solve the stiffness equation.

8. Calculate strain, stress and reaction force.




5.3. Finite Element Programming (Linear static case)

Program Structure

Data Structure

- Control Data :

# of nodes, # of elements, # of support, # of forces applied at nodes...

- Geometry Data :

Nodal Coordinates

Element information (Type, Material Properties, Incidencies)

- Material Properties

- Boundary Condition

Traction boundary condition including forces applied at nodes

Displacement boundary condition

- Miscellaneous options

INPUT

Preprocessing

Calculation of

Element stiffness matrix and Load Vector

Assembling E.S.M and E.L.V.

Solve Global SM

Postprocessing

Loop over all elements

Global Stiffness Matrix

and Load Vector

- Assemble Nodal Load Vector

- Cal. of Destination Array

- Cal. of Band width

Cal. Strain & Stress

Cal of Reaction Force

Dsiplay

Gauss Elimination

Band solver

Decomposition, etc




52

Differential Equation & Boundary Conditions


i

S

ii

V

ii

V

ijij udSTudVbudV

t

admissible allfor 2

1


i

S

ii

V

ii

V

ij

j

i udSTudVbudVx

u

t

admissible allfor 0

Weighted Residual Method (Galerkin Method)

ii

S

iii

V

jiji vdSTTvdVbv

t

admissible allfor 0)()( ,

Minimization of Error

iij

V

e

ij

j

i

j

e

i

error vdVx

v

x

u admissible allfor ))((

2

1


h

i

S

i

h

i

V

i

h

i

V

h

ij

h

ij udSTudVbudV

t

admissible allfor 2

1


h

i

S

i

h

i

V

i

h

i

V

h

ij

j

h

i udSTudVbudVx

u

t

admissible allfor 0


0- admissible allfor -2

1

admissible allfor )(2

1

fUKUfUUKUTT

h

i

e S

i

h

i

V

i

h

i

V

h

ij

h

ij udSTudVbudVrt

ee


0 admissible allfor 0

admissible allfor 0)(

fUKUfUUKUTT

h

i

e S

i

h

i

V

i

h

i

V

h

ij

j

h

i udSTudVbudVx

u

et

ee

Strong form

Weak form

Exact Solution

Approximate Solution Rayleigh-Ritz type Discretization

Finite Element Discretization




53

Chapter6

Two-Dimensional Elasticity

Problems

6.1. Plane Stress

6.2. Plane Strain

6.3. Axisymmetry

x y

z




54

6.1. Plane Stress

Stress : 33 13 23 0

Strain :

33 11 22 13 231

0 0

v

( ), ,

Modified Stress-strain Relation

1212

2211222

2211233221111

)1(2

)(1

)(1

))(1

()21)(1(

)1(

E

vE

vEE

e

e

e

e

e

e

e

e E D

12

22

11

2

12

22

11

2

100

01

01

1

Interpolation of Displacement

ee

e

n

n

n

n

e

e

e

v

u

v

u

v

u

N

N

N

N

N

N

v

uUNu

2

2

1

1

2

2

1

1

0

0

0

0

0

0

Strain-Displacement Relation

e

e

n

n

nn

n

n

ee

e

e

e

e

e

e

v

u

v

u

v

u

x

N

y

N

y

Nx

N

x

N

y

N

y

Nx

N

x

N

y

N

y

Nx

N

x

v

y

u

y

vx

u

BU

2

2

1

1

22

2

2

11

1

1

12

22

11

0

0

0

0

0

0

Principle of Virtual work

e S

T

e A

T

e A

T

y

h

S

x

h

y

h

A

x

hhhhh

A

hh

eee

t

tdStdAtdA

tdSTvTutdAbvbutdA

TNbNUBDB

)()()( 121222221111




55

6.2. Plane Strain

Strain : 33 13 230 0 0 , ,

Stress : 13 23 33 11 220 ( )

Stress-strain Relation

e

e

e

e

e

e

e

e

v

v

vvE

D

12

22

11

12

22

11

)1(2

2100

011

01

1

)21)(1(

)1(


ee

e

n

n

n

n

e

e

e

v

u

v

u

v

u

N

N

N

N

N

N

v

uUNu

2

2

1

1

2

2

1

1

0

0

0

0

0

0


e

e

n

n

nn

n

n

ee

e

e

e

e

e

e

v

u

v

u

v

u

x

N

y

N

y

Nx

N

x

N

y

N

y

Nx

N

x

N

y

N

y

Nx

N

x

v

y

u

y

vx

u

BU

2

2

1

1

22

2

2

11

1

1

12

22

11

0

0

0

0

0

0


e S

T

e A

T

e A

T

y

h

S

x

h

y

h

A

x

hhhhh

A

hh

eee

t

tdStdAtdA

tdSTvTutdAbvbutdA

TNbNUBDB

)()()( 121222221111




56

6.3. Axisymmetry

Strain : 0 , 0 , , , ,

zrrzzzrr

r

v

z

u

r

u

z

v

r

u

Stress : r z 0

Stress-strain Relation

e

e

rz

e

e

zz

e

rr

e

rz

e

e

zz

e

rr

e

v

vv

vv

vv

vE

D

)1(2

21000

0111

01

11

011

1

)21)(1(

)1(


ee

e

n

n

n

n

e

e

e

v

u

v

u

v

u

N

N

N

N

N

N

v

uUNu

2

2

1

1

2

2

1

1

0

0

0

0

0

0


e

e

n

nnn

n

n

n

ee

e

e

e

e

rz

e

e

zz

e

rr

e

v

u

v

u

v

u

r

N

z

Nr

Nz

Nr

N

r

N

z

Nr

Nz

Nr

N

r

N

z

Nr

Nz

Nr

N

r

v

z

ur

uz

vr

u

BU

2

2

1

1

22

2

2

2

11

1

1

1

0

0

0

0

0

0

0

0

0


( ) ( ) ( )

rr

h

rr

h

A

zz

h

zz

h h h

r

h

r

h h

r

A

h

z

h

r

S

h

z

T

Ae

T

Ae

T

Se

rdA u b v b rdA u T v T rdS

rdA rdA rdS

t

e e e

2 2 2

2 2 2B D B U N b N T




57

Chapter 7

Various Types of Elements

7.1. Constant Strain Triangle (CST) Element

7.2. Isoparametric Formulation

7.3. Bilinear Isoparametric Element

7.4. Higher Order Rectangular Element - Lagrange Family

7.5. Higher Order Rectangular Element - Serendipity Family

7.6. Triangular Isoparametric Element




58

7.1. Constant Strain Triangle (CST) Element

Displacement Interpolation

yxyxvyxyxu ee

654321 ),( , ),(

33321333

23221222

13121111

),(

),(

),(

yxuyxu

yxuyxu

yxuyxu

e

e

e

→

3

2

1

33

22

11

3

2

1

1

1

1

yx

yx

yx

u

u

u

3

2

1

321

321

321

3

2

1

33

22

11

3

2

1

2

1

1

1

11

u

u

u

ccc

bbb

aaa

u

u

u

yx

yx

yx

where a x y x y b y y c x xi j m m j i j m i m j , ,

333322221111

333322221111

)(2

1)(

2

1 +)(

2

1),(

)(2

1)(

2

1 +)(

2

1),(

vycxbavycxbavycxbayxv

uycxbauycxbauycxbayxu

e

e

Strain Components

e

e

ee

e

e

e

e

e

e

v

u

v

u

v

u

bc

c

b

bc

c

b

bc

c

b

x

v

y

u

y

vx

u

BU

3

3

2

2

1

1

33

3

3

22

2

2

11

1

1

12

22

11

0

0

0

0

0

0

2

1

Stiffness Matrix

K B D B U B D Be T T e

A

tdA te

1u

1v

2u

2v

3u

3v

( , )x y1 1

( , )x y2 2

( , )x y3 3




59

Homework 6

- Cantilever Beam with three load cases -

Implement your own finite element program for 2-D elasticity problems using CST element.

When you build your program, consider expandability so that you can easily add other types

of elements to your program for next homeworks.

a) Discuss how to simulate two boundary conditions given in the Timoshenkos’s book

(equation(k) and (l) in page 44)

b) For end shear load case, solve the problem for both boundary conditions with 40 CST

elements.

c) For other load cases, use one boundary condition of your choice with 40 CST

elements

d) Perform the convergence test with at least 5 different mesh layouts for the end shear

load case. Use the boundary condition of your choice.

e) Discuss local effects, St-Venant effect, Poisson effect stress concentration, etc.

Present suitable plots and tables of displacement and stress to justfy or clearfy your

discussions.

f) Comparision of your results with other solutions such as analytic solutions, one-

dimensional solutions is strongly recommended for your discussion . Assume E = 1.0,

= 0.3.

10

2

1

1

1

1




60

7.2. Isoparametric Formulation

Interpolation of Geometry

m

i

iimm

m

i

iimm

yNyNyNyNy

xNxNxNxNx

1

2211

1

2211

),(),(),(),(

),(),(),(),(

ee

e

m

m

m

m

e

e

e

y

x

y

x

y

x

N

N

N

N

N

N

y

xXNx

2

2

1

1

2

2

1

1

0

0

0

0

0

0




61

Interpolation of Displacement in a Parent Element

ee

e

e

e

v

uUNu ),(

Derivatives of the Displacement Shape Functions

y

Nx

N

yx

yx

N

N

y

y

Nx

x

NN

y

y

Nx

x

NN

i

i

i

i

iii

iii

NNN

N

N

N

yx

yx

y

Nx

N

ixi

i

i

i

i

i

11

1

or JJ

XNJ

mm

m

m

m

i

i

im

i

i

i

m

i

i

im

i

i

i

yx

yx

yx

N

N

N

N

N

N

yN

xN

yN

xN

yx

yx

22

11

2

2

1

1

11

11

m > n N N : Superparametric element

m = n N N : Isoparametric element

m < n N N : Subparametric element

1

1

1

1

||),(),( ddJttdA T

A

T

e

BDBBDB




62

7.3. Bilinear Isoparametric Element

Shape functions in the parent coordinate system.

4321)),(),,(( yxu

)),(),,(( 8765 yxv

4321444434214444444

4321334333213333333

4321224232212222222

4321114131211111111

)),(),,((),(

)),(),,((),(

)),(),,((),(

)),(),,((),(

yxuyxuu

yxuyxuu

yxuyxuu

yxuyxuu

4

3

2

1

4

3

2

1

1111

1111

1111

1111

u

u

u

u

),( , ),( 4

4

3

3

2

2

1

1

4

4

3

3

2

2

1

1 vNvNvNvNyxvuNuNuNuNyxu ee

N N N N1 2 3 4

1

41 1

1

41 1

1

41 1

1

41 1 ( )( ) ( )( ) ( )( ) ( )( ) , , ,

1 2

3

4

1 2

3 4

N1 N2

N4 N3




63

7.4. Higher Order Rectangular Element - Lagrange Family

Shape Function of m-th Order for k-th Node in One Dimension

lk

m k k m

k k k k k k k m

( )( )( ) ( )( ) ( )

( )( ) ( )( ) ( )

1 2 1 1 1

1 2 1 1 1

lk

m k k m

k k k k k k k m

( )( )( ) ( )( ) ( )

( )( ) ( )( ) ( )

1 2 1 1 1

1 2 1 1 1

N N l li IJ I

m

J

m ( ) ( )




64

Q9 Element

Total number of nodes in an element : (m+1)(m+1)

Total number of terms in m-th order polynomials : ( )( )m m 2 1

2

Total number of the parasitic terms : ( )m m1

2

The Pascal Polynomials

1

2 2

3 2 2 3

1 1

1 1

m m m m

m m m m

m m

Complete Polynomial

Parasitic Term




65

7.5. Higher Order Rectangular Element - Serendipity Family

Q8 Element

)1)(1(2

1 2

5 N

)1)(1(2

1 2

8 N

)1)(1(4

1ˆ1 N

512

1ˆ NN

85112

1

2

1ˆ NNNN




66

7.6. Triangular Isoparametric Element

Total number of nodes on sides of a triangle element for m-th order S.F.: 3m

Total number of terms in m-th order polynomials : ( )( )m m 2 1

2

Total number of internal nodes : ( )( )m m 2 1

2-3m =

( )( )m m 2 1

2

The Pascal Polynomials

mmmm

11

3223

22

1

Area Coordinate System

11

22

33

1 2 3 1 A

A

A

A

A

A , , ,

Shape functions

- CST Element

N N N1 1 2 2 3 3 , ,

- LST Element

316325214

333222111

444

)1(2)1(2)1(2

NNN

NNN

A3

A1 A2

1

2

3

constant line

constant line

constant line




67

Interpolation of Geometry

n

i

ii

n

i

ii xNxNx1

21

1

321 ),(~

),,(

n

i

ii

n

i

ii yNyNy1

21

1

321 ),(~

),,(

)(]~

[~0

0~

~0

0~

~0

0~

)( 2

2

1

1

2

2

1

1 ee

e

m

m

n

n

e

e

e XN

y

x

y

x

y

x

N

N

N

N

N

N

y

x

X

Interpolation of Displacement in a Parent Element

( ) [ ( , , )] ( ) [~

( , )] ( )uu

vN U N Ue

e

e

e e e e

1 2 3 1 2

Derivatives of the Displacement Shape Functions

~

~

~

~

~

~

~

~

~

~

~~~

~~~

2

11

2

1

1

22

11

22

11

2

1

222

111

i

i

i

i

i

i

i

i

i

i

iii

iii

N

N

N

N

yx

yx

y

Nx

N

y

Nx

N

yx

yx

N

N

y

y

Nx

x

NN

y

y

Nx

x

NN

J

iix NN~~

or

1

J

XNJ

~

~

~

~

~

~

~

~~

~~

22

11

2

1

2

2

1

2

2

1

1

1

1 21 2

1 11 1

22

11

mm

n

n

m

i

ii

m

i

ii

m

i

ii

m

i

ii

yx

yx

yx

N

N

N

N

N

N

yN

xN

yN

xN

yx

yx

Homework 7

Drive shape functions qubic serendipity element and triangle element in the parent coordinate.




68

Chapter8

Numerical Integration

8.1 Gauss Quadrature Rule

8.2. Reduced Integration

8.3. Spurious Zero Energy mode

8.4. Selective Integration




69

8.1. Gauss Quadrature Rule

One Dimension

f d W fi i

i

n

( ) ( )

11

1

If the given function f ( ) is a polynomial, it is possible to construct the quadrature rule

that yields the exact integration.

- f ( ) is constant: 0)( af

2 1 2)( 1

1

1 1

00

W n=Waadfn

i

i

- f ( ) is first order: 10)( aaf One point rule is good enough.

- f ( ) is second order: 2

210)( aaaf

2 2 , 0 , 3

2

23

2)(

111

2

1

1 1

0

1

1

1

2

202

nWWW

WaWaWaaadf

n

i

i

n

i

ii

n

i

ii

n

i

i

n

i

ii

n

i

ii

2121

1

1 , 0

WWWn

i

ii

W W W W Wi i

i

2

1

2

1 1

2

2 2

2

2 2

2

2 2

222

3

1

3

W W W W Wi

i

1

2

1 2 2 2 22 2 1 057735 = 1/ 3 02691 89626 .

- f ( ) is third order: 3

3

2

210)( aaaaf Two point rule is enough.

- f ( ) is fourth order: 4

4

3

3

2

210)( aaaaaf

3 2 , 0 , 3

2 , 0 ,

5

2

23

2

5

2)(

111

2

1

3

1

4

1

0

1

1

1

2

2

1

3

3

1

4

4

1

1

024

nWWWWW

WaWaWaWaWa

aaadf

n

i

i

n

i

ii

n

i

ii

n

i

ii

n

i

ii

n

i

i

n

i

ii

n

i

ii

n

i

ii

n

i

ii

W W W Wi i

i

n

i i

i

n

3

1 1

1 3 1 3 20 0 0

, , ,

W W W W Wi i

i

4

1

3

1 1

4

3 3

4

3 3

4

3 3

422

5

1

5




70

W W W W W W

W

i i

i

2

1

2

1 1

2

3 3

2

3 3

2

3 3

2

3 3

3 3

22

3

1

33 5

5

9

0 77459 055555 55555 55555

, =

66692 41483 ,

/

. .

W W W W W W Wi

i

1

2

1 2 3 3 2 22 28

9 = 0.88888 88888 88888

- Because of the symmetry condition, we need to decide only n unknowns for n-points

G.Q..

- We can integrate 2n-1-th polynomials exactly with n-points G.Q. Since for 2m-th order

polynomials we have 2m conditions for G.Q. -which means we can determine (m+1)-

point G.Q..

- Stiffness Equation

B D B B D B B D B B D BT

V

T

x

T

i

T

i i i i i

i

n

dV Adx A J d W A Je e

( ) ( ) ( ) ( )

1

1

1

N b N b N b N bT

V

T

x

T

i

T

i i i i

i

n

dV Adx A J d W A Je e

( ) ( ) | | ( ) ( ) | |

1

1

1

Two-dimensional Case – Rectangular Elements

- Quadrature rule

f d d W f d W W f WW fi i

i

n

j

j

m

i i j

i

n

i j i j

i

n

j

m

( , ) ( ) ( ) ( ) 1

1

1

1

11

1

1 1 11

- Stiffness equation

n

i

m

j

ijijjiijji

T

ji

T

A

T

JtWW

ddJttdAe

1 1

1

1

1

1

||),(),(

||),(),(

BDB

BDBBDB

n

i

ijijijpi

T

ip

T

S

T

n

i

m

j

ijijijji

T

ji

T

A

T

KtWdKttdS

JtWWddJttdA

e

e

1

1

1

1 1

1

1

1

1

||),(||),(

||),(||),(

TNTNTN

bNbNbN

Two-dimensional Case – Triangular Elements

n

i

ii

ii

ij

iiT

i

T

A

T

JtW

ddJttdAe

1

2121

1

0

1

0

122121

||),(),(2

1

||),(),(1

BDB

BDBBDB




71

8.2. Reduced Integration

Q8 element

2

7

2

6

2

54

2

3210

2

7

2

6

2

54

2

3210

bbbbbbbbv

aaaaaaaau

2

76431

2

76431 22 , 22

bbbbb

vaaaaa

u

7

2

65427

2

6542 22 , 22 bbbbbv

aaaaau

Terms in stiffness matrix

- From complete polynomials: 22 , , , , , 1

- From parasitic terms: 4433223322 , , , , , , , ,

Reduced Integration

Reduce the integration order by one to eliminate the effect of parasitic terms in the stiffness

matrix

Exact average

shear stress Nodal values extrapolated

from Gauss point

Gauss point value by the

reduced integration




72

8.3. Spurious Zero Energy mode

Independent Displacement Modes of a Bilinear Element

Spurious Zero Energy Mode

Hour glass mode

Rigid Body motion – zero energy mode




73

Zero Energy mode of Q8 Element

Zero Energy Modes of Q9 Element

Near Zero Energy Modes




74

8.4. Selective Integration

SN

EE

E

DD

D

100000

010000

001000

000000

000000

000000

)1(2

000000

000000

000000

000111

0001

11

00011

1

)21)(1(

)1(

)1(2

2100000

0)1(2

210000

00)1(2

21000

000111

0001

11

00011

1

)21)(1(

)1(

nintegratio Reducednintegratio Full

)( dVdVdVdVeeee V

S

T

V

N

T

V

SN

T

V

T

BDBBDBBDDBDBB

Homework 8

Solve the cantilever beam problem in homework 6 for the end shear load case. Use 40 LST

elements and 20 Q8 elements. For Q8 element, try the full and the reduced integration

scheme. Compare your results including the displacement field and the stress field with those

by CST elements and analytical solution.




75

Chapter 9

Convergence Criteria in the

Isoparametric Element




76

The interelement Continuity condition

This condition is satisfied because the geometry and the displacement field are defined

uniquely on the interelement boundary with the coordinates and displacement field on the

element boundary, respectively.

The Completeness Condition

( , ) ( ) ( ) ( ) ( , , )

( ) ( ) ( ) ( , , )

( ) ( ) ( ) ( , , )

x y N a a a H

x N x a x a x a x H x

y N y a y a y a y H y

i

i

i k k k k

i

i

i k k k k

i

i

i k k k k

0 1 2

0 1 2

0 1 2

1 2 2

1 1

0

0

1 2 1 2

| |

( ) ( )

( ) ( )

( ) ( , , )

( ) ( , , )

| | ( ) ( ) ( ) ( )

A

a y a x

a y a x

x a x H x

y a y H y

A a x a y a y a x

k k

k k

k k

k k

k k k k

where

( , ) ( )

( )

| |( ( ) ( ) ( ) ( ))

( )

| |( ( ) ( ) ( ) ( ))

( ( ) ( ) ( ) ( ))| |

( ( ) ( )

x y a

a

Aa x a y a y a x

a

Aa x a y a y a x

a a y a y ax

Aa a x

k

kk k k k

kk k k k

k k k k k k

0

10 2 0 2

20 1 0 1

1 2 1 2 1 2

a x ay

Ak k1 2( ) ( ))

| | H.O.T.




77

Chapter 10

Miscellaneous Topics

10.1 Stress Evaluation, Smoothing and Loubignac Iteration

10.2. Incompatible Element - Q6 and QM 6

10.3. Static Condensation & Substructuring

10.4. Symmetry of Structure

10.5. Constraints in Discrete Problems

10.6. Constraints in Continuous Problems




78

10.1. Stress Evaluation, Smoothing and Loubignac Iteration

Stress evaluation

- Stress components should be evaluated at the GP’s in each element, not at nodes.

- The stress field is not uniquely determined on inter-element boundaries.

Stress smoothing at nodes

- Continuous stress field can be obtained by extrapolating stresses at the GP’s to nodes,

and averaging them.

- The bilinear shape function or the Q9 shape function may be utilized for extrapola-

tion of stress to nodes depending on the integration schemes.

- Mid-side nodes may be treated as either independent nodes or dependent nodes for

the stress field

Loubignac iteration

FUKBFUDBB

UDBBBBBUDBB

FTNbNUDBB

e

e V

T

e

e

V

T

e

e

V

T

e V

T

e V

T

e V

T

e

e

V

T

e

T

e V

T

e

e

V

T

ee

eeeee

et

ee

dVdV

dVdVdVdVdV

ddVdV

~

~)~(

where ~ denotes the extrapolated and averaged stress field.




79

10.2. Incompatible Element - Q6

Deformed shape of Bilinear Element

for pure bending Correct deformed shape in pure bending

Behaviors of Bilinear Element for Pure Bending

- Displacement field

u u u u u u

v

1

41 1

1

41 1

1

41 1

1

41 1

0

( )( ) ( )( ) ( )( ) ( )( )

- Strain & Stress field

ab

xuE

ab

yuE

ab

yuE

ab

xu

x

v

y

u

y

u

ab

yu

a

u

x

u

xyyx

xyyx

)1(2 ,

1 ,

1

, 0 , =

22

-Strain Energy - Full Integration

2

2

22

2

))1(2(13

1

))()1(2

)(1

(2

1

)(2

1

ub

a

a

bE

dydxab

xuE

ab

yuE

dydx

a

a

b

b

xyxyyy

a

a

b

b

xxb

-Strain Energy - Selective Integration

ua

bE

dydxab

xuE

ab

yuEa

a

b

b

s

b

2

22

2

13

2

))()1(2

)(1

(2

1




80

Real Behaviors for Pure Bending

- Strain & Stress field

x y xy

x y xy

ky

k

Ey

kv

Ey

, ,

, ,

0 0

0

- Displacement field

uk

Exy f y v

kv

Ey +g x

k

Ex f y g x

g xk

Ex f y C g x

k

Ex Cx C f y Cy C

xy

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

, ,

, ,

1

20

1

2

2

2

1 2

a

ub

b

y

b

ua

a

xCx

ab

uy

ab

vuvxy

ab

uu

C

ab

uEkuubyax

CCCCyux

CCxxE

ky

E

kvvCCyxy

E

ku

2))(1(

2))(1(

2

1

2

1 ,

Arbitrary

,

00 0

2

1

2

1 ,

22

1

22

1

22

1

22

2

-Strain Energy – Exact Solution

R x x

b

b

a

a

y y xy xy

b

b

a

a

dydx Euy

abdydx E

b

au

1

2

1

2

2

3

2 2( ) ( )

Ratio of strain Energy

b

R

b

s

R

a

bv

1

1

1

1

1

2

1

111 0 32

2 ( ( ) ) . . , for

The effect of parasitic shear becomes disastrous as the aspect ratio of bilinear element is large.

Q6 Incompatible Element

- Shape function

u N u a a

v N v a a

i

B

i

i

i

B

i

i

1

2

2

2

3

2

4

2

1 1

1 1

( ) ( )

( ) ( )

a1, a2, a3, a4 are called as the “nodeless degrees of freedom”.




81

- Element Stiffnes Equation

K K

K K

u

a

f

f

bb bi

ib ii

b

i

( )

( )

( )

( )

- Static Condensation

)()]([)(][)()])([]][[]([

))]([)((][)()()]([)]([

6611

1

QQi

ii

b

ibiibibb

ib

i

ii

i

iiib

fuKfKfuKKKK

uKfKafaKuK

10.3. Static Condensation & Substructuring

Static Condensation: Eliminate some DOFs prior to a main analysis.

r

e

r

e

rrre

eree

P

P

u

u

KK

KK

)()( 1

rereeeeeeeerer uKPKuPuKuK

eeererrereererr

rrereeererrr

rererrr

PKKPuKKKK

PuKPKKuK

PuKuK

11

1

)())((

)()(

- From Gauss elimination point of view

eeerer

e

r

e

ereererr

eree

PKKP

P

u

u

KKKK

KK11 )()(0

ue, Pe ur, Pr




82

Substructuring

- Substructure 1

1

11

1

1

111

iii

i

ii P

P

u

u

KK

KK

1

1

111

1

1

1

111

1 )())(( PKKPuKKKK iiiiiii

- Substructure 2

2

22

2

2

222

iii

i

ii P

P

U

U

KK

KK

2

1

222

2

2

1

222


- Assembling

2

1

2221

1

111

21

2

1

2221

1

111

21 )()())()(( PKKPKKPPuKKKKKKKK iiiiiiiiiiiii

or

2

1

2221

1

1112

1

2221

1

111

21 )()())()(( PKKPKKPuKKKKKKKK iiiiiiiiiiii

u1, P1 u2, P2

ui, Pi




83

Partial Substructuring

- Substructure 1

1

11

1

1

111

iiiii

i

P

P

u

u

KK

KK

1

1

111

1

1

1

111


- Substructure 2

2

22

2

2

222

iiiii

i

P

P

u

u

KK

KK

- Assembling

1

1

111

21

22

1

1

111

12

2

222

)()( PKKPP

P

u

u

KKKKKK

KK

iiiiiiiiiii

i

Pi

u1, P1 u2, P2

ui, Pi




84

10.4. Symmetry of Structure

Symmetry

In plane displacement 0

Out of plane displacement 0

P P

C L




85

Anti-Symmetry

In plane displacement 0

Out of plane displacement 0

P P

C L




86

Symmetric structures with Non-Symmetric loading

- General loading

- Symmetric loading

- Antisymmetric loading

2

2

21

21

2

1

PPP

PPP

P

P

P

P

P

P

A

S

A

A

S

S

P

P/2 P/2

P/2 P/2




87

Cyclic Symmetry

1u

iu

2u




88

- Structural Resistance force in a segment

2

1

22221

21

12111

2

1

u

u

u

KKK

KKK

KKK

F

F

F

i

i

iiii

i

i

- Compatibility

21 uu

2

2

1

u

u

I0

0I

0

u

u

ui

i

- Equilibrium

122 FFP

FP

T

ii

2

1

2F

F

F

I0

0I0

P

PiT

i

2

22221

21

12111

2 u

u

I0

0I

0

KKK

KKK

KKK

I0

0I0

P

P i

i

iiii

i

T

i

- Final Equation

22221121121

21

2

22212

21

12111

2

u

u

KKKKKK

KKK

u

u

KKK

KKK

KKK

I0

0I0

P

P

i

TT

ii

T

iiii

i

i

iiii

i

T

i




89

10.5. Constraints in Discrete Problems

Minimization Problem

PUKUUu

TT 2

1Min subject to 0UA )(

r

f

U

UU ,

rrrf

frff

KK

KKK

r

f

P

PP

1st order Optimality condition

0

U 0PKU

Usual homogenous support conditions

0rU

ffff

T

fff

T

ff

PUKPUUKUU

2

1Min

Non-homogenous support conditions (support Settlement)

rr UU

r

T

rf

T

frrr

T

rrfr

T

ffff

T

f

r

T

rf

T

frrr

T

rrfr

T

ffrf

T

rfff

T

f

PUPUUKUUKUUKU

PUPUUKUUKUUKUUKU

)2(2

1

)(2

1

0MinMin frfrffff

PUKUKUU

rfrffff UKPUK

or

00

f

f

f

r

r

f

f UU

UU

UU

UU

U

General Non-homogenous Linear Constraints

0

1

0 0)( i

ndof

j

jij rur

rRUUA , i =1 ,…, ncon

00

11

0 CCUrRURRUrURURRU

frffrrffrr

0

0

CU

C

I

U

UU f

r

f




90

)(2

1)(

][2

1

2

1

0000 r

T

f

T

frr

T

rr

T

fr

T

f

frr

T

rf

T

frff

T

f

TT

PCPUCKCCKCCKU

UCKCKCCKKU

PUKUU

)()(][ 00 CKCCKPCPUCKCKCCKK rr

T

frr

T

ffrr

T

rf

T

frff

Lagrange Multiplier

)(2

1Min 0

,rRUPUKUU

U

TTT 0

U, 0

0

0 0)(

0)(

r

PU

0R

RK

rRUUA

RPKUU

UAPKU

U

TTT

)()(0)(

)(0

0

111

0

1

0

1

rPRKRRKrRPRKrRU

RPKURPKU

TT

TT

0

11111111

0

1111

)))((

))()((

rRRKRKPRKRRKRKK

rPRKRRKRPKU

TTTT

TT

10.6. Constraints in Continuous Problems

Lagrange Multiplier

u

Min subject to 0uA )(

where is the original functional derived from the minimization principle or equivalent,

and 0uA )( denotes an additional set of constraints, which may be defined in some

volume, on some surface, or at some points. For the simplicity of derivation, only con-

straints specified along a surface is considered in this note.

0)( 0 rLuuA on S

dSdSTudVbudV

dSuu

SS

i

h

i

V

i

h

i

h

ij

V

h

ij

S

iu

t

i

)(2

1

)()(),(Min

0

0,

rLu

rLu




91

By discretizing eΛNλ in an element and the displacement field in usual way,

qΛRUΛfUKUU

rΛULNΛfUKUU

rΛULNΛfUKUU

TTTT

S

TT

S

TTTT

e S

TTe

e

e

S

TTeTT

i

ee

ee

dSdS

dSdSu

2

1

2

1

)()(2

1),(

0

0

Therefore, the stationary condition for modified energy functional becomes

q

f

Λ

U

0R

RK

qRUΛ

ΛRfKUU

TT

0

0

The solution procedure from this point is exactly the same as that of discrete problems.

The last and the most important question is what kind of interpolation function has to be

employed for the Lagrange multiplier.

Penalty Function

S

T

udS)()(

2)()(Min uAuAuu

)2(22

1

}()(2)(){(22

1

)()(22

1)(

000

00

CqUUKUfUKUU

rrrLUULNLUfUKUU

rLurLuu

T

R

TTT

S

T

S

TTe

e

e

S

TTeTT

S

T

S

i

h

i

V

i

h

i

h

ij

V

h

ij

eee

t

dSdSdS

dSdSTudVbudV

qfUKKU

u

)(0)(Min R

u




92

Chapter 11

Problems with Higher

Continuity Requirement

– Beams –

4 = ??

11.1. C1 Formulation

11.2. C0 Formuation

11.3. Timoshenko Beam




93

11.1. C1-Formulation

Governing Equation

EId w

dxq

4

4

Weak Form of the governing equation

( ) w EId w

dxq dx w

l 4

4

0

0 for all admissible

wEId w

dx

dw

dxEI

d w

dx

d w

dxEI

d w

dxdx wqdx

l l ll3

3

0

2

2

0

2

2

2

2

00

0

d w

dxEI

d w

dxdx wqdx wV M

d w

dxEI

d w

dxdx wqdx

ll

ll

ll2

2

2

2

0

0

00

2

2

2

2

00

Continuity Requirement

The second derivative of the displacement field has to be piecewise-continuous for the valid

finite element formulation. Therefore, w has to be continuous up to the first derivatives.

d w

dxEI

d w

dxdx

d w

dxEI

d w

dxdx

le

l

e

2

2

2

2

2

2

2

2

0

Hermitian Shape Functions

elx

rer

x

ll

rrll

dx

dwlww

dx

dwww

NwNNwNw

, )( , , )0( where

0

4321

2

32

43

3

2

2

3

2

32

23

3

2

2

1

, 23

2 , 231

eeee

eeee

l

x

l

xN

l

x

l

xN

l

x

l

xxN

l

x

l

xN




94

11.2. C0-Formulation

Energy Consideration

w: total deflection, : rotation , dw

dx : shear deformation

1

20 0

0

0

( ( ) ( ) )d

dxEI

d

dxdx

dw

dxGA

dw

dxdx wqdx

l l l

i

i

ii

i

i NwNw ,

T

nn

e www ][ 2211

e

M

eni

i

i

dx

dN

dx

dN

dx

dN

dx

dN

dx

d B

)](0 0 0[ 21

e

S

e

nn

i

i

ii

i

i Ndx

dNN

dx

dNN

dx

dNNw

dx

dN

dx

dw B ] [ 2

21

1

e

nNNNw ]0 0 0[ 21

eTTeT

SM

T

e

e

Teee

S

e

M

e

Te

l

e

Tee

S

l

T

SM

l

T

M

e

Teeee

qdxdxGAdxEI

fKfKK

fKK

NBBBB

2

1)(

2

1

)()()(2

1

][)()()(2

1

0

0

00

0

efK




95

Problems –Functions Space Interlocking

EI

d

dx

d

dxdx

GA l

EI l

dw

dx

dw

dxdx

wq

EIdx

l l l

1

2

1

0

0

2

2

0 0

( ( )( ) )

beam retangularfor )1(

62

22

0

t

l

vEI

lGA

.

As , 0)()(1

0

2 dx

dx

dw

dx

dw

l

l

or dx

dw to keep the total potential energy

finite. This condition may be satisfied in case the entire, exact function spaces for w and

are used. For the finite element method, the energy expression is expressed as

e

eee

e

e

e

h

dEI

wqld

d

dw

ld

dw

ll

l

EI

lGAd

d

d

d

d

EI

l1

0

2

1

0

2

22

0

1

0

))1

()1

((2

1

To keep the energy expression finite as 0t , there are two alternatives:

1) tl e , 2) dx

dw. The first condition requires a fine mesh layout because the element

size should be almost the same as the thickness of beam. The second condition requires

the derivative of displacement field have to be exactly the same as the rotation flied,

which is very difficult to satisfy. For example, both the displacement and the rotation

field are interpolated by linear shape functions, ie, baw , dc . The second

condition becomes dca for all 10 , which is equivalent to 0c and

da . Therefore, the rotation field should be constant within an element. This condi-

tion can be satisfied only when the rotation field is constant for the entire beam because

of the continuity requirement of 0C , which allows only rigid body rotation of the

beam, and yields a meaningless solution. In case you try to analyze beams or plates,

do not use linear shape functions, which merely results in meaningless solutions.

The aforementioned difficulty may be avoided by using higher order shape functions. In

case shape functions of the same order for both fields are employed, the highest order

term of the rotation field should always vanish, and the function space for the rotation

fields is always a subspace of the displacement field. Since, however, the rotation and

the displacement field are totally independent fields as shown in the next section, the so-

lution space of the rotation field has to be defined independently, and thus the aforemen-

tioned function space constraint yields sub-optimal convergence and unfavorable results.

Be always very careful when you use C1-formulation for the analysis of beams or

plates.




96

11.3. Timoshenko Beam

Governing Equation

EId

dxGA

dw

dx

2

2 0 0

( ) , GAd w

dx

d

dxp0

2

2( )

Week Form

w

ll

wdxpdx

d

dx

wdGAwdx

dx

dwGA

dx

dEI

& 0))(())((

2

2

0

0

02

2

0

0)())(( 0

000

02

2

0

dxdx

dwGAdx

dx

dEI

dx

d

dx

dEIdx

dx

dwGA

dx

dEI

llll

0)()())((0

0

00

02

2

0

0

pdxwdxdx

dwGA

dx

wd

dx

dwwGAdxp

dx

d

dx

wdGAw

llll

w

lllll

w

pdxwdx

dwwGA

dx

dEIdx

dx

dwGA

dx

wddx

dx

dEI

dx

d

&

)()()(00

0

0

0

00

By assuming homogeneous boundary conditions,

w

lll

wpdxwdxdx

dwGA

dx

wddx

dx

dEI

dx

d

& )()(

0

0

00

Boundary Conditions

0 0 0 00 or or or EId

dxw GA

dw

dx( )

Elimination of the displacement – Beginning of Nightmare

Differentiation of the first equation and substitution of the second equation into the first

equation

03

3

pdx

dEI (Oh, My God !!!)

Unfortunately, we have an odd order differential equation, which does not have the min-

imum characteristics, and thus is very difficult to solve. At this point, we have to con-

sider the Petrov-Galerkin method seriously !!!




97

Chapter 12

Mixed Formulation




98

What is the mixed formulation??? - Definition

Stress or strain fields are treated and interpolated independently!!!

Governing Equations and Boundary Conditions

Equilibrium Equation : 0 b in V

Constitutive Law : :D in V

Strain-Displacement Relationship : ))((2

1 Tuu in V

Displacement Boundary condition : 0 uu on uS

Traction Boundary Condition : 0TT on tS

Cauchy’s Relation on the Boundary : nT on S

Weak statement.

iuii

S

iiijjiij

V

iji

V

jiji udSTTudVuudVbu

t

ˆ ˆ 0 )(ˆ))(2

1(ˆ)(ˆ

,,,

iuii

S

iikllkklijkl

V

iji

V

jiji udSTTudVuuDdVbu

t

ˆ ˆ 0 )(ˆ))(2

1(ˆ)(ˆ

,,,

iuikllkklijkl

V

ij

S

ii

V

ii

V

ij

j

i udVuuDdSTudVbudVx

u

t

ˆ ˆ 0))(2

1(ˆˆˆ

ˆ,,

iuikllkklijkl

V

ij

S

ii

V

ii

V

klijkl

j

i udVuuDdSTudVbudVDx

u

t

ˆ ˆ 0))(2

1(ˆˆˆ

ˆ,,

Interpolations

eeeUNu , eee

EN

Finite element discretization

eee

V

T

e

e

V

TTe

T

V

T

e

e

V

TTe

dVdV

ddVdV

e

te

EUUDBNEDNNE

TNbNEDNBU

ˆ and ˆ admissible allfor 0 )()ˆ(

)()ˆ(

FEQ T , 0QUME

0

F

E

U

MQ

Q0T

FQUMQ 1T




99

Important Question

What are the admissible function spaces for the displacement field and the strain field??

Can we choose the interpolation shape functions for the displacement and the strain inde-

pendently ?? Unfortunately, the answer is “No”. In case we choose the function spac-

es arbitrarily, the solutions of the mixed formulation become very unstable, which is

caused by so called “function space interlocking”. The Babuzuka-Brezzi condition (BB

condition) states the required relationship between the individual function spaces. This

issue is out of scope for this class.

Please be very careful when you use the FEM based on the mixed formulation!!!

Possible choices of function spaces

1. 11 ˆ , ˆ HHu ii

2. 2

1 ˆ , ˆ LHu ii

3. ˆ , ˆ22 LLu ii

Which one do you like? Can you give a proper explanation for your choice?

Documents

Introduction to Finite Element Methodstrana.snu.ac.kr/lecture/fem_2018/Note18_FEM.pdf · 1. Introduction 2. Approximation of Functions & Variational Calculus 3. Differential Equations