Basic Theory of Finite Element Method.pdf

7/27/2019 Basic Theory of Finite Element Method.pdf

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Basic Theory of Finite Element Method

2013/06

Taiki SAITO

Professor, Toyohashi University of Technology


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1. INTRODUCTION1-1. Section

1-2. Stress and Strain1) One-Dimensional Problem

bDA D

bb

Area Moment of Inertia

12

3bD

Ix

12

3Db

Iy

L

bD N

A

N

L

E

E

Stress Strain Hooks Law

E: Youngs Modulus

L

EAN

Force Deformation Relationship


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2) Two-Dimensional Problem

xy

y

x

xy

y

x

G

EE

EE

100

011

011

or

xy

y

x

xy

y

xE

2

100

01

01

1 2

E

E

x

y

x

x

: Poisson Ratio

x

x x x

y

y

EE

EE

xy

y

yx

x

xy

xyxy

xy

xy

Normal Stress and Strain

Shear Stress and Strain

xyxy G

EG)1(2

1

: Shear Modulus


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1-3. Beam TheoryP

P

Q

QM

M

Q : Shear Force

M : Moment

P

M(x)

x

P

y

)(1

2

2

xMEIdx

yd

)()(

xQ

dx

xdM

Example )

21

3

1

2

2

2

2

2

6

1

2

1

)(),(1

cxcxEI

Py

cxEI

P

dx

dy

xEI

P

dx

yd

PxxMxMEIdx

yd

EI

PLc

EI

PLc

Therefore

yanddxdyLxat

3

2

2

13

1,

2

,

:00,

L

EI

PLx

EI

PLx

EI

Py

323

3

1

2

1

6

1


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1.4 Properties of Reinforced Concrete StructureUnit Weight

Concrete TypeNominal Strength

N/mm2 = MPa)Unit Weight kN/m3

Normal Concrete Fc36 24

Material Parameters

Youngs Modulus

N/mm2 = MPa)Poissons Ratio

Thermal

Expansion

Coefficient1/

Steel Bar 200 000 1/4 1 x 10-5

Concrete

22 000 ( Fc = 18 )

25 000 ( Fc = 24 )

28 000 ( Fc = 30 )

1/6 1 x 10-5


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2. SIMPLE EXAMPLE FOR FEM FORMULATIONStep.1: Description of the ProblemThe problem is to obtain the deformation of a simple supported beam under various load

conditions.

If you change the load condition, you will get the different deformation pattern. Actually,

there are infinite variations for the deformation pattern.

Step.2: Assumption of deformation functionWe assume a particular function for the deformation pattern to fix the variation, such

as the following function:

)sin()( xL

axv

(2-1)

Step.3: Relation between nodal displacement and element deformationFrom Equation (2-1), The displacement at the center node A is calculated as

aLv )5.0( (2-2)

The relation between nodal displacement and element deformation is then expressed as,

)sin()( xL

xv

(2-3)

A

x = 0 x = Lx

v

etc.


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Step.4: Stiffness equation at the nodeWe obtain the relation between the nodal force and the nodal displacement, for example,

by using the Principle of Virtual Work Method.

KP (2-4)

The process is summarized as follows:

Translate external forces into

equivalent nodal force, P.

Calculate nodal displacement, ,

from the stiffness equation,

PK1

Obtain the element deformation

from the nodal displacement.

)sin()( xL

xv

The above example tells the essence of the finite element analysis, which is:

Assume the deformation pattern to reduce the degree of freedom of the element, then,

obtain the deformation from the limited number of nodal displacements.

v

P

P

A

P


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3. TRIANGULAR ELEMENT FOR PLANE ANALYSISStep.1: Description of the ProblemThe problem is to obtain the deformation of a simple triangular element.

There are infinite variations for the deformation patterns.

Step.2: Assumption of deformation functionTo fix the variation for the deformation patterns, we assume a linear function for the

deformation pattern.

yxyxv

yxyxu

654

321

),(

),(

(3-1)

In a matrix form,

6

5

4

3

2

1

1000

0001

yx

yx

v

u(3-2)

Step.3: Relation between nodal displacement and element deformationThe displacements of the element nodes are expressed as,

etc.

x

y


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Node 1:

6

5

4

3

2

1

11

11

1

1

10000001

yxyx

vu (3-3)

Node 2:

6

5

4

3

2

1

22

22

2

2

1000

0001

yx

yx

v

u (3-4)

Node 3:

6

5

4

3

2

1

33

33

3

3

1000

0001

yx

yx

v

u (3-5)

It is summarized as,

6

5

4

3

2

1

33

22

11

33

22

11

3

2

1

3

2

1

1000

1000

1000

0001

0001

0001

yx

yx

yx

yx

yx

yx

v

v

v

u

u

u

(3-6)

U = A

xx1 x2 x3

y3

u1

v1u2

v2

u3

v3

1

2

3

y

y2

y1


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We can obtain the coefficients 61 , from the nodal displacements as,

=A -1 U (3-7)

Substituting Equation (3-7) into Equation (3-2), the relation between nodal

displacement and element deformation is,

3

2

1

3

2

1

1

1000

0001

v

v

v

u

u

u

Ayx

yx

v

u(3-8)

u(x,y) = H(x,y) U

Step.4: Stiffness equation at the nodeWe obtain the relation between the nodal force and the nodal displacement, for example,

by using the Principle of Virtual Work Method.

3

2

1

3

2

1

3

2

1

3

2

1

v

v

v

u

uu

K

Q

Q

Q

P

PP

(3-9)

F = K U


(1) Translate external forces into equivalent nodal force,

F = {P1, P2, P3, Q1, Q2, Q3}T(2) Calculate the nodal displacements from the stiffness equation,

U = K -1 F(3) Obtain the element deformation from the nodal displacement.

u(x,y) =H(x,y)U

P1

Q1P2

Q2

P3

Q3

1

2

3


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4. STIFFNESS MATRIX FOR TRIANGULAR ELEMENTStiffness matrix in Equation (3-9) can be obtained from the Principle of Virtual Work

Method, which is expressed in the following form:

V

TTFUdv (4-1)

where, is a virtual strain vector, is a stress vector, U is a virtual displacement

vector and F is a load vector, respectively.

In case of the plane problem, the strain vector is defined as,

x

v

y

u

y

vx

u

xy

y

x

(4-2)

Substituting Equation (3-8) into Equation (4-2), the strain vector is calculated from the

nodal displacement vector as,

3

2

1

3

2

1

1

010100

100000

000010

v

v

v

u

uu

A

x

v

y

u

y

vx

u

xy

y

x

(4-3)

= B U

In the plane stress problem, the stress-strain relationship is expressed as,

xy

y

x

xy

y

xE

2

100

01

01

1 2(4-4)

= D


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Substituting Equation (4-3) into Equation (4-4),

= D B U (4-5)

From the Principle of Virtual Work Method,

FUUDBdvBUdvDBUUB TV

TTT

V

(4-6)

Therefore, the stiffness equation is obtained as,

V

TDBdvBKKUF , (4-7)


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5. FROM ELEMENT STIFFNESS MATRIX TO GLOBAL STIFFNESS MATRIX

Element Stiffness Matrix:

Element (1)

4

2

1

4

2

1

)1(

66

)1(

65

)1(

64

)1(

63

)1(

62

)1(

61

)1(

56

)1(

55

)1(

54

)1(

53

)1(

52

)1(

51

)1(

46

)1(

45

)1(

44

)1(

43

)1(

42

)1(

41

)1(

36

)1(

35

)1(

34

)1(

33

)1(

32

)1(

31

)1(

26

)1(

25

)1(

24

)1(

23

)1(

22

)1(

21

)1(

16

)1(

15

)1(

14

)1(

13

)1(

12

)1(

11

4

2

1

4

2

1

v

v

v

u

u

u

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

Q

Q

Q

P

P

P

(5-1)

Element (2)

4

3

1

4

3

1

)2(

66

)2(

65

)2(

64

)2(

63

)2(

62

)2(

61

)2(

56

)2(

55

)2(

54

)2(

53

)2(

52

)2(

51

)2(

46

)2(

45

)2(

44

)2(

43

)2(

42

)2(

41

)2(

36

)2(

35

)2(

34

)2(

33

)2(

32

)2(

31

)2(

26

)2(

25

)2(

24

)2(

23

)2(

22

)2(

21

)2(

16

)2(

15

)2(

14

)2(

13

)2(

12

)2(

11

4

3

1

4

3

1

v

v

v

u

u

u

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

Q

Q

Q

P

P

P

(5-2)

Global Stiffness Matrix:

4

3

4

3

)2(

66

)1(

66

)2(

65

)2(

63

)1(

63

)2(

62

)2(

56

)2(

55

)2(

53

)2(

52

)2(

36

)2(

36

)2(

35

)2(

33

)1(

33

)2(

32

)2(

26

)2(

25

)2(

23

)2(

22

4

3

4

3

v

v

u

u

kkkkkk

kkkk

kkkkkk

kkkk

Q

Q

P

P

(5-3)

F = K U

P

(2)

(1)

4

3

4

3

4

3

2

1

4

3

2

1

v

v

u

u

ConditionBoundary

v

v

v

v

u

u

uu

fixed u1=v1=0

fixed u2=v2=0


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Set the load condition,

0

0

0

4

3

4

3

P

Q

Q

P

P

(5-4)

The displacement vector is then obtained by solving the stiffness equation,

0

0

0

1

4

3

4

3

PK

v

v

u

u

(5-5)


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6. HIGHER ORDER ELEMENTThe linear triangular element assumes the

deformation pattern to be a linear functionbetween two nodes.

It requires a large number of elements at the

place where deformation changes largely.

To reduce the number of elements, we

introduce the higher order elements, such as

the following second order elements where

the deformation pattern is assumed to be the

second order function of coordinate.

2

1211

2

10987

2

65

2

4321

yxyxyxv

yxyxyxu

(6-1)

In a matrix form,

12

2

1

22

22

1000000

0000001

yxyxyx

yxyxyx

v

u(6-2)

In order to define the second order function, we need

an additional node in the middle of each side of the

triangle. At the result, the total number of nodes in

one element is 6.

Beforedeformation

Afterdeformation

Before

After

Before

After

v3

2

6

3

4

5

1

u3

v2

u2


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The displacement of the element nodes are then expressed as,

12

8

7

6

2

1

2

666

2

666

2

222

2

222

2

111

2

111

2

666

2

666

2

222

2

222

2

111

2

111

6

2

1

6

2

1

1|

|0

1|

1|

|

|1

|

0|1

|1

yyxxyx

yyxxyx

yyxxyx

yyxxyx

yyxxyx

yyxxyx

v

v

v

u

u

u

(6-3)

u = A

From Equations (6-1) and (6-2), we obtain

6

2

1

6

2

1

1

22

22

1000000

0000001

v

vv

u

u

u

A

yxyxyx

yxyxyx

v

u

(6-4)

u(x,y) = H(x,y) U

As the same as the linear triangular element, the stiffness equation is obtained as

6

2

1

6

2

1

6

2

1

6

2

1

v

v

v

u

uu

K

Q

Q

Q

P

PP

(6-5)

F = K U


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(1) Translate external forces into equivalent nodal force,

F = {P1, , P6, Q1, , Q6}T(2) Calculate the nodal displacements from the stiffness equation,U = K -1 F(3) Obtain the element deformation from the nodal displacement.

u(x,y) =H(x,y)U


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7. INTERPOLATION FUNCTIONSuppose we have one dimensional element under loading. As discussed before, we

assume a linear function for the deformation pattern after loading,

xaaxu 10)(

or

1

01)(

a

axxu (7-1)

The next step is to obtain the coefficients, a0, a1, from the nodal displacements. Fromthe relations:

1101 xaau

2102 xaau

or

1

0

2

1

2

1

1

1

a

a

x

x

u

u(7-2)

U = A

The coefficients are obtained as, = A -1 U. Then, the relation between the deformationand the nodal displacements is,

2

111)(u

uAxxu (7-3)

Instead of the previous procedure, we introduce the interpolation functions to expressthe deformation directly from the nodal displacements:

2211 )()()( uxhuxhxu (7-4)

The interpolation functions, h1and h2, have the following characteristics:

1

1

1,0

,1)(

ux

uxxh ,

2

2

2,0

,1)(

ux

uxxh (7-5)

x1 x2

x1 2u1

u2

l


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From these characteristics, the functions are easily obtained as,

l

xxxh

21 )( ,

l

xxxh 12 )(

(7-6)

One of the advantages of using interpolation functions is to reduce the burden to

calculate the inverse matrix of A in Equation (7-3).

In the same manner, if we assume a second order function for the deformation pattern,

the deformation can be directly expressed using interpolation functions as follows:

332211 )()()()( uxhuxhuxhxu (7-7)

x1 x2

x1 2u1

u2

l

x1 x2x

u1

x1 x2

x

u2

x1 x2

x1 2u1

u2

l

x1 x2

xu1

x1 x2x

u2

x1 x2

x

u3

h1(x)u1

h2(x)u2

3

h3(x)u3

h1(x)u1

h2(x)u2

Second order interpolation functionFirst order interpolation function


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8. NATURAL COORDINATE1) Natural coordinate

When we measure the coordinate ofthe pencil, the result is different

depending on the scale we use. In

this example, the coordinate of the

head of the pencil is 5.0 in x-scale

and 9.5 in t-scale.

As long as we have one-to-one

relationship between two scales,we can translate the value in one

scale to the value in another scale

anytime.

The total weight of the pencil will be calculated in x-axis as,

5

0

)( dxxwW (8-1)

To translate it into t-axis, we use the following relationships:

Global relationship:

)7(2 tx (8-2)

Local relationship:

dtdx 2 (8-3)

Substituting Equations (8-2) and (8-3) into (8-1), the total weight is expressed as,

5.9

5.7

)(2 dttxwW (8-4)

x

t = 7 + 0.5 x

t

x = 2 ( t 7 )

1 2 3 4 5 60

x

8 9 107

t6

w x : distribution of wei ht

x

x

w x

x x+dx

dx

t

1

2

3

4

5

6 7 8 9 10

dt

2

x

0


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Next we consider a more complicated scale to measure the total weight of the pencil.

The relationships between x-axis and t-axis are:

Global relationship: )(txx (8-5)

Local relationship: dtdt

tdxdx

)( (8-6)

Where dx(t)/dt represents the first derivative of x(x) by the variable t, which correspond

to the slope of x(t) at t. Substituting Equations (8-5) and (8-6) into (8-1), the total weight

will be expressed in t-axis as,

dtdt

tdxtxwW

)()( (8-7)

Setting =-1, =1,

dt

tdxtxwtfdttfW

)()()(,)(

1

1

(8-9)

Such coordinate is called natural coordinate.

1 2 3 4 5 60

x

t

w x : distribution of wei ht

x

x

w x

x x+dx

dx

t

1

2

3

4

5

dt

0 t

x = x(t)

x


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2) Gaussian quadrature ruleIf the integration range is [-1, 1], the integration can be evaluated approximately by

n-point Gaussian quadrature rule which is generally expressed in the following form:

)()()()(

1

1

2211 nn tftftfdttf

(8-8)

where, n ,,, 21 are the weighting coefficients. This formula requires a limited

number of function values, )(,),(),( 21 ntftftf , at the sampling points, nttt ,,, 21 , to

evaluate the integration.

For example, the 3 point Gaussian quadrature rule is defined as:

)()()(

)7746.0(5556.0)0(8889.0)7746.0(5556.0)(

332211

1

1

tftftf

fffdttf

(16-1)

where, 5556.0,8889.0,5556.0 321

7746.0,0,7746.0 321 ttt

-1 -0.7746 0 +0.7746 +1

f(t)

f(-0.7746)f(0)

f(0.7746)

t


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9. ISOPARAMETRIC ELEMENTWe now introduce the natural coordinate for the example of one dimensional element.

If we assume the linear transfer function x(t) between x-axis and t-axis, x(t) will be

expressed as

2211 )()()( xthxthtx (9-1)

where

)1(2

1)(),1(

2

1)( 21 tthtth (9-2)

Actually, it satisfies the fact that

21 )1(,)1( xxxx (9-3)

The deformation of the element is also

expressed as,

2211 )()()( uthuthtu (9-4)

Therefore, the functions )(),( 21 thth are the

interpolation functions we introduced before.

The element where both the coordinate

transfer function x(t) and the deformation

function u(t) are expressed using the same

interpolation functions on the natural

coordinate is called Isoparametric element.

x1 x2

x1 2u1

u2

-1 +1

t-1 +1

t

x

x1

x2 x(t)

t

-1 +1

u1u2

u1

u2

-1 +1

-1 +1

h1(t)u1

h2(t)u2

t

t


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Advantages of using isplarametric elements are summarized below:

(1) The relation

n

i

iiuthtu

1

)()( does not require the calculation of inverse matrix.

(2) The relation

n

i

iixthtx

1

)()( enables to use the numerical integration method.

(3) Both functions u(t) and x(t) are expressed using the same interpolation functions.


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10. SYSTEMATIC FORMULATION OF INTERPOLATION FUNCTION(1) One dimensional element

2 Node

rh 12

11

1r 1r

1 rh 12

12

1r 1r1

)1(2

1

)1(2

1

2

1

rh

rh

3 Node)1(

2

1

)1(2

1

2

1

rh

rh

)1(2

1

)1(2

1

2

2

r

r

2

3 1 rh

As presented here, if you increase a node to define the second order function for the

deformation, the interpolation function changes in the following manners:

- Modify the existing interpolation functions, h1 and h2,

- Define a new interpolation function, h3.


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(2) Two dimensional element


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(3) Three dimensional element


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28

11. STIFFNESS MATRIX FOR ISOPARAMETRIC ELEMENTUsing a two dimensional isoparametric element, we will see the procedure to derive the

stiffness matrix.

The coordinate transfer function {x, y} is expressed using the interpolation functions as

follows:

4321

4

1

4321

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1),(),(

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1),(),(

ysrysrysrysrysrhsry

xsrxsrxsrxsrxsrhsrx

i

ii

i

ii

(11-1)

The deformation function {u, v} is also expressed using the same interpolation functions.

4321

4

1

4321

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1),(),(

)1)(1(4

1

)1)(1(4

1

)1)(1(4

1

)1)(1(4

1

),(),(

vsrvsrvsrvsrvsrhsrv

usrusrusrusrusrhsru

i

ii

iii

(11-2)

Stiffness matrix can be obtained from the Principle of Virtual Work Method, which is

expressed in the following form:

V

TT FUdv (11-3)

x, u

y, v

x4

y4

r

s

Node 1

Node 2

Node 3

Node 4


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29

where, is a virtual strain vector, is a stress vector, U is a virtual displacement

vector and F is a load vector, respectively.

In case of the plane problem, the strain vector is defined as,

x

v

y

u

y

vx

u

xy

y

x

(11-4)

Substituting Equation (11-2) into Equation (11-4), the strain vector is calculated from

the nodal displacement vector as,

4

4

3

3

2

2

1

1

44332211

4321

4321

4

1

4

1

4

1

4

1

0000

0000

v

u

v

u

v

u

v

u

x

h

y

h

x

h

y

h

x

h

y

h

x

h

y

h

y

h

y

h

y

h

y

hxh

xh

xh

xh

vx

hu

y

h

vy

h

ux

h

x

v

y

u

y

vx

u

i

i

i

i

i

i

i

i

i

i

i

i

xy

y

x

= B U (11-5)

In the plane stress problem, the stress-strain relationship is expressed as,

xy

y

x

xy

y

xE

2

100

01

01

1(11-6)

= D


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Substituting Equation (11-5) into Equation (11-6),

= D B U (11-7)

From the Principle of Virtual Work Method,

FUUDBdvBUdvDBUUB TV

TTT

V

(11-8)

Therefore, the stiffness equation is obtained as,

V

TDBdvBKKUF , (11-9)

If we assume the constant thickness of the plate (= t), using the relation tdxdydv ,

),( yxV

TDBdxdyBtK (11-10)

Since this integration is defined in x-y coordinate, we must transfer the coordinate into

r-s coordinate to use the numerical integration method. Introducing the Jacobianmatrix,

MatrixJacobian

s

y

s

xr

y

r

x

J ;

(11-11)

the above integration is expressed in r-s coordinate as,

1

1

1

1),(

),(,,,,,, drds

sr

yxsrysrxDBsrysrxBtK

T(11-12)

where

s

y

s

xr

y

r

x

Jsr

yx

det

),(

),((11-13)


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31

1) Evaluation of Jacobian Matrix

4

1

4

1

4

1

4

1

i

i

i

i

i

i

i i

i

i i

i

ys

hx

s

h

yr

hx

r

h

s

y

s

xr

y

r

x

J (11-14)

2) Evaluation of the matrix B

x

h

y

h

x

h

y

h

x

h

y

h

x

h

y

h

yh

yh

yh

yh

x

h

x

h

x

h

x

h

B

44332211

4321

4321

0000

0000

(11-15)

The derivativesy

h

y

h

x

h

x

h

4141 ,,,,, are calculated as,

y

s

s

h

y

r

r

h

y

h

y

s

s

h

y

r

r

h

y

h

x

s

s

h

x

r

r

h

x

h

x

s

s

h

x

r

r

h

x

h

444111

444111

,,

,,,

In a matrix form,

s

h

s

h

s

h

s

h

r

h

r

h

r

h

r

h

ys

yr

x

s

x

r

yh

yh

yh

yh

x

h

x

h

x

h

x

h

4321

4321

4321

4321

s

h

s

h

s

h

s

hr

h

r

h

r

h

r

h

J4321

4321

1 (11-16)


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32

3) Evaluation of partial derivatives of the interpolation functions

)1(4

1

)1(4

1

)1(4

1

)1(4

1

4

3

2

1

sr

h

sr

h

sr

h

sr

h

,

)1(4

1

)1(4

1

)1(4

1

)1(4

1

4

3

2

1

ss

h

rs

h

rs

h

rs

h

(11-17)

4) Numerical integrationUsing the 3 point Gaussian quadrature rule, the stiffness matrix is calculated

numerically as follows:

3

1

3

1

1

1

1

1

1

1

1

1

),(

),(

),(

),(,,,,,,

i j

jiji

T

srFt

drdssrFt

drdssr

yxsrysrxCBsrysrxBtK

(16-2)

where

),(

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

sr

yxsrysrxCBsrysrxBsrF

T

5556.0,8889.0,5556.0 321

7746.0,0,7746.0 332211 srsrsr

5) Assemble of finite element

To total stiffness matrix can be obtained to assemble the element stiffness matrix over

the areas of all finite elements.

m

mKK (11-20)

where mdenotes the m-th element.


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33

12. STRESS-STRAIN AT GAUSSIAN POINTS1) Stress and strain at Gaussian point

If you use the 3-points Gaussian Integration Method, there are nine Gaussian points

3,2,1,3,2,1, jisr ji in an element.

The stress and strain at the Gaussian point, ji sr, , is obtained from Equations (11-5)

and (11-7) as

UDBij

ijxy

y

x

ij

(12-1)

UBij

ijxy

y

x

ij

(12-2)

s = -1

s = +1

s = s2 = 0

r = -1

r = +1

r = r2 = 0

r = r3

s = s3

s = s1

: Gaussian points

x, u

y, v

x4

y4

r

s

Node 1

Node 2

Node 3

Node 4r = r1


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34

2) Principal stress at Gaussian point

2

2

122

xy

yxyx

(12-3)

2

2

222

xy

yxyx

(12-4)

yx

xy

p

2arctan

2

1(12-5)

x

y

xy 12

x

y

xy

1

2


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35

3) Displacement at Gaussian point

After obtaining the nodal displacement, the displacement at the Gaussian point

jisr, , is obtained from Equation (11-2) as

4

1

4

1

),(),(

),(),(

i

ijiiji

i

ijiiji

vsrhsrv

usrhsru

(12-6)

x

y

xy

1

2

x

y

xy

1

2


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36

13. MASS MATRIX FOR ISOPARAMETRIC ELEMENT1) Formulation

Under dynamic loading, the Principle of Virtual Work Method in dynamic problem =DAlemberts principle is expressed in the following form:

0 WQ (13-1)

V

TdvQ (13-2)

V

T dvt

u

t

uFuW

2

2

(13-3)

where F: body force, T: surface force, : density, : damping coefficient

Substitute following relationships into above equations:

u(x,y) =H(x,y)U=B U= D B U

UDBdvBUdvDBUUBQV

TTT

V

(13-4)

UHdvHUUHdvHUdvFHU

dvt

u

t

uFuW

V

TT

V

TT

V

TT

V

T

2

2

(13-5)

Therefore, from Equation (13-1),

V

TT

V

T

V

T

V

T dvFHUUCBdvBUHdvHUHdvH (13-5)

That is, the equilibrium equation is expressed as,

V

T

V

T

V

T

V

T dvFHRDBdvBKHdvHCHdvHM

RKUUCUM

,,,

(13-6)


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37

2) Evaluation of the matrix H and HTH

4321

4321

0000

0000

hhhh

hhhhH (13-7)

24

2

4

43

2

3

43

2

3

4232

2

2

4232

2

2

413121

2

1

413121

2

1

4321

4321

4

4

3

3

2

2

1

1

0.

0

00

00

000

000

0000

0000

0000

0

0

0

0

0

0

0

0

h

hsym

hhh

hhh

hhhhh

hhhhh

hhhhhhh

hhhhhhh

hhhh

hhhh

h

h

h

h

h

h

h

h

HHT

(13-8)

3) Numerical integration

The integration for mass matrix can be expressed in r-s coordinate as,

1

1

1

1

),(

det drdsJHHt

HdxdyHt

HdvHM

T

yxV

T

V

T

(13-9)

The integration can be evaluated by the Gaussian Integration Formula as,

ijij

T

ijij

i j

ijji

JHHG

GtM

det

3

1

3

1

(13-10)


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38

4) Lumped mass model

The mass matrix obtained from the density of material is called the consistent massmatrix using the same interpolation functions for stiffness matrix, mass matrix andload vectors. Instead of performing the integrations, we may evaluate an approximate

mass matrix by lumping equal parts of the total element mass to the nodal points which

is called the lumped mass matrix. An important advantage of using a lumped massmatrix is that the matrix is diagonal and the numerical operations for the solution of

the dynamic equations are reduced significantly.

Suppose that the consistent mass matrix is expressed as,

CnnCnCn

nCCC

nCCC

C

mmm

mmm

mmm

M

21

22221

11211

(13-11)

The lumped matrix can be evaluated from the consistent mass matrix from the

following formula:

Lnn

L

L

L

m

m

m

M

00

00

00

22

11

,

n

j

CijLii mm1

(13-12)


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39

14. EIGEN VALUE PROBLEMThe free vibration equilibrium equation without damping is

0 KUUM (14-1)

where Kis the stiffness matrix and Mis the lumped mass matrix in the form,

nm

m

m

M

00

00

00

2

1

(14-2)

The solution can be postulated to be in the form

tieU (14-3)

where is a vector of order n, is a frequency of vibration of the vector .

Then, the generalized eigenproblem is,

MK 2 (14-4)

This eigenproblem yields the neigensolutions nn ,,,,,, 2222121 where the

eigenvevtors are M-orthonormalized as,

ji

jimM

n

k

kjkikj

T

i;0

;1

1

,, (14-5)

22

2

2

10 n

(14-6)

The vector i is called the i-th mode shape vector, and i is the corresponding

frequency of vibration.

Defining a matrix whose columns are the eigenvectors and a diagonal matrix 2

which stores the eigenvalues on its diagonal as,


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40

n 21 ,

2

2

2

2

1

2

n

(14-7)

We introduce the following transformation on the displacement vector of the

equilibrium equation (13-6),

)()( tXtU (14-8)

Then,

RXKXCXM (14-9)

Multiplying T ,

RXKXCXM TTTT (14-10)

Using 2, KIM TT ,

RXXCX TT 2 (14-11)

A damping matrix that is diagonalized by is called a classical damping matrix.

nn

T

h

h

h

CC

2

2

2

22

11

(14-12)

wherei

h is the modal damping ratio of the i-th mode.

Then, Equation (14-11) reduce to n- equations of the form

)()()(2)(2

trtxtxhtxiiiii

(14-13)

where )()( tRtrT

ii

The initial conditions on X(t) are obtained from Equation (14-8) as,

0000 , tT

tt

T

t UMXMUX (14-14)


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41

15. CLASSICAL DAMPINGThree procedures for constructing a classical damping matrix are described as follow:

1) Proportional damping

Consider first mass-proportional damping and stiffness-proportional damping,

MaC 0 and KaC 1 (15-1)

where the constants 10 , aa have units of sec-1 and sec, respectively.

For a system with mass-proportional damping, the generalized damping for the i-th

mode is,

iimac 0 , iiii hmc 2/ (15-2)

Therefore,

iiha 20 ,

i

i

ah

1

2

0 (15-3)

Similarly, for a system with stiffness-proportional damping, the generalized damping

for the i-th mode is,

iiimac

2

1 , iiii hmc 2/ (15-4)

Therefore,

i

ih

a

21 , ii

ah

2

1 (15-5)

ih

KaC 1

ii

ah

2

1

MaC 0

i

i

ah

1

2

0


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42

2) Rayleigh damping

A Rayleigh damping matrix is proportional to the mass and stiffness matrices as,

KaMaC 10 (15-6)

The modal damping ratio for the i-th mode of such a system is,

i

i

i

aah

2

1

2

10 (15-7)

The coefficients 10 , aa can be determined from specified damping ratios 21, hh modes,

respectively. Expressing Equation (15-3) for these two modes in matrix form leads to:

2

1

1

0

22

11

/1

/1

2

1

h

h

a

a

(15-8)

Solving the above system, we obtain the coefficients 10 , aa :

2221

12211

2

2

2

1

1221210

2

2

hha

hha

(15-9)

ih

KaMaC 10

i

i

i

aah

2

1

2

10


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43

3) Modal damping

It is an alternative procedure to determinate a classical damping matrix from modal

damping ratios. From the definition of a classical damping matrix,

nn

T

h

h

h

CC

2

2

2

22

11

11 CC T (15-10)

Since ,IM

T

MT 1 , MT 1 (15-11)

Therefore,

MCMC T (15-12)


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44

16. EQUATION OF MOTION UNDER EARTHQUAKE GROUND MOTION

1) Equation of motion under earthquake ground motion

Earthquake ground motions are defined as two components acceleration; 0X and 0Y

, in X and Y

directions. The inertia forces at node i are defined as,

0

0

0

0

0

0

10

01

Y

XMIUM

Y

XM

v

uM

YvM

XuM

i

i

ii

ii

(16-1)

where

10

01,

00

00

00

,2

1

I

m

m

m

Mv

uU

n

i

i(16-2)

Equilibrium condition of the structure under earthquake ground motion is:

Finally the equation of motion is obtained as:

RY

XMIKUUCUM

0

0

(16-3)

Inertia forceDamping force

Restoring force

0

0

Y

X

MIUMKUUC


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45

2) Numerical integration by Newmark- method

The incremental formulation for the equation of motion of a structural system is,

iiii pdKvCaM (16-4)

where, M , C and K are the mass, damping and stiffness matrices. id , iv , ia and ip are the increments of the displacement, velocity, acceleration and external force

vectors, that is,

iii ddd 1 , iii vvv 1 ,

iii aaa 1 , iii ppp 1 (16-5)

Using the Newmark- method,

tatav iii 2

1(16-6)

222

1tatatvd iiii (16-7)

From Equation (16-7), we obtain

iiii av

td

ta

2

1112

(16-8)

Substituting Equation (16-7) into Equation (16-6) gives

tavdt

v iiii

4

11

2

1

2

1(16-9)

Equations (16-8) and (16-9) are substituted into Equation (16-4), and we obtain

tavCavt

Mp

KCt

Mt

d

iiiii

i

14

1

2

1

2

11

2

112

(16-10)

The equation can be rewritten as,

ii

pdK (16-11)

where,

Mt

Ct

KK2

1

2

1

(16-12)

tavCav

tMpp iiiiii 1

4

1

2

1

2

11

(16-13)


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46

17. INDEPENDENT FREEDOM1) Freedom Vector

Exercise 1)Please obtain the freedom vector of the following structure.

21

3 4

X

Y

200 mm

P = 500 N

1000 mm

P

P

50 mm

E = 22000 N/mm/mm, = 0.1666

Initial Restrained freedom = 1 Numbering

1X 0 1 0

1Y 0 1 0

2X 0 0 1

{ F } = 2Y 0 0 2

3X 0 1 0

3Y 0 1 0

4X 0 0 3

4Y 0 0 4

21 3X

Y

54 6


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47

2) Location Matrix

21

3 4

X

Y

200 mm

P = 500 N

1000 mm

P

P

50 mm

E = 22000 N/mm/mm, = 0.1666

1X 1 Y 2 X 2 Y 3 X 3 Y 4 X 4 Y

P1 K11 K12 K13 K14 K15 K16 K17 K18 U1

Q1 K21 K22 K23 K24 K25 K26 K27 K28 V1

P2 K31 K32 K33 K34 K35 K36 K37 K38 U2

Q2 = K41 K42 K43 K44 K45 K46 K47 K48 V2

P3 K51 K52 K53 K54 K55 K56 K57 K58 U3

Q3 K61 K62 K63 K64 K65 K66 K67 K68 V3

P4 K71 K72 K73 K74 K75 K76 K77 K78 U4

Q4 K81 K82 K83 K84 K85 K86 K87 K88 V4

Element stiffness matrix

Total stiffness matrix

U2 V2 U3 V3

P2 K33 K34 K35 K36 U2

Q2 = K43 K44 K45 K46 V2

P3 K53 K54 K55 K56 U3

Q3 K63 K64 K65 K66 V3

1 2

34

0

0

1

2

3

4

0

0

Locationmatrix

Element node number should bein anti-clockwise order.


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48

Exercise 2)Please obtain the location matrix of the element of the following structure.

Freedom vector

1 2

34

21

3 4

1X 0

1Y 0

2X 1

{ F } = 2Y 2

3X 0

3Y 0

4X 3

4Y 4

Node number

1 > 1

2 > 2

3 > 4

4 > 3

1X 0

1Y 0

2X 1

2Y 2

3X 3

3Y 4

4X 0

4Y 0

Location matrix

21 3X

Y

54 6


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49

Answers 1) and 2)

1) Freedom vector

2) Location matrix

21 3X

Y

4 6

5

2

1

3

4

Initial Restrained freedom = 1 Numbering

1X 0 1 01Y 0 1 0

2X 0 0 1

{ F } = 2Y 0 0 2

3X 0 1 0

3Y 0 1 0

4X 0 1 0

4Y 0 1 0

5X 0 0 35Y 0 0 4

6X 0 1 0

6Y 0 1 0

1 2

34

1 2

34

1X 0

1Y 0

2X 1

2Y 2

3X 3

3Y 4

4X 0

4Y 0

1X 1

1Y 2

2X 0

2Y 0

3X 0

3Y 0

4X 3

4Y 4


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50

Total stiffness matrix

88

7877

282722

18171211

66

5655

464544

36353433

.. ksym

kk

kkk

kkkk

ksym

kk

kkk

kkkk

K

Total mass matrix (in diagonal form)

88

77

22

11

66

55

44

33

.. msym

m

m

m

msym

m

m

m

M

Inertial force

10

01

10

01

,0

0 IY

XMIUM

1 3X

Y

4 6

2

5

2 1

34


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51

18. SKYLINE METHODUsually, the total stiffness matrix [ K ] is symmetric and sparse as shown below.

Therefore, to save memory size and to reduce calculation time for linear equation solver,the elements in the upper triangular part of the matrix under the Skyline (thick line)are stored in an one-dimension vector.

1 K11

2 K22

3 K214 K33

5 K23

6 K44

7 K43

8 0

9 K14

10 K55

11 K45

12 K35

13 K25

14 K66

15 K56

16 K77

17 K67

18 K57

19 K47

20 K37

21 K27

22 K17

23 K88

24 K78

25 K68

26 K58

27

0 1 1 3 3 1 6 3

Skyline height

1 2 4 6 10 14 16 23 27

Diagonal element order

Total stiffness matrix [ K ] Stiffness vectorSkyline

6

Band width ( = the maximum skyline height)


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Exercise 3)Using the same structure in Exercise 1), please compare

1) skyline height,2) diagonal element order and3) band width

between the following two cases with different node order.

Case 1

Case 2

21 4X

Y

3 5

76 98 10

31 7X

Y

5 9

42 86 10

Basic Theory of Finite Element Method.pdf

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