FINITE ELEMENT METHOD - Direktori File UPI

FINITE ELEMENT METHOD

Kartika yulianti

Jurusan Pendidikan Matematika

FPMIPA – UPI

Pendahuluan

Mencari sehingga u ( u ) 0 , : operator differensial Tujuan :

1 N

j j j=

u(x) u(x) =  a Alat :

Finite Element Method ??

Aproksimasi

crit £

u { u u M & u S }, S M

1 N

j j

j

u( x ) a

1

) (

j

j

aj

x u

crit

u £ u u M , M = u(x) x a,b

1

xj x_j x_{j 1}

j x

y

Basis:

) , ( ,

1

) , ( ,

0 )

(

1 1

1 1

j j j

j j

j x x x

h x x

x x x x

u

x y

a b

u₁ u₂ u₃

N N

j

j j

N a a a

S ₁,...,

1

N

N

u

a

u a

u a

2 2

1 1

Cara 1.

1

1 2

£

£ £

£

£ crit





  

 



N

j j j

b

x a

N

u crit{ u u M }

u x u x a x

u u

u F u, u, x dx

u L( a ) L( a ,a ,...,a ) L a

0 u

Review

u u

u

0

£

?

£

crit

1

crit

0

0

0

j

N

L( a ) L( a )

a

L( a ) a

L( a ) a







 

a

L 0

a

u





£ £ £

_j

0

j j j

u( a )

L( a ) u

u , u ,

a a a

   

 

Cara 2: Ritz-galerkin

1

1

0

0

0 1



N

j j j

N

j j k

j

( u )

( u ) a

a , , k ,...,N

Contoh

a Q a

a P a

L  _jk   _k 

.

) (

2_x

u x, u' 0 0 , u 1 1

N

j j j=1

u = a

Misal

2

2

2

1 2 1 2

x

1 0 x

N N

1

x j j j j

0 j=1 j=1

u + x = 0

£(u) = - u + xu dx

£(u) = - ∂ a + x a dx

Cara 1

1

0 1

0

dx x

Q

dx P

k k

k x j x jk

0 0

)

( a P

_jk

a Q

_k

L  

Cara 2

2 1

, 0

N

x j j k

j

a x

1 2 0 1

0

N

x j j k k

j

a x dx

1 1

1 0 1

0 0

N N

j x k k j x j x k k

j j

a a x dx

xu 0( ) 0 dan j( )1 0 untuk j, 0 1, ,...,N 1

1 N

j x j x k k

0 j 1

a x dx 0 , k = 1, 2, …, N-1 dan u

~ ( 1 ) 1

k

0

jk

a Q

P 

0 0.2 0.4 0.6 0.8 1 x

1

y

0

1 X 0 X 1

0

1 (1 )

)

~(X Xa X a

u

a h a x

xu

)1 (

)

~(

0 1

x h X X

x

) 1 (

1 ) (

1

) 0

1 ( )

~(X X a Xa

u

a h a x

xu

)1 (

)

~(

1 0

x h X X

x

) 1 (

1 ) (

V_-1 V₁

j h 0

(j+1)h 1 (j-1)h

-1

V₂ V_-2

x:

X:

V_-1= 1 + X V₁= 1 –X V_-2= -X V₂= X

h k X x

Basis :

) , ( ,

1

) , ( ,

0 ) (

1 1

1 1

j j j

j j

j x x x

h x x

x x x x

1

1

1

0 1

1

1

1 1

( ) ( )

( )

1 ( )

1 2

j

j

N x

j x j x k x x k

j x

j

j j j

a dx u x x dx

d X

du X dX

h dX dX

a a a

h





j h dX

j X X dX

j X X h

dx x x

j

j j

j j

2 1

1 2 1

0

) )(

1 ( )

)(

1 ( )

(

untuk

j = 1, 2, …, N-1

0 untuk 6 ,

1

1 ₂

1 a h j

h a^{j j}

2 1

1 2

1 ) 1 (

6

1 0 0 0

0 0

1 2

1 0

0 0

0 0 0 1 2

1

0 0 0 0

1 1

1

h h N

h h

a a

a a

h

N N

o

 







Matriks discretisasi:

Solusi eksak 6 7 6

) 1

(x x³ u

TERIMA KASIH