# Elementos finitos

Solo disponible en BuenasTareas
• Páginas : 11 (2712 palabras )
• Descarga(s) : 0
• Publicado : 25 de diciembre de 2011

Vista previa del texto
Capitulo 1

!

" %

#

\$

&

'

(

& ' & '

)

(

& ' & ' ) )

*

(

+

&\$&

,

&\$&

[ ke ] = [ Be ] [ D ][ Be ] dVe
Ve

T

-

[ D] = E
dVe = A( x)dx

.

[ ke ] =
L

E [ Be ] [ Be ] A( x)dx

T

(

[ Be ] =
Ne = .

dN e dx

/

# 2 #

0+ \$ &\$'1 # \$

+

&\$'

.
ui = 1 xi

Ni

i = 0,1, 2,3,........., n

!

\$

# #
N1u2 = 0 N2 ( x1 ,1)

+

&\$'

#

u1

u2 N1 ( x2 , 0) u1 = 1

-

N1

#
m= ( y2 − y1 ) ( x2 − x1 )

( y − yi ) = m( x − xi )

y = Ni y1 = u1 y2 = u 2

.
m= (u2 − u1 ) ( x2 − x1 )

( N i − ui ) = m( x − xi ) N i = m( x − xi ) + ui

m= (0 − 1) −1 = ( x2 − x1 ) ( x2 − x1 ) x2 − x1 = L −1 L

m=

N1 =

−1 ( x − x1 ) + 1 L ( x1 − x) +1 L

N1 =

( .
( x2 ,1)

N2
N2u2 = 1 u2 u1 = 0

!
( x1 , 0)

m=

(1 − 0) 1 = ( x2 − x1 ) ( x2 − x1 ) x2 − x1 = L 1 L

m=

N2 =

1 ( x − x2 ) + 1 L ( x − x2 ) +1 L

N2 =

(
N1 = ( x1 − x) +1 L ( x − x2 ) +1 L

N2 =

-

&

x1 = 0 x2 = L

N1 = 1 −

x L

N2 =

x L

3
x L x L

Ne = 1 −

( .
Ne
Be =

%

&\$&

!

Be

dN e dx

/
Ne = 1 − x L x L

d 1− Be = dx

x L

d

xL dx

Be =

−1 L

1 L

3
−1 L

[ Be ]

T

= 1 L

−1 L

[ Be ] [ Be ] =
1 L
T

T

−1 L

1 L

[ Be ] [ Be ] =
4

1 1 −1 L2 −1 1

[ ke ] =
L

E [ Be ] [ Be ] A( x)dx

T

/

[ ke ] = EA [ Be ] [ Be ] dx
L

T

[ ke ] = EA
L

1 1 −1 dx L2 −1 1

3

#

[ Be ] [ Be ] =

T

1 1 −1 L2 −1 1

[ ke ] =

EA 1 −1 L2 −1 1

dx
L

[ ke ] =

EA1 −1 L L2 −1 1

[ ke ] =

EA 1 −1 L −1 1

5

# ! 6 29 x10 psi \$ 0/
7

6
62 p lg 2

#

1

&

'

( (
x1 = 0 x2 = 5 ft = 60 p lg N1

& '
u1 = 1 u2 = 0

8
( x1 , u1 ) = (0,1) ( x2 , u2 ) = (60, 0)

N1

m=

(u2 − u1 ) ( x2 − x1 )

( N i − ui ) = m( x − xi ) N i = m( x − xi ) + ui

m=

(0 − 1) −1 = ( x2 − x1 ) (60 − 0) −1 60

m=

N1 =

−1 ( x − 0) + 1 60−x +1 60 x 60

N1 =

N1 = 1 −

(

N2

' &
u1 = 0 u2 = 1

8
( x1 , u1 ) = (0, 0) ( x2 , u2 ) = (60,1)

N2

m=

(1 − 0) 1 = ( x2 − x1 ) (60 − 0) x2 − x1 = 60 1 60

m=

N2 =

1 ( x − 60) + 1 60 ( x − 60) +1 60

N2 =

x 60 x N1 = 1 − 60 N2 =

N2 =

x 60

Ne = 1 −

x 60

x 60

(

Be

Ne

d 1− Be =

x 60 dx

d

x 60 dx

Be =

−1 60

1 60

−160

[ Be ]

T

= 1 60

−1 60

[ Be ] [ Be ] =
1 60
T

T

−1 60

1 60

[ Be ] [ Be ] =

1 −1 1 −1 1 1 = 2 (60) −1 1 3, 600 −1 1

4

[ ke ] =
L

E [ Be ] [ Be ] A( x)dx

T

/

[ ke ] = EA [ Be ] [ Be ] dx
L

T

[ ke ] = EA

1 −1 1 dx 3, 600 −1 1 60

3

#

[ Be ] [ Be ] =

T

1 −1 1 3, 600 −1 1

[ ke ] =

EA 1 −1 L2 −1 1

dx =
L

EA 1 −13, 600 −1 1

dx
60

[ ke ] =

EA 1 −1 EA 1 −1 60 L= 2 3, 600 −1 1 L −1 1

[ ke ] =
EA = (29 x106 )(62) = 1798 x106 lbs

EA 1 −1 60 −1 1

[ ke ] =

1798 x106 1 −1 −1 1 60

.

[ ke ] = (29,966, 666.67)
!" # \$ \$

1 −1

−1 1

.

[ ke ] =
5 # 2

kij = EA( x) [ Bi ]
L

T

B j dx

+

&\$)

(
L2

& # ! ! \$

'

L1

'

)

9

+

&\$*

+

&\$6

+&\$:

. %
N1

Bi

Bj

N kn
N
2 1

Nk
N1

'

&\$
N kn

. ! 9

N1

& &\$*

x1 = 0, x2 = L1 , x3 = L2 (0,1)( L1 , 0)( L2 , 0)

N1

1 N1

m=

(u2 − u1 ) ( x2 − x1 )

( N kn − un ) = m( x − xn ) N kn = m( x − xn ) + un

m= (0 − 1) −1 = ( x2 − x1 ) ( L1 − 0) −1 L1

m=

1 N1 =

−1 ( x − 0) + 1 L1 −x +1 L1

1 N1 =

N1

N12
m= (u3 − u2 ) ( x3 − x2 )( N kn − un ) = m( x − xn ) N kn = m( x − xn ) + un
m= (0 − 0) =0 ( x2 − x1 ) m=0 N12 = 0( x − x2 ) + 0 N12 = 0

N2

/

'
(0, 0)( L1 ,1)( L2 , 0)

N2

1 & N2

m=

(u2 − u1 ) ( x2 − x1 )

( N kn − un ) = m( x − xn ) N kn = m( x − xn ) + un

m= (1 − 0) 1 = ( x2 − x1 ) ( L1 − 0) 1 L1

m=

1 N2 =

1 ( x − L1 ) + 1 L1 ( x − L1 ) +1 L1 x L1

1 N2 =

1 N2 =

N2

'...