ELEMENTO TRIANGULAR DE 10 NODOS

Comenzando por los nodos de las esquinas, la función de forma se calcula de la siguiente manera:

N1:

N1=(L1−L1

( 4−9 )) (L1−L1(5−8 ) )(L1−L1

(2−3 ))(L1

( 1)−L1( 4−9) ) (L1

(1)−L1( 5−8) ) (L1

( 1)−L1(2−3 ))

¿(L1−

23 )(L1−

13 )(L1−0 )

(1−23 )(1−1

3 ) (1−0 ) ¿L1

19

(3 L1−2 ) (3 L1−1 )

29

N1=12L1 (3L1−2 ) ( 3L1−1 )

N2:

N2=(L2−L2

(5−6) ) (L2−L2(4−7 ) ) (L2−L2

(1−3 ))(L2

( 2)−L2(5−6 ) ) (L2

(2 )−L2(4−7) ) (L2

( 2)−L2(1−3 ))

¿(L2−

23 )(L2−

13 )(L2−0 )

(1−23 )(1−1

3 ) (1−0 ) ¿L2

19

(3 L2−2 ) (3 L2−1 )29

N2=12L2 (3 L2−2 ) (3L2−1 )

N3:

N3=(L3−L3

(7−8 )) (L3−L3(6−9 ) ) (L3−L3

(2−1 ))(L3

( 3)−L3( 7−8) ) (L3

( 3)−L3( 6−9) ) (L3

( 3)−L3( 2−1 ))

¿(L3−

23 )(L3−

13 ) (L3−0 )

(1−23 )(1−1

3 )(1−0 ) ¿L3

19

(3 L3−2 ) (3 L3−1 )

29

N3=12L3 (3 L3−2 ) (3 L3−1 )

Para los siguientes puntos se calcula de la siguiente manera:

N4 (L1 ) (L2 )

N4 (L1 )=(L1−L1

(5−8 ) )(L1−L1(2−3 ))

(L1(4 )−L1

(5−8 ) )(L1(4 )−L1

( 2−3 ) )

¿(L1−

13 ) (L1−0 )

( 23−1

3 )( 23−0)

¿L1(L1−

13 )

29

=L1

13

(3L1−1 )29

N4 (L1 )=32L1 (3 L1−1 )

N4 (L2 )=(L2−L2

( 1−3 ) )(L2

(4 )−L2(1−3 ))

¿

(L2−0 )

( 13−0)

=3 L2

N4 (L2 )=3L2

Luego reemplazando valores, tenemos:

N4 (L1 ) (L2 )=32L1 (3 L1−1 ) 3L2

N4 (L1 ) (L2 )=92L1 L2 (3 L1−1 )

N5 (L1 ) (L2 )

N5 (L1 )=(L1−L1

(2−3) )(L1

(5 )−L1(2−3 ))

¿

(L1−0 )

( 13−0)

=3 L1

N5 (L1 )=3 L1

N5 (L2 )=(L2−L2

( 4−7) ) (L2−L2(1−3) )

(L2(5 )−L2

( 4−7) ) (L2( 5)−L2

( 1−3 ) )

¿(L2−

13 ) (L2−0 )

( 23−1

3 )( 23−0)

¿L2(L2−

13 )

29

=L2

13

(3 L2−1 )29

N5 (L2 )=32L2 (3L2−1 )

Luego reemplazando valores, tenemos:

N5 (L1 ) (L2 )=92L1L2 (3 L2−1 )

N6 (L2 ) (L3 )

N6 (L2 )=(L2−L2

( 4−7 )) (L2−L2(1−3) )

(L2( 6)−L2

( 4−7 )) (L2(6 )−L2

(1−3) )

¿(L2−

13 ) (L2−0 )

( 23−1

3 )( 23−0)

¿L2(L2−

13 )

29

=L2

13

(3 L2−1 )29

N6 (L2 )=32L2 (3 L2−1 )

N6 (L3 )=(L3−L3

(1−2) )(L3

( 6)−L3( 1−2 ) )

¿

(L3−0 )

( 13−0)

=3 L3

N6 (L3 )=3 L3

Luego reemplazando valores, tenemos:

N6 (L2 ) (L3 )=92L2 L3 (3 L2−1 )

N7 (L2 ) ( L3 )

N7 (L2 )=(L2−L2

(1−3) )(L2

(7 )−L2(1−3 ) )

¿

(L2−0 )

( 13−0)

=3 L2

N7 (L2 )=3 L2

N7 (L3 )=(L3−L3

( 6−9) ) (L3−L3(1−3) )

(L3( 7)−L3

( 6−9) ) (L3( 7)−L3

( 1−3 ) )

¿(L3−

13 ) (L3−0 )

( 23−1

3 )( 23−0)

¿L3(L3−

13 )

29

=L3

13

(3 L3−1 )29

N7 (L3 )=32L3 (3 L3−1 )

Luego reemplazando valores, tenemos:

N7 (L2 ) ( L3 )=92L2 L3 (3 L3−1 )

N8 (L1 ) ( L3 )

N8 (L1 )=(L1−L1

(2−3) )(L1

( 8)−L1( 2−3 ) )

¿

(L1−0 )

( 13−0)

=3 L1

N8 (L1 )=3 L1

N8 (L3 )=(L3−L3

( 6−9) ) (L3−L3(1−2) )

(L3( 8)−L3

( 6−9) ) (L3( 8)−L3

( 1−2) )

¿(L3−

13 ) (L3−0 )

( 23−1

3 )( 23−0)

¿L3(L3−

13 )

29

=L3

13

(3 L3−1 )29

N8 (L3 )=32L3 (3 L3−1 )

Reemplazando valores, tenemos:

N8 (L1 ) ( L3 )=92L1L3 (3 L3−1 )

N9 (L1 ) ( L3 )

N9 (L1 )=(L1−L1

(5−8) ) (L1−L1( 2−3 ) )

(L1( 9)−L1

(5−8) ) (L1( 9)−L1

(2−3 ) )

¿(L1−

13 ) (L1−0 )

( 23−1

3 )( 23−0)

¿L1(L1−

13 )

29

=L1

13

(3L1−1 )29

N9 (L1 )=32L1 (3L1−1 )

N9 (L3 )=(L3−L3

(1−2) )(L3

( 5)−L3(1−2 ))

¿

(L3−0 )

( 13−0)

=3 L3

N9 (L3 )=3 L3

Reemplazando valores, tenemos:

N9 (L1 ) ( L3 )=92L1L3 (3 L1−1 )

N10 ( L1) (L2 ) (L3 )

N10 ( L1) (L2 ) (L3 )=(L1−L1

(2−3 )) (L2−L2(1−3) ) (L3−L3

(1−2) )(L1

( 10)−L1(2−3 )) (L2

(10 )−L2(1−3) ) (L3

( 10)−L3(1−2 ))

¿(L1−0 ) (L2−0 ) ( L3−0 )

( 13−0)( 1

3−0)( 1

3−0)

¿L1 L2L3

127

N10 ( L1) (L2 ) (L3 )=27 L1 L2L3

En resumen se tiene los siguientes resultados:

N1=12L1 (3L1−2 ) ( 3L1−1 )

N2=12L2 (3 L2−2 ) (3L2−1 )

N3=12L3 (3 L3−2 ) (3 L3−1 )

N4 (L1 ) (L2 )=92L1 L2 (3 L1−1 )

N5 (L1 ) (L2 )=92L1L2 (3 L2−1 )

N6 (L2 ) (L3 )=92L2 L3 (3 L2−1 )

N7 (L2 ) ( L3 )=92L2 L3 (3 L3−1 )

N8 (L1 ) ( L3 )=92L1L3 (3 L3−1 )

N9 (L1 ) ( L3 )=92L1L3 (3 L1−1 )

N10 ( L1) (L2 ) (L3 )=27 L1 L2L3