elemento triangular de 10 nodos.docx
TRANSCRIPT
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