Post on 22-Apr-2015
ELEMENTO RECTANGULAR SIMPLE MEJORADONodo central (5 nodos)
figura 1.1
Una manera de mejorar el comportamiento de un elemento finito, consiste en agregar a suexpansión para desplazamientos, una función que se anula sobre el contorno del elemento en la figura 1.1
Adoptaremos la siguiente aproximación para los desplazamientos:
Teniendo nuestra matriz [N] de funciones de forma y considerando el estado plano de tensiones podemosobtener nuestra matriz [B]
B1
a b⋅( )
b y−( )−
0
a x−( )−
0
a x−( )−
b y−( )−
b y−( )
0
x−
0
x−
b y−( )
y
0
x
0
x
y
y−
0
a x−( )
0
a x−( )
y−
π
acos π
x
a⋅
⋅ sin πy
b⋅
⋅
0
π
bsin π
x
a⋅
⋅ cos πy
b⋅
⋅
0
π
bsin π
x
b⋅
⋅ cos πy
b⋅
⋅
π
acos π
x
a⋅
⋅ sin πy
b⋅
⋅
:=a
DE
1 v2−
1
v
0
v
1
0
0
0
1 v−( )
2
⋅:=E
K BT D⋅ B⋅:= B
Sabemos que la matriz K está definida:
Sin embargo el programa no fué capaz de desarrollarla debido a la cantidad de operaciones y la extension de ella ya quecomo sabemos es una matriz de 10x10. Ante esta dificultad nos propusimos desarrollarla por partes ya que el programa sisería capaz de hacerlo de esta forma.El siguiente diagrama representa las sub matrices que se desarrollaron y qué posicion ocupan dentro de la matriz K
A continuación se presentan los algoritmos que utilizamos para hallar cada sub matriz.
Ke1 p
Ke1i j,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 1 4..∈for
i 1 5..∈for
Ke1
:=
K
Ke1
E a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−−
E a2
v⋅ a2− 4 b
2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E a2
a2
v⋅− 2 b2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E b2
v⋅ a2+ b
2−( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E a2
v⋅ a2− 4 b
2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E a2
v⋅ a2− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E b2
v⋅ a2+ b
2−( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
→
Ke2 p
Ke2i j 4−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 5 8..∈for
i 1 5..∈for
Ke2
:=
K
Ke2
E a2
a2
v⋅− 2 b2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−
E a2
v⋅ a2− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E b2
v⋅ 4 a2⋅+ b
2−( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E a2
v⋅ a2− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E a2
a2
v⋅− 2 b2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E a2
v⋅ a2− 4 b
2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E b2
v⋅ 4 a2⋅+ b
2−( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
→
Ke3 p
Ke3i j 8−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 9 10..∈for
i 1 1..∈for
Ke3
:=
K
Ke3 0
2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−
→
Ke4 p
Ke4i 1− j 8−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 9 10..∈for
i 2 2..∈for
Ke4
:=
K
Ke42 E⋅
π2
a⋅ b⋅ v 1−( )⋅0
→
Ke5 p
Ke5i 2− j 8−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 9 10..∈for
i 3 3..∈for
Ke5
:=
K
Ke5 0
4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−
→
Ke6 p
Ke6i 3− j 8−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 9 10..∈for
i 4 5..∈for
Ke6
:=
K
Ke6
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅−
0
0
2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−
→
Ke7 p
Ke7i 5− j,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 1 4..∈for
i 6 8..∈for
Ke7
:=
K
Ke7
E
8 v⋅ 8−
E a2
v⋅ a2− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E b2
v⋅ 4 a2⋅+ b
2−( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E a2
a2
v⋅− 2 b2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E b2
v⋅ 4 a2⋅+ b
2−( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E
8 v⋅ 8−−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
→
Ke8 p
Ke8i 5− j 4−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 5 8..∈for
i 6 8..∈for
Ke8
:=
K
Ke8
E
8 v⋅ 8−−
E a2
v⋅ a2− 4 b
2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E b2
v⋅ a2+ b
2−( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E 3 v⋅ 1−( )⋅
8 v2⋅ 8−
E a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E b2
v⋅ a2+ b
2−( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
E
8 v⋅ 8−
E 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−
→
Ke9 p
Ke9i 5− j 8−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 9 10..∈for
i 6 8..∈for
Ke9
:=
K
Ke9
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅
0
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅−
0
4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−
0
→
Ke10 p
Ke10i 8− j,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 1 4..∈for
i 9 10..∈for
Ke10
:=
K
Ke10
0
2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅
0
0
4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅−
0
→
Ke11 p
Ke11i 8− j 4−,
0
b
y
0
a
xKi j,
⌠⌡
d⌠⌡
d←
j 5 10..∈for
i 9 10..∈for
Ke11
:=
K
Ke11T
0
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅
0
2 E⋅
π2
a⋅ b⋅ v 1−( )⋅−
π2
E⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
8 a3⋅ b
3⋅ v2
1−( )⋅−
0
2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−
0
4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−
0
0
2 π2⋅ E⋅ a
2⋅ π2
E⋅ b2⋅+ π
2E⋅ b
2⋅ v⋅− π E⋅ a⋅ b⋅ sin2 π⋅ a⋅
b
⋅−
8 a3⋅ b
3⋅ 8 a3⋅ b
3⋅ v2⋅−
→
Ke
wi j, Ke1
i j, ←
i 1 5..∈for
j 1 4..∈for
wi j Ke7i 5− j, ←,
i 6 8..∈for
j 1 4..∈for
wi j, Ke10
i 8− j, ←
i 9 10..∈for
j 1 4..∈for
wi j, Ke2
i j 4−, ←
i 1 5..∈for
j 5 8..∈for
wi j, Ke8
i 5− j 4−, ←
i 6 8..∈for
j 5 8..∈for
wi j, Ke11
i 8− j 4−, ←
i 9 10..∈for
j 5 8..∈for
wi j, Ke3
i j 8−, ←
i 1 1..∈for
j 9 10..∈for
wi j, Ke4
i 1− j 8−, ←
i 2 2..∈for
j 9 10..∈for
wi j, Ke5
i 2− j 8−, ←
i 3 3..∈for
j 9 10..∈for
wi j, Ke6
i 3− j 8−, ←
i 4 5..∈for
j 9 10..∈for
wi j, Ke9
i 5− j 8−, ←
i 6 8..∈for
j 9 10..∈for
wi j, Ke11
i 8− j 8−, ←
i 9 10..∈for
j 9 10..∈for
w t⋅
:=
Ke1
Habiendo ya calculado todas las sub matricesprogramaremos una sub rutina paraensamblarlas en una matriz que llamaremos Ke(matriz de rigidez sin condensar)
A continuación mostraremos algunos elementos de la matriz Ke para demostrar que es correcto el procedimientohecho
Ke1 1,
E t⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke2 2,
E t⋅ 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke3 3,
E t⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke4 4,
E t⋅ 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke5 5,
E t⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke7 7,
E t⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke8 8,
E t⋅ 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Ke9 9, 0→
Ke10 10, 0→
Ahora procederemos a condensar la matriz a fin de obtener la matriz Keq de 8X8. Para alcanzar dicho objetivodesarrollaremos una rutina que nos permita obtener la matriz condensada.
Antes, definimos nuestro vector de cargas nodales
P NT Px
Py
⋅:=Px
P
Pxx1
a1−
⋅y1
b1−
⋅
Pyx1
a1−
⋅y1
b1−
⋅
Pxx1
a1−
⋅y1
b1−
⋅
Pyx1
a1−
⋅y1
b1−
⋅
Px x1⋅ y1⋅a b⋅
Py x1⋅ y1⋅a b⋅
Px y1⋅x1
a1−
⋅
b−
Py y1⋅x1
a1−
⋅
b−
Px sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅
Py sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅
→
Donde Px y Py son las componentes de una carga aplicada en un punto del elemento x1,y1
Kee r 0←
ri j, Ke
i j, ←
j 1 8..∈for
i 1 8..∈for
r
:=
Ke
Kei r 0←
ri j, Ke
i j 8+, ←
j 1 2..∈for
i 1 8..∈for
r
:=
Ke
Pe r 0←
ri
Pi
←
i 1 8..∈for
r
:=
P
Kie r 0←
ri j, Ke
i 8+ j, ←
j 1 8..∈for
i 1 2..∈for
r
:=
Ke
Kii r 0←
ri j, Ke
i 8+ j 8+, ←
j 1 2..∈for
i 1 2..∈for
r
:=
Ke
Pi r 0←
ri
Pi 8+←
i 1 2..∈for
r
:=
P
Keq Kee Kei Kii1−⋅ Kie⋅−:= Kee
Peq Pe Kei Kii1−⋅ Pi⋅−:= Pe
Finalmente mediante condensacion estática obtenemos la matriz Keq y Peq.No obstante como ya hemosmencionado el programa mathcad no nos permite visualizar la matriz Keq y Peq debido a la extension de ellas alusar notación algebraica, por este motivo sólo mostraremos algunos elementos de dichas matrices
Keq1 1,
E t⋅ a2
a2
v⋅− 2 b2⋅+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→ Keq
4 6, E t⋅ b
2v⋅ 4 a
2⋅+ b2−( )⋅
12 a⋅ b⋅ v2
1−( )⋅→ Keq
3 1, E t⋅ a
2v⋅ a
2− 4 b2⋅+( )⋅
12 a⋅ b⋅ v2
1−( )⋅→
Keq2 1,
2 E⋅ t2⋅ 2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cos
π a⋅2 b⋅
2
1−
⋅+
⋅
π2
a⋅ b⋅ v 1−( )⋅ 2 E⋅ a⋅ t⋅ 4 E⋅ b⋅ t⋅ v⋅ cosπ a⋅2 b⋅
2
⋅− 2 E⋅ a⋅ t⋅ v⋅− 4 E⋅ b⋅ t⋅ v⋅+
⋅
E t⋅8 v⋅ 8−
−→
Keq7 8,
E t⋅8 v⋅ 8−
t π2
a⋅ b⋅ π2
a⋅ b⋅ v⋅−( )⋅ 4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
⋅
π2
a⋅ b⋅ π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−( )⋅ v 1−( )⋅
−→
Keq6 8,
E t⋅ b2
v⋅ a2+ b
2−( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Keq8 8,
E t⋅ 2 a2⋅ b
2v⋅− b
2+( )⋅
6 a⋅ b⋅ v2
1−( )⋅−→
Peq1 1, Px
x1
a1−
⋅y1
b1−
⋅Px sin
π x1⋅a
⋅ sinπ y1⋅
b
⋅ π2
a⋅ b⋅ π2
a⋅ b⋅ v⋅−( )⋅ 2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
⋅
2 E⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅+→
Peq2 1, Py
x1
a1−
⋅y1
b1−
⋅2 E⋅ Py⋅ t⋅ sin
π x1⋅a
⋅ sinπ y1⋅
b
⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅
π2
a⋅ b⋅ v 1−( )⋅ 2 E⋅ a⋅ t⋅ 4 E⋅ b⋅ t⋅ v⋅ cosπ a⋅2 b⋅
2
⋅− 2 E⋅ a⋅ t⋅ v⋅− 4 E⋅ b⋅ t⋅ v⋅+
⋅
+→
Peq3 1, Px
x1
a1−
⋅y1
b1−
⋅Px sin
π x1⋅a
⋅ sinπ y1⋅
b
⋅ π2
a⋅ b⋅ π2
a⋅ b⋅ v⋅−( )⋅ 4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
⋅
2 E⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅+→
Peq4 1, Py
x1
a1−
⋅y1
b1−
⋅2 E⋅ Py⋅ t⋅ sin
π x1⋅a
⋅ sinπ y1⋅
b
⋅ π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−( )⋅
π2
a⋅ b⋅ v 1−( )⋅ 2 E⋅ a⋅ t⋅ 4 E⋅ b⋅ t⋅ v⋅ cosπ a⋅2 b⋅
2
⋅− 2 E⋅ a⋅ t⋅ v⋅− 4 E⋅ b⋅ t⋅ v⋅+
⋅
−→
Peq5 1,
Px x1⋅ y1⋅a b⋅
Px sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅ π2
a⋅ b⋅ π2
a⋅ b⋅ v⋅−( )⋅ 2 E⋅ a⋅ v⋅ 2 E⋅ a⋅− 4 E⋅ b⋅ v⋅ cosπ a⋅2 b⋅
2
1−
⋅+
⋅
2 E⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅+→
Peq6 1,
Py x1⋅ y1⋅a b⋅
2 E⋅ Py⋅ t⋅ sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅
π2
a⋅ b⋅ v 1−( )⋅ 2 E⋅ a⋅ t⋅ 4 E⋅ b⋅ t⋅ v⋅ cosπ a⋅2 b⋅
2
⋅− 2 E⋅ a⋅ t⋅ v⋅− 4 E⋅ b⋅ t⋅ v⋅+
⋅
+→
Peq7 1,
Px sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅ π2
a⋅ b⋅ π2
a⋅ b⋅ v⋅−( )⋅ 4 E⋅ b⋅ v⋅ sinπ a⋅2 b⋅
2
⋅ 2 E⋅ a⋅+ 2 E⋅ a⋅ v⋅−
⋅
2 E⋅ π2
a2⋅ b⋅ π
2a2⋅ b⋅ v
2⋅−( )⋅
Px y1⋅x1
a1−
⋅
b−→
Peq8 1,
Py y1⋅x1
a1−
⋅
b−
2 E⋅ Py⋅ t⋅ sinπ x1⋅
a
⋅ sinπ y1⋅
b
⋅ π2
a2⋅ b⋅ π
2a
2⋅ b⋅ v2⋅−( )⋅
π2
a⋅ b⋅ v 1−( )⋅ 2 E⋅ a⋅ t⋅ 4 E⋅ b⋅ t⋅ v⋅ cosπ a⋅2 b⋅
2
⋅− 2 E⋅ a⋅ t⋅ v⋅− 4 E⋅ b⋅ t⋅ v⋅+
⋅
−→
N
1x1
a−
1y1
b−
0
0
1x1
a−
1y1
b−
1x1
a−
1y1
b−
0
0
1x1
a−
1y1
b−
x1 y1⋅a b⋅
0
0
x1 y1⋅a b⋅
y1
b1
x1
a−
0
0
y1
b1
x1
a−
:=
x1
sin πx1
a⋅
sin πy1
b⋅
⋅
0
0
sin πx1
a⋅
sin πy1
b⋅
⋅