Elementos finitos CIV - 313

Tema 5: Elementos Bidimensionales

Formulas a emplear:
1. Modelos Constitutivos
Tensión plana σz ＝ 0 Deformación plana ξz ＝ 0
⎡ μ ⎤
⎢ 1 ―― 0 ⎥
⎢ 1-μ ⎥
⎡1 μ 0 ⎤
E ⎢ E ⋅ ((1 - μ)) ⎢ μ ⎥
D ＝ ―――μ 1 0 ⎥ D ＝ ―――――――― 1 0
1 - μ2 ⎢ ⎥ ((1 + μ)) ⋅ ((1 - 2 μ)) ⎢ 1 - μ ⎥
1-μ ⎥ ⎢
⎢ 0 0 ―― ((1 - 2 μ)) ⎥
⎢⎣ 2 ⎥⎦ ⎢ 0 0 ―――― ⎥
⎢⎣ 2 ⋅ ((1 - μ)) ⎥⎦
2. Jacobiano J
⎡x y ⎤
J ＝ ⎢ 13 13 ⎥ ||J|| ＝ 2 ⋅ A
⎣ x23 y23 ⎦
3. Matriz de deformación [[ B ]]
⎡ y23 0 y31 0 y12 0 ⎤
1 ⎢
[[ B ]] ＝ ― 0 x32 0 x13 0 x21 ⎥
||J|| ⎢ ⎥
⎢⎣ x32 y23 x13 y31 x21 y12 ⎥⎦
4. Matriz de rigidez
K＝BT ⋅ D ⋅ B ⋅ t ⋅ A
5. Ecuación de equilibrio
[[ K ]] ⋅ [[ a ]] + [[ b ]] + [[ fe ＝ t ]] + ⎡⎣ fi ＝ qi ⎤⎦ ＝ 0 [[ b ]] ＝ Peso_propio
[[ fe ]] ＝ Carga_distribuido
[[ fi ]] ＝ Cargas_nodales

Peso propio: se distribuye de manera equitativa a todos los nudos

Carga distribuida:
6. Tensor de deformaciones unitarias ξ＝B ⋅ a
7. Tensor de esfuerzos σ＝D ⋅ ξ

Ejemplo n1: Determinar ξ y σ E ≔ 2000000

t ≔ 0.15 μ ≔ 0.25
Element N1 N2 N3

1 1 2 3
2 1 3 4

Si existe restricción (Rx, Ry) 1, si no existe 0.

N Cx Cy Fix Fiy Rx Ry

1 0 0 0 0 1 1
2 1.5 0 0 0 0 0
3 0.6 2 0 0 0 0
4 0 2 0 0 1 1

Auxiliar: Univ.: Ramos Rodríguez Juan Pablo

Elementos finitos CIV - 313

0. Cargas equivalentes
Fe ≔ [[ 0 0 0 -0.2742 0 -0.2193 0 0 ]]
Fb ≔ [[ 0 0 0 0 0 0 0 0 ]]
1. Tensión plana (TP) o Deformación plana TP ≔ “si”
⎡ 2.133 ⋅ 10 6 5.333 ⋅ 10 5 0 ⎤
⎢ ⎥
D = ⎢ 5.333 ⋅ 10 5 2.133 ⋅ 10 6 0 ⎥
⎢⎣ 0 0 8 ⋅ 10 5 ⎥⎦
Elemento No ≔ 1 Nudos ((No)) = [[ 1 2 3 ]]

⎡0 0⎤
⎡ x1 - x3 y1 - y3 ⎤ ⎡ -0.6 -2 ⎤ ||J ((No))||
Coord ((No)) = ⎢ 1.5 0 ⎥ J＝⎢ J ((No)) = ⎢ A ≔ ―――
⎢ ⎥ ⎣ x2 - x3 y2 - y3 ⎥⎦ ⎣ 0.9 -2 ⎥⎦ 2
⎣ 0.6 2 ⎦

⎡ y2 - y3 0 y3 - y1 0 y1 - y2 0 ⎤ ⎡ -0.66667 0 0.66667 0 0 0 ⎤
1
B ＝ ――⋅ ⎢ 0 x3 - x2 0 x1 - x3 0 x2 - x1 ⎥ B ((No)) = ⎢ 0 -0.3 0 -0.2 0 0.5 ⎥
2 ⋅ A ⎢ x3 - x2 y2 - y3 x1 - x3 y3 - y1 x2 - x1 y1 - y2 ⎥ ⎢ ⎥
⎣ ⎦ ⎣ -0.3 -0.66667 -0.2 0.66667 0.5 0 ⎦

⎡ 229533.3333 60000 -202533.3333 -20000 -27000 -40000 ⎤ ⎡1⎤

⎢ 60000 123200 0 -51200 -60000 -72000 ⎥ ⎢2⎥
⎢ ⎥ ⎢ ⎥
-202533.3333 0 220533.3333 -40000 -18000 40000 ⎥ 3
k＝BT ⋅ D ⋅ B ⋅ t ⋅ A k ((No)) = ⎢ GDL ((No)) = ⎢ ⎥
⎢ -20000 -51200 -40000 99200 60000 -48000 ⎥ ⎢4⎥
⎢ -27000 -60000 -18000 60000 45000 0⎥ ⎢5⎥
⎢⎣ -40000 -72000 40000 -48000 0 120000 ⎥⎦ ⎢⎣ 6 ⎥⎦

⎡ 229533.333 60000 -202533.333 -20000 -27000 -40000 0 0 ⎤

⎢ 60000 123200 0 -51200 -60000 -72000 0 0 ⎥
⎢ ⎥
⎢ -202533.333 0 220533.333 -40000 -18000 40000 0 0 ⎥
T
-20000 -51200 -40000 99200 60000 -48000 0 0 ⎥
k1 ≔ ( ) ( ) ⋅ ( ) = ⎢
P (No) ⋅ k (No) P (No)
⎢ -27000 -60000 -18000 60000 45000 0 0 0⎥
⎢ -40000 -72000 40000 -48000 0 120000 0 0 ⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 0 0⎥
⎣ 0 0 0 0 0 0 0 0⎦

Elemento No ≔ 2 Nudos ((No)) = [[ 1 3 4 ]]

⎡0 0⎤
⎡ x1 - x3 y1 - y3 ⎤ ⎡0 -2 ⎤ ||J ((No))||
Coord (No) = ⎢ 0.6 2 ⎥
( ) J＝⎢ J ((No)) = ⎢ A ≔ ―――
⎢ ⎥ ⎣ x2 - x3 y2 - y3 ⎥⎦ ⎣ 0.6 0 ⎥⎦ 2
⎣0 2⎦

⎡ y2 - y3 0 y3 - y1 0 y1 - y2 0 ⎤ ⎡ 0 0 1.66667 0 -1.66667 0 ⎤
1
B ＝ ――⋅ ⎢ 0 x3 - x2 0 x1 - x3 0 x2 - x1 ⎥ B ((No)) = ⎢ 0 -0.5 0 0 0 0.5 ⎥
2 ⋅ A ⎢ x3 - x2 y2 - y3 x1 - x3 y3 - y1 x2 - x1 y1 - y2 ⎥ ⎢ ⎥
⎣ ⎦ ⎣ -0.5 0 0 1.66667 0.5 -1.66667 ⎦

⎡ 18000 0 0 -60000 -18000 60000 ⎤ ⎡1⎤

⎢ 0 48000 -40000 0 40000 -48000 ⎥ ⎢2⎥
⎢ ⎥ ⎢ ⎥
0 -40000 533333.3333 0 -533333.3333 40000 ⎥ 5
k＝BT ⋅ D ⋅ B ⋅ t ⋅ A k ((No)) = ⎢ GDL ((No)) = ⎢ ⎥
⎢ -60000 0 0 200000 60000 -200000 ⎥ ⎢6⎥
⎢ -18000 40000 -533333.3333 60000 551333.3333 -100000 ⎥ ⎢7⎥
⎢⎣ 60000 -48000 40000 -200000 -100000 248000 ⎥⎦ ⎢⎣ 8 ⎥⎦

⎡ 18000 0 0 0 0 -60000 -18000 60000 ⎤

⎢ 0 48000 0 0 -40000 0 40000 -48000 ⎥
⎢ ⎥
⎢ 0 0 0 0 0 0 0 0⎥
T
0 0 0 0 0 0 0 0⎥
k2 ≔ P ((No)) ⋅ k ((No)) ⋅ P ((No)) = ⎢
⎢ 0 -40000 0 0 533333.333 0 -533333.333 40000 ⎥
⎢ -60000 0 0 0 0 200000 60000 -200000 ⎥
⎢ ⎥
⎢ -18000 40000 0 0 -533333.333 60000 551333.333 -100000 ⎥
⎣ 60000 -48000 0 0 40000 -200000 -100000 248000 ⎦

Auxiliar: Univ.: Ramos Rodríguez Juan Pablo

Elementos finitos CIV - 313

2. Matriz ensamblada e inversa

⎡ 247533.333 60000 -202533.333 -20000 -27000 -100000 -18000 60000 ⎤ ⎡ 0 ⎤

⎢ 60000 171200 0 -51200 -100000 -72000 40000 -48000 ⎥ ⎢ 0 ⎥
⎢ ⎥ ⎢ ⎥
⎢ -202533.333 0 220533.333 -40000 -18000 40000 0 0⎥ ⎢ 0 ⎥
-20000 -51200 -40000 99200 60000 -48000 0 0⎥ -0.2742 ⎥
Ke = ⎢ F=⎢
⎢ -27000 -100000 -18000 60000 578333.333 0 -533333.333 40000 ⎥ ⎢ 0 ⎥
⎢ -100000 -72000 40000 -48000 0 320000 60000 -200000 ⎥ ⎢ -0.2193 ⎥
⎢ ⎥ ⎢ ⎥
⎢ -18000 40000 0 0 -533333.333 60000 551333.333 -100000 ⎥ ⎢ 0 ⎥
⎣ 60000 -48000 0 0 40000 -200000 -100000 248000 ⎦ ⎣ 0 ⎦

⎡1 0 0 0 0 0 0 0⎤ ⎡ 0 ⎤
⎢0 1 0 0 0 0 0 0⎥ ⎢ 0 ⎥
⎢ ⎥ ⎢ -7 ⎥
⎢0 0 4.9276666 ⋅ 10 -6 1.8459711 ⋅ 10 -6 -3.814455 ⋅ 10 -8 -3.3906267 ⋅ 10 -7 0 0⎥ ⎢ -4.3180882 ⋅ 10 ⎥
⎢0 0 1.8459711 ⋅ 10 -6 1.2349905 ⋅ 10 -5 -1.2238043 ⋅ 10 -6 1.6217394 ⋅ 10 -6 0 0⎥ ⎢ -3.7419914 ⋅ 10 -6 ⎥
Ks = ⎢ ⎥ ∆=⎢ -7 ⎥
⎢0 0 -3.814455 ⋅ 10 -8 -1.2238043 ⋅ 10 -6 1.8548847 ⋅ 10 -6 -1.7880258 ⋅ 10 -7 0 0⎥ ⎢ 3.7477855 ⋅ 10 ⎥
⎢0 0 -3.3906267 ⋅ 10 -7 1.6217394 ⋅ 10 -6 -1.7880258 ⋅ 10 -7 3.4106437 ⋅ 10 -6 0 0⎥ -6
⎢ -1.1926351 ⋅ 10 ⎥
⎢0 0 0 0 0 0 1 0⎥ ⎢ 0 ⎥
⎢0 0 0 0 0 0 0 1 ⎥⎦ ⎢ 0 ⎥
⎣ ⎣ ⎦

3. Tensor de deformación unitaria y esfuerzos

Elemento No ≔ 1 ⎡ 0 ⎤
⎢ 0 ⎥
⎡ -0.6667 ⎢ ⎥ ⎡ -2.8787 ⋅ 10 -7 ⎤
0 0.6667 0 0 0 ⎤ -7
T
⎢ -4.3181 ⋅ 10 ⎥
T ⎢ ⎥
B ((No)) = ⎢ 0 -0.3 0 -0.2 0 0.5 ⎥ P ((No)) ⋅ ∆ = ξ1 ≔ B ((No)) ⋅ P ((No)) ⋅ ∆ = ⎢ 1.5208 ⋅ 10 -7 ⎥
⎢ ⎥ ⎢ -3.742 ⋅ 10 ⎥-6
⎣ -0.3 -0.6667 -0.2 0.6667 0.5 0 ⎦ ⎢ -7 ⎥
-6
⎢⎣ -2.2209 ⋅ 10 ⎥⎦
⎢ 3.7478 ⋅ 10 ⎥
-6
⎣⎢ -1.1926 ⋅ 10 ⎦⎥
⎡ 2133333.3333 533333.3333 0⎤ ⎡ -0.533 ⎤
⎢
D = 533333.3333 2133333.3333 0⎥ σ1 ≔ D ⋅ ξ1 = ⎢ 0.1709 ⎥
⎢ ⎥ ⎢ ⎥
⎣ 0 0 800000 ⎦ ⎣ -1.7767 ⎦

Elemento No ≔ 2 ⎡ 0 ⎤
⎢ 0 ⎥
⎡ 0 0 1.6667 0 -1.6667 0 ⎤ ⎢ ⎥ ⎡ 6.2463 ⋅ 10 -7 ⎤
-7 T
T 3.7478 ⋅ 10 ⎢ ⎥
B ((No)) = ⎢ 0 -0.5 0 0 0 0.5 ⎥ P ((No)) ⋅ ∆ = ⎢ ⎥ ( ) ( )
ξ1 ≔ B (No) ⋅ P (No) ⋅ ∆ = ⎢ 0
⎢ ⎥ ⎢ -1.1926 ⋅ 10 ⎥-6 ⎥
⎣ -0.5 0 0 1.6667 0.5 -1.6667 ⎦ ⎢ 0 ⎥
-6
⎣ -1.9877 ⋅ 10 ⎦
⎢ ⎥
⎣ 0 ⎦
⎡ 2133333.3333 533333.3333 0⎤ ⎡ 1.3325 ⎤
D = ⎢ 533333.3333 2133333.3333 0⎥ σ1 ≔ D ⋅ ξ1 = ⎢ 0.3331 ⎥
⎢ ⎥ ⎢ ⎥
⎣ 0 0 800000 ⎦ ⎣ -1.5902 ⎦

Auxiliar: Univ.: Ramos Rodríguez Juan Pablo