examen elementos finitos(limpio)
TRANSCRIPT
2º EXAMEN PARCIALELEMENTOS FINITOS
CIV - 313
Arena :
1700
30 º
Tajmar de HºCº :
E = 180000
u = 0.25
a) DISCRETIZACION
1. Calcular : s , e y (u, v) para la siguiente estructura :
ga = kg/m3
=
kg/cm2
P
Arena
Roca
P
Y
X
1
2
3
4
5
1
2
3
4 5
6
7
c) DETERMINACION DE "P" (para 1m de ancho)
P = 22950 kg
d) ELEMENTOS
Nº i j k1 1 6 22 2 6 33 3 5 44 3 6 55 1 7 6
e) COORDENADAS DE NODOS
Nodo x y1 0 02 0 1003 0 2004 0 4005 60 4006 150 1007 150 0
2
)2/45(22 g =
TgHP a
Elemento 5:Nodo Nº x y
i 1 0 0j 7 150 0k 6 150 100
Cálculo de Area
1 0 02A = 1 150 0 = 15000
1 150 100
-100 0
100 -150
0 150
-100 0 100 0 0 06.7E-05 0 0 0 -150 0 150
0 -100 -150 100 150 0
1 4 2-0.0066667 0 0.00666667 0 0 0
0 0 0 -0.01 0 0.010 -0.0066667 -0.01 0.00666667 0.01 0
-0.0066667 0 00 0 -0.0066667
0.00666667 0 -0.010 -0.01 0.006666670 0 0.010 0.01 0
Mariz constitutiva
E = 180000 1 0.25 0u = 0.25 D = 192000 0.25 1 0t = 100 cm 0 0 0.375
v = Axt = 750000 192000 48000 0D = 48000 192000 0
0 0 72000
-1280.00 -320.00 0.00
y23 = x32 =
y31 = x13 =
y12 = x21 =
Be =
[Be ]=
[Be ]T=
kg/cm2
0.00 0.00 -480.001280.00 320.00 -720.00-480.00 -1920.00 480.00
0.00 0.00 720.00480.00 1920.00 0.00
8.53 0.00 -8.53 3.20 0.00 -3.200.00 3.20 4.80 -3.20 -4.80 0.00-8.53 4.80 15.73 -8.00 -7.20 3.203.20 -3.20 -8.00 22.40 4.80 -19.200.00 -4.80 -7.20 4.80 7.20 0.00-3.20 0.00 3.20 -19.20 0.00 19.20
1 7 66400000.0 0.0 -6400000.0 2400000.0 0.0 -2400000.0
10.0 2400000.0 3600000.0 -2400000.0 -3600000.0 0.0
-6400000.0 3600000.0 11800000.0 -6000000.0 -5400000.0 2400000.0 72400000.0 -2400000.0 -6000000.0 16800000.0 3600000.0 -14400000.0
0.0 -3600000.0 -5400000.0 3600000.0 5400000.0 0.06
-2400000.0 0.0 2400000.0 -14400000.0 0.0 14400000.0
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
0
0 0
0 = = 0
0 0
0
0
0
= 0
0
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Ejemplo de tensión plana
Nodos x y Elem. i j1 0 30 1 1 42 120 30 2 1 33 0 04 120 0
k24
2º EXAMEN PARCIALELEMENTOS FINITOS
CIV - 313
Arena :
1700
30 º
Tajmar de HºCº :
E = 180000
u = 0.25
a) DISCRETIZACION
1. Calcular : s , e y (u, v) para la siguiente estructura :
ga = kg/m3
=
kg/cm2
P
Arena
Roca
P
Y
X
1
2
3
4
5
1
2
3
4 5
6
7
c) DETERMINACION DE "P" (para 1m de ancho)
P = 22950 kg
d) ELEMENTOS
Nº i j k1 1 6 22 2 6 33 3 5 44 3 6 55 1 7 6
e) COORDENADAS DE NODOS
Nodo x y1 0 02 0 1003 0 2004 0 4005 60 4006 150 1007 150 0
2
)2/45(22 g =
TgHP a
Elemento 3:Nodo Nº x y
i 3 0 200j 5 60 400k 4 0 400
Cálculo de Area
1 0 2002A = 1 60 400 = 12000
1 0 400
0 -60
200 0
-200 60
0 0 200 0 -200 08.3E-05 0 -60 0 0 0 60
-60 0 0 200 60 -200
1 4 20 0 0.01666667 0 -0.0166667 00 -0.005 0 0 0 0.005
-0.005 0 0 0.01666667 0.005 -0.0166667
0 0 -0.0050 -0.005 0
0.01666667 0 00 0 0.01666667
-0.0166667 0 0.0050 0.005 -0.0166667
Mariz constitutiva
E = 180000 1 0.25 0u = 0.25 D = 192000 0.25 1 0t = 100 cm 0 0 0.375
v = Axt = 600000 192000 48000 0D = 48000 192000 0
0 0 72000
0.00 0.00 -360.00
y23 = x32 =
y31 = x13 =
y12 = x21 =
Be =
[Be ]=
[Be ]T=
kg/cm2
-240.00 -960.00 0.003200.00 800.00 0.00
0.00 0.00 1200.00-3200.00 -800.00 360.00240.00 960.00 -1200.00
1.80 0.00 0.00 -6.00 -1.80 6.000.00 4.80 -4.00 0.00 4.00 -4.800.00 -4.00 53.33 0.00 -53.33 4.00-6.00 0.00 0.00 20.00 6.00 -20.00-1.80 4.00 -53.33 6.00 55.13 -10.006.00 -4.80 4.00 -20.00 -10.00 24.80
3 5 41080000.0 0.0 0.0 -3600000.0 -1080000.0 3600000.0
30.0 2880000.0 -2400000.0 0.0 2400000.0 -2880000.0
0.0 -2400000.0 32000000.0 0.0 -32000000.0 2400000.0 5-3600000.0 0.0 0.0 12000000.0 3600000.0 -12000000.0-1080000.0 2400000.0 -32000000.0 3600000.0 33080000.0 -6000000.0
43600000.0 -2880000.0 2400000.0 -12000000.0 -6000000.0 14880000.0
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
-0.00085
-0.00132 2.8712E-07
-0.00253 = = -9.24E-07
-0.00093 1.1174E-06
-0.00255
-0.00151
0.01077634
= -0.1636182
0.08045209
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Ejemplo de tensión plana
Nodos x y Elem. i j1 0 30 1 1 42 120 30 2 1 33 0 04 120 0
k24
2º EXAMEN PARCIALELEMENTOS FINITOS
CIV - 313
Arena :
1700
30 º
Tajmar de HºCº :
E = 180000
u = 0.25
a) DISCRETIZACION
1. Calcular : s , e y (u, v) para la siguiente estructura :
ga = kg/m3
=
kg/cm2
P
Arena
Roca
P
Y
X
1
2
3
4
5
1
2
3
4 5
6
7
c) DETERMINACION DE "P" (para 1m de ancho)
P = 22950 kg
d) ELEMENTOS
Nº i j k1 1 6 22 2 6 33 3 5 44 3 6 55 1 7 6
e) COORDENADAS DE NODOS
Nodo x y1 0 02 0 1003 0 2004 0 4005 60 4006 150 1007 150 0
2
)2/45(22 g =
TgHP a
Elemento 4:Nodo Nº x y
i 3 0 200j 6 150 100k 5 60 400
Cálculo de Area
1 0 2002A = 1 150 100 = 36000
1 60 400
-300 -90
200 -60
100 150
-300 0 200 0 100 02.8E-05 0 -90 0 -60 0 150
-90 -300 -60 200 150 100
1 4 2-0.0083333 0 0.00555556 0 0.00277778 0
0 -0.0025 0 -0.0016667 0 0.00416667-0.0025 -0.0083333 -0.0016667 0.00555556 0.00416667 0.00277778
-0.0083333 0 -0.00250 -0.0025 -0.0083333
0.00555556 0 -0.00166670 -0.0016667 0.00555556
0.00277778 0 0.004166670 0.00416667 0.00277778
Mariz constitutiva
E = 180000 1 0.25 0u = 0.25 D = 192000 0.25 1 0t = 100 cm 0 0 0.375
v = Axt = 1800000 192000 48000 0D = 48000 192000 0
0 0 72000
-1600.00 -400.00 -180.00
y23 = x32 =
y31 = x13 =
y12 = x21 =
Be =
[Be ]=
[Be ]T=
kg/cm2
-120.00 -480.00 -600.001066.67 266.67 -120.00-80.00 -320.00 400.00533.33 133.33 300.00200.00 800.00 200.00
13.78 2.50 -8.59 -0.33 -5.19 -2.172.50 6.20 0.33 -2.53 -2.83 -3.67-8.59 0.33 6.13 -1.11 2.46 0.78-0.33 -2.53 -1.11 2.76 1.44 -0.22-5.19 -2.83 2.46 1.44 2.73 1.39-2.17 -3.67 0.78 -0.22 1.39 3.89
3 6 524810000.0 4500000.0 -15460000.0 -600000.0 -9350000.0 -3900000.0
34500000.0 11160000.0 600000.0 -4560000.0 -5100000.0 -6600000.0
-15460000.0 600000.0 11026666.7 -2000000.0 4433333.3 1400000.0 6-600000.0 -4560000.0 -2000000.0 4960000.0 2600000.0 -400000.0-9350000.0 -5100000.0 4433333.3 2600000.0 4916666.7 2500000.0
5-3900000.0 -6600000.0 1400000.0 -400000.0 2500000.0 7000000.0
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
-0.00085
-0.00132 2.7376E-08
0 = = -5.664E-07
0 3.5112E-09
-0.00253
-0.00093
-0.0219319
= -0.107438
0.0002528
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Ejemplo de tensión plana
Nodos x y Elem. i j1 0 30 1 1 42 120 30 2 1 33 0 04 120 0
k24
2º EXAMEN PARCIALELEMENTOS FINITOS
CIV - 313
Arena :
1700
30 º
Tajmar de HºCº :
E = 180000
u = 0.25
a) DISCRETIZACION
1. Calcular : s , e y (u, v) para la siguiente estructura :
ga = kg/m3
=
kg/cm2
P
Arena
Roca
P
Y
X
1
2
3
4
5
1
2
3
4 5
6
7
c) DETERMINACION DE "P" (para 1m de ancho)
P = 22950 kg
d) ELEMENTOS
Nº i j k1 1 6 22 2 6 33 3 5 44 3 6 55 1 7 6
e) COORDENADAS DE NODOS
Nodo x y1 0 02 0 1003 0 2004 0 4005 60 4006 150 1007 150 0
2
)2/45(22 g =
TgHP a
Elemento 2:Nodo Nº x y
i 2 0 100j 6 150 100k 3 0 200
Cálculo de Area
1 0 1002A = 1 150 100 = 15000
1 0 200
-100 -150
100 0
0 150
-100 0 100 0 0 06.7E-05 0 -150 0 0 0 150
-150 -100 0 100 150 0
1 4 2-0.0066667 0 0.00666667 0 0 0
0 -0.01 0 0 0 0.01-0.01 -0.0066667 0 0.00666667 0.01 0
-0.0066667 0 -0.010 -0.01 -0.0066667
0.00666667 0 00 0 0.006666670 0 0.010 0.01 0
Mariz constitutiva
E = 180000 1 0.25 0u = 0.25 D = 192000 0.25 1 0t = 100 cm 0 0 0.375
v = Axt = 750000 192000 48000 0D = 48000 192000 0
0 0 72000
-1280.00 -320.00 -720.00
y23 = x32 =
y31 = x13 =
y12 = x21 =
Be =
[Be ]=
[Be ]T=
kg/cm2
-480.00 -1920.00 -480.001280.00 320.00 0.00
0.00 0.00 480.000.00 0.00 720.00
480.00 1920.00 0.00
15.73 8.00 -8.53 -4.80 -7.20 -3.208.00 22.40 -3.20 -3.20 -4.80 -19.20-8.53 -3.20 8.53 0.00 0.00 3.20-4.80 -3.20 0.00 3.20 4.80 0.00-7.20 -4.80 0.00 4.80 7.20 0.00-3.20 -19.20 3.20 0.00 0.00 19.20
2 6 311800000.0 6000000.0 -6400000.0 -3600000.0 -5400000.0 -2400000.0
26000000.0 16800000.0 -2400000.0 -2400000.0 -3600000.0 -14400000.0
-6400000.0 -2400000.0 6400000.0 0.0 0.0 2400000.0 6-3600000.0 -2400000.0 0.0 2400000.0 3600000.0 0.0-5400000.0 -3600000.0 0.0 3600000.0 5400000.0 0.0
3-2400000.0 -14400000.0 2400000.0 0.0 0.0 14400000.0
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
0
0 0
0 = = -1.323E-05
0 -8.481E-06
-0.00085
-0.00132
-0.6351141
= -2.5404562
-0.6106298
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Ejemplo de tensión plana
Nodos x y Elem. i j1 0 30 1 1 42 120 30 2 1 33 0 04 120 0
k24
2º EXAMEN PARCIALELEMENTOS FINITOS
CIV - 313
Arena :
1700
30 º
Tajmar de HºCº :
E = 180000
u = 0.25
a) DISCRETIZACION
1. Calcular : s , e y (u, v) para la siguiente estructura :
ga = kg/m3
=
kg/cm2P
Arena
Roca
P
Y
X
1
2
3
4
5
1
2
3
4 5
6
7
c) DETERMINACION DE "P" (para 1m de ancho)
P = 22950 kg
d) ELEMENTOS
Nº i j k1 1 6 22 2 6 33 3 5 44 3 6 55 1 7 6
e) COORDENADAS DE NODOS
Nodo x y1 0 02 0 1003 0 2004 0 4005 60 4006 150 1007 150 0
Elemento 1:Nodo Nº x y
i 1 0 0j 6 150 100k 2 0 100
Cálculo de Area
1 0 02A = 1 150 100 = 15000
1 0 100
0 -150
100 0
-100 150
0 0 100 0 -100 0
y23 = x32 =
y31 = x13 =
y12 = x21 =
P=γa¿H
2⋅Tg2( 45+θ /2 )2
6.7E-05 0 -150 0 0 0 150-150 0 0 100 150 -100
1 4 20 0 0.00666667 0 -0.0066667 00 -0.01 0 0 0 0.01
-0.01 0 0 0.00666667 0.01 -0.0066667
0 0 -0.010 -0.01 0
0.00666667 0 00 0 0.00666667
-0.0066667 0 0.010 0.01 -0.0066667
Mariz constitutiva
E = 180000 1 0.25 0u = 0.25 D = 192000 0.25 1 0t = 100 cm 0 0 0.375
v = Axt = 750000 192000 48000 0D = 48000 192000 0
0 0 72000
0.00 0.00 -720.00-480.00 -1920.00 0.001280.00 320.00 0.00
0.00 0.00 480.00-1280.00 -320.00 720.00480.00 1920.00 -480.00
7.20 0.00 0.00 -4.80 -7.20 4.800.00 19.20 -3.20 0.00 3.20 -19.200.00 -3.20 8.53 0.00 -8.53 3.20-4.80 0.00 0.00 3.20 4.80 -3.20-7.20 3.20 -8.53 4.80 15.73 -8.004.80 -19.20 3.20 -3.20 -8.00 22.40
1 6 25400000.0 0.0 0.0 -3600000.0 -5400000.0 3600000.0
10.0 14400000.0 -2400000.0 0.0 2400000.0 -14400000.0
0.0 -2400000.0 6400000.0 0.0 -6400000.0 2400000.0 6-3600000.0 0.0 0.0 2400000.0 3600000.0 -2400000.0-5400000.0 2400000.0 -6400000.0 3600000.0 11800000.0 -6000000.0
23600000.0 -14400000.0 2400000.0 -2400000.0 -6000000.0 16800000.0
Be =
[Be ]=
[Be ]T=
kg/cm2
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
0
0 0
0 = = 0
0 0
0
0
0
= 0
0
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Ejemplo de tensión plana
Nodos x y Elem. i j1 0 30 1 1 42 120 30 2 1 33 0 04 120 0
k24
Ejemplo de tensión plana
Nodos x y Elem. i j k1 0 30 1 1 4 22 120 30 2 1 3 43 0 04 120 0
Elemento 2Nodo Nº x y
i 1 0 30j 3 0 0k 4 120 0
Cálculo de Area
1 0 302A = 1 0 0 = 3600
1 120 0
0 120
-30 -120
30 0
0 0 -30 0 30 00.00028 0 120 0 -120 0 0
120 0 -120 -30 0 30
0 0 -0.008333 0 0.008333 00 0.033333 0 -0.033333 0 0
0.033333 0 -0.033333 -0.008333 0 0.008333
0 0 0.0333330 0.033333 0
-0.008333 0 -0.0333330 -0.033333 -0.008333
0.008333 0 00 0 0.008333
Mariz constitutivaE = 2E+06 k/cm2 1 0.3 0u = 0.3 D = 2307692.3 0.3 1 0t = 4 cm 0 0 0.35
v = Axt = 7200 2307692.3 692307.69 0D = 692307.69 2307692.3 0
0 0 807692.308
0 0 26923.07723076.923 76923.077 0
-19230.769 -5769.2308 -26923.077-23076.923 -76923.077 -6730.769219230.769 5769.2308 0
0 0 6730.7692
897.4359 0 -897.4359 -224.35897 0 224.358970 2564.1026 -192.30769 -2564.1026 192.30769 0
-897.4359 -192.30769 1057.6923 416.66667 -160.25641 -224.35897-224.35897 -2564.1026 416.66667 2620.1923 -192.30769 -56.089744
0 192.30769 -160.25641 -192.30769 160.25641 0224.35897 0 -224.35897 -56.089744 0 56.089744
1 3 4
6461538.5 0.0 -6461538.5 -1615384.6 0.0 1615384.61
0.0 18461538.5 -1384615.4 -18461538.5 1384615.4 0.0
-6461538.5 -1384615.4 7615384.6 3000000.0 -1153846.2 -1615384.63
-1615384.6 -18461538.5 3000000.0 18865384.6 -1384615.4 -403846.2
0.0 1384615.4 -1153846.2 -1384615.4 1153846.2 0.04
1615384.6 0.0 -1615384.6 -403846.2 0.0 403846.2
Una vez resuelto el sistema obtenemos las tensiones y las deformaciones unitarias
#REF!
#REF! #REF!
0 = = #REF!
0 #REF!
#REF!
#REF!
#REF!
= #REF!
#REF!
y23 = x32 =
y31 = x13 =
y12 = x21 =
Be =
[Be ]=
[Be ]T=
[Be ]T[D]=
[Be ]T[D][Be ]=
Ke =
ex
e = [B][u] = [B}x ey
gxy
sx
s = [D][e] = sy
txy
Elemento 1:1 6 2
5400000 0 0 -3600000 -5400000 36000001
0 14400000 -2400000 0 2400000 -144000000 -2400000 6400000 0 -6400000 2400000
6-3600000 0 0 2400000 3600000 -2400000-5400000 2400000 -6400000 3600000 11800000 -6000000
23600000 -14400000 2400000 -2400000 -6000000 16800000
Elemento 2:1 3 4
6461538.4615 0 -6461538.462 -1615384.615 0 1615384.61541
0 18461538.462 -1384615.385 -18461538.46 1384615.3846 0-6461538.462 -1384615.385 7615384.6154 3000000 -1153846.154 -1615384.615
3-1615384.615 -18461538.46 3000000 18865384.615 -1384615.385 -403846.1538
0 1384615.3846 -1153846.154 -1384615.385 1153846.1538 04
1615384.6154 0 -1615384.615 -403846.1538 0 403846.15385
Elemento 3:3 5 4
1080000.0 0.0 0.0 -3600000.0 -1080000.0 3600000.03
0.0 2880000.0 -2400000.0 0.0 2400000.0 -2880000.00.0 -2400000.0 32000000.0 0.0 -32000000.0 2400000.0
5-3600000.0 0.0 0.0 12000000.0 3600000.0 -12000000.0-1080000.0 2400000.0 -32000000.0 3600000.0 33080000.0 -6000000.0
43600000.0 -2880000.0 2400000.0 -12000000.0 -6000000.0 14880000.0
Elemento 4:3 6 5
24810000.0 4500000.0 -15460000.0 -600000.0 -9350000.0 -3900000.03
4500000.0 11160000.0 600000.0 -4560000.0 -5100000.0 -6600000.0-15460000.0 600000.0 11026666.7 -2000000.0 4433333.3 1400000.0
6-600000.0 -4560000.0 -2000000.0 4960000.0 2600000.0 -400000.0
-9350000.0 -5100000.0 4433333.3 2600000.0 4916666.7 2500000.05
-3900000.0 -6600000.0 1400000.0 -400000.0 2500000.0 7000000.0
Elemento 5:1 7 6
6400000.0 0.0 -6400000.0 2400000.0 0.0 -2400000.01
0.0 2400000.0 3600000.0 -2400000.0 -3600000.0 0.0-6400000.0 3600000.0 11800000.0 -6000000.0 -5400000.0 2400000.0
72400000.0 -2400000.0 -6000000.0 16800000.0 3600000.0 -14400000.0
0.0 -3600000.0 -5400000.0 3600000.0 5400000.0 0.06
-2400000.0 0.0 2400000.0 -14400000.0 0.0 14400000.0
ENSAMBLAJE DE LA MATRIZu1 v1 u2 v2 u3 v3 u4 v4 u5 v5 u6 v6 u7 v7
1 2 3 4 5 6 718261538.5 0.0 -12861538.5 -2815384.6 -6461538.5 -1615384.6 0.0 1615384.6 0.0 0.0 0.0 -3600000.0 -6400000.0 2400000.0
1u1
0.0 35261538.5 -184615.4 -20861538.5 -1384615.4 -18461538.5 1384615.4 0.0 0.0 0.0 -2400000.0 0.0 3600000.0 -2400000.0 v1-5400000 2400000 11800000 -6000000 0 0 0 0 0 0 -6400000 3600000 0 0
2u2
3600000 -14400000 -6000000 16800000 0 0 0 0 0 0 2400000 -2400000 0 0 v2-6461538.462 -1384615.385 0 0 7615384.6154 3000000 -2233846.154 1984615.3846 -9350000 -7500000 -15460000 -600000 0 0
3u3
-1615384.615 -18461538.46 0 0 3000000 18865384.615 1015384.6154 -3283846.154 -7500000 -6600000 600000 -4560000 0 0 v30 1384615.3846 0 0 -2233846.154 1015384.6154 34233846.154 -6000000 -32000000 3600000 0 0 0 0
4u4
1615384.6154 0 0 0 1984615.3846 -3283846.154 -6000000 15283846.154 2400000 -12000000 0 0 0 0 v40 0 0 0 -9350000 -7500000 -32000000 2400000 36916666.667 2500000 4433333.3333 2600000 0 0
5u5
0 0 0 0 -7500000 -6600000 3600000 -12000000 2500000 19000000 1400000 -400000 0 0 v50 -6000000 -6400000 2400000 -15460000 600000 0 0 4433333.3333 1400000 22826666.667 -2000000 -5400000 3600000
6u6
-6000000 0 3600000 -2400000 -600000 -4560000 0 0 2600000 -400000 -2000000 21760000 2400000 -14400000 v6-6400000 3600000 0 0 0 0 0 0 0 0 -5400000 2400000 11800000 -6000000
7u7
2400000 -2400000 0 0 0 0 0 0 0 0 3600000 -14400000 -6000000 16800000 v7
Vector de fuerzas: f1 0f2 0f3 0f4 0f5 22950f6 0f7 0f8 = 0f9 0f10 0f11 0f12 0f13 0f14 0
Condiciones de bordeu1 = 0v1 = 0u2 = 0v2 = 0u6 = 0v6 = 0u7 = 0v7 = 0
luego la matriz se reduce a:
7615384.6 3000000 -2233846 1984615.4 -9350000 -7500000 u3 229503000000 18865385 1015384.6 -3283846 -7500000 -6600000 v3 0
-2233846 1015384.6 34233846 -6000000 -32000000 3600000 u4 01984615.4 -3283846 -6000000 15283846 2400000 -12000000 v4 = 0-9350000 -7500000 -32000000 2400000 36916667 2500000 u5 0-7500000 -6600000 3600000 -12000000 2500000 19000000 v5 0
Resolviendo el sistema siguiente tenemos:
-3.695E-08 -5.765E-08 -1.112E-07 -6.571E-08 -1.104E-07 -4.052E-08-5.765E-08 7.443E-08 -4.123E-08 5.318E-08 -4.207E-08 5.003E-08-1.112E-07 -4.123E-08 -1.648E-08 -6.678E-08 -4.025E-08 -9.197E-08-6.571E-08 5.318E-08 -6.678E-08 1.683E-07 -8.295E-08 1.224E-07-1.104E-07 -4.207E-08 -4.025E-08 -8.295E-08 -3.224E-08 -9.873E-08-4.052E-08 5.003E-08 -9.197E-08 1.224E-07 -9.873E-08 1.617E-07
luego los desplazamientos son:
u3 -0.000848v3 -0.001323u4 -0.002552v4 = -0.001508u5 -0.002534v5 -0.00093
los desplazamientos finales son:
u1 0v1 0u2 0v2 0u3 -0.000848v3 -0.001323u4 -0.002552v4 = -0.001508u5 -0.002534v5 -0.00093u6 0v6 0u7 0v7 0
