ejercicio nro 02. rididecez metodo matricial
DESCRIPTION
Ejercicio Resuelto De Vigas METODO RIGIDEZ O METODO MATRICAL, Fuente Doc. Ing. Alexander HumpiriTRANSCRIPT
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a.- Mdulo de Elasticidad del Concreto [f'c]
b.- Momentos de Inercia [ I ]
c. Rigidez Axial [AE] y a la Flexin [EI]
d. Determinacin de Constantes de Rigideces
L
4
4
3
e. Constantes de Carga
M'ij (Kg-m) M'ji (Kg-m)
V 1-2 4 2285.714 761.905
V 3-4 4 761.905 2285.714
C 1-2 3
f. Planteamiento de Ecuaciones
Viga 1-2 u1 v1 1 u2 v2 2
X 1-2 0.00 0.00 0.00 -50000.00 0.00 0.00 u1 0.0000000 0.000000 0.000 0.000
Y 1-2 0.00 0.00 0.00 0.00 -500.00 1000.00 v1 0.000000 0.685714 5.714 6.400
M 1-2 = 0.00 0.00 0.00 0.00 -1000.00 1333.33 X 1 0.000000 1.537662 + 2.286 = 3.823
X 2-1 0.00 0.00 0.00 50000.00 0.00 0.00 u2 -2.040E-14 0.000000 0.000 0.000
Y 2-1 0.00 0.00 0.00 0.00 500.00 -1000.00 v2 -1.870E-03 -0.685714 0.686 0.000
M 2-1 0.00 0.00 0.00 0.00 -1000.00 2666.67 2 -2.494E-04 1.205195 -0.762 0.443
Viga 3-4 u3 v3 3 u4 v4 4
X 3-4 50000.00 0.00 0.00 0.000 0.000 0.000 u3 2.040E-14 0.000000 0.000 0.000
Y 3-4 0.00 500.00 1000.00 0.000 0.000 0.000 v3 -1.870E-03 -0.685714 0.686 0.000
M 3-4 = 0.00 1000.00 2666.67 0.000 0.000 0.000 X 3 2.494E-04 -1.205195 + 0.762 = -0.443
X 4-3 -50000.00 0.00 0.00 0.000 0.000 0.000 u4 0.00000 0.000000 0.000 0.000
Y 4-3 0.00 -500.00 -1000.00 0.000 0.000 0.000 v4 0.00000 0.685714 5.714 6.400
M 4-3 0.00 1000.00 1333.33 0.000 0.000 0.000 4 0.00000 -1.537662 -2.286 -3.823
Col 2-3 u2 v2 2 u3 v3 3
X 2-3 1185.19 0.00 -1777.78 -1185.19 0.00 -1777.78 u2 -2.040E-14 0.000000 0.000 0.000
Ec (Kg/cm2)
200000.000
200000.000
Elemento
Vigas
I Cm^4
133333.333
133333.333
Area(cm^2)
1000.000
2666.667
AE Ton
200000
200000Rectangular
1000.000
Seccin EI Ton-m2
2666.667
Columnas
Seccin
Rectangular
Rectangular
Elemento
Viga
Columnas
Rectangular
685.714
5714.285
1777.778
Reaccin Total
1185.185
6EI/L^2
C 2-3
1333.333
1333.333
50000.000
50000.000
66666.667
AE/L 2EI/L 4EI/L
Tramo LMom. Resistentes
1000.000
1777.778
Tramo
2666.667
2666.667
3555.556
V 1-2
V 3-4
ANALISIS DE UNA VIGA CONTINUA POR EL METODO DE RIGIDECES [SECCION CONTINUA]
5714.285
685.714
12EI/L^3
500.000
500.000
1000.000
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Y 2-3 0.00 66666.67 0.00 0.00 -66666.67 0.00 v2 -1.870E-03 0.000000 0.000 0.000
M 2-3 = -1777.78 0.00 3555.56 1777.78 0.00 1777.78 X e2 -2.494E-04 -0.443290 + 0.000 = -0.443
X 3-2 -1185.19 0.00 1777.78 1185.19 0.00 1777.78 u3 2.040E-14 0.000000 0.000 0.000
Y 3-2 0.00 -66666.67 0.00 0.00 66666.67 0.00 v3 -1.870E-03 0.000000 0.000 0.000
M 3-2 -1777.78 0.00 1777.78 1777.78 0.00 3555.56 3 2.494E-04 0.443290 0.000 0.443
g. Condiciones de Continuidad
X1 = X1-2 = 0.000 Ton
Y1 = Y1-2 = 6.400 Ton
M1 = M1-2 = 3.823 Ton-m
X2 = X2-1 + X2-3 = 0.000 Verificacin
Y2 = Y2-1 + Y2-3 = 0.000
M2 = M2-1 + M2-3 = 0.000
X3 = X3-4 +X 3-2 = 0.000 Ton
Y3 = Y3-4 + Y 3-2 = 0.000 Ton
M3 = M3-2+ M3-4 = 0.000 Ton-m
X4 = X4-3 = 0.000
Y4 = Y4-3 = 6.400
M4 = M4-3 = -3.823
Por consiguiente:
u2 v2 2 u3 v3 3
X2=0 51185.185 0.000 -1777.778 -1185.185 0.000 -1777.778 u2 0.000
Y2=0 0.000 67166.667 -1000.000 0.000 -66666.667 0.000 v2 0.686
M2=0 -1777.778 -1000.000 6222.223 1777.778 0.000 1777.778 2 -0.762
X3=0 = -1185.185 0.000 1777.778 51185.185 0.000 1777.778 u3 + 0.000
Y3=0 0.000 -66666.667 0.000 0.000 67166.667 1000.000 v3 0.686
M3=0 -1777.778 0.000 1777.778 1777.778 1000.000 6222.223 3 0.762
INVERSA 1.98447E-05 3.27237E-08 4.37952E-06 1.5528E-07 -3.27237E-08 4.37952E-06 0.000
3.27237E-08 0.001821922 0.000409573 -3.2724E-08 0.001814443 -0.00040861 -0.686
4.37952E-06 0.000409573 0.000269052 -4.3795E-06 0.000408609 -0.00014004 0.762
1.5528E-07 -3.27237E-08 -4.37952E-06 1.98447E-05 3.27237E-08 -4.3795E-06 0.000
-3.2724E-08 0.001814443 0.000408609 3.27237E-08 0.001821922 -0.00040957 -0.686
4.37952E-06 -0.000408609 -0.000140039 -4.3795E-06 -0.000409573 0.000269052 -0.762
u2 -2.0399184E-14
v2 -1.8701297E-03
2 = -2.4935052E-04 Luego reemplace en las ecuaciones del tem f
u3 2.0399185E-14
v3 -1.8701297E-03
3 2.4935052E-04