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EJEMPLO
DATOS
0°
1
4.80 m
BARRA 1-2
θ - °
A 0.0005 m
E 2.00E+08 kg/m2
L 4.80 m
SOLUCION
BARRA 1-2
-
1.00
AE/L 20833
K
2.08E+04 0.00E+00 -2.08E+04
0.00E+00 0.00E+00 0.00E+00
-2.08E+04 0.00E+00 2.08E+04
0.00E+00 0.00E+00 0.00E+00
K
2.08E+04
senθ
cosθ
4.80E-05
F
0.00E+00
0.00E+00
1.00E+01
0.00E+00
0.00E+00
0.00E+00
4.80E-04
0.00E+00
-10.00
-
10.00
-
K -1
F X1
F Y1
F X2
F Y2
∆ = K -1 F
u1
v1
u2
v2
X = K ∆
X1
Y1
X2
Y2
DATOS
MATRIZ DE FUERZAS F
2
P
SOLUCION
0.00E+00
0.00E+00
0.00E+00
0.00E+00
F X1
F Y1
F X2
F Y2
DATOS
MATRIZ DE FUERZAS F MATRIZ DE DESPLAZAMIENTOS
? -
? -
10.00 ?
- -
SOLUCION
∆ X1
∆ Y1
∆ X2
∆ Y2
EJEMPLO
DATOS
0°
1
4.80 m
BARRA 1-2
θ - °
A 0.0005 m
E 2.00E+08 kg/m2
L 4.80 m
SOLUCION
BARRA 1-2
-
1.00
AE/L 20833
MATRIZ DE RIGIDEZ K - PARCIAL
2.08E+04 0.00E+00 -2.08E+04
senθ
cosθ
0.00E+00 0.00E+00 0.00E+00
-2.08E+04 0.00E+00 2.08E+04
0.00E+00 0.00E+00 0.00E+00
K
2.08E+04 0.00E+00 -2.08E+04
0.00E+00 0.00E+00 0.00E+00
-2.08E+04 0.00E+00 4.17E+04
0.00E+00 0.00E+00 0.00E+00
0.00E+00 0.00E+00 -2.08E+04
0.00E+00 0.00E+00 0.00E+00
K
4.17E+04
-2.08E+04
4.80E-05
4.80E-05
F
0.00E+00
0.00E+00
1.00E+01
0.00E+00
1.00E+01
K -1
F X1
F Y1
F X2
F Y2
F X3
0.00E+00
0.00E+00
0.00E+00
9.60E-04
0.00E+00
1.44E-03
0.00E+00
-20.00
-
10.00
-
10.00
-
F Y3
∆ = K -1 F
u1
v1
u2
v2
u3
v3
X = K ∆
X1
Y1
X2
Y2
X3
Y3
DATOS
0°
2 10
4.80 m
SOLUCION
0.00E+00
0.00E+00
0.00E+00
0.00E+00
0.00E+00 0.00E+00 0.00E+00
0.00E+00 0.00E+00 0.00E+00
0.00E+00 -2.08E+04 0.00E+00
0.00E+00 0.00E+00 0.00E+00
0.00E+00 2.08E+04 0.00E+00
0.00E+00 0.00E+00 0.00E+00
-2.08E+04
2.08E+04
4.80E-05
9.60E-05
DATOS
MATRIZ DE FUERZAS F MATRIZ DE DESPLAZAMIENTOS
?
?
10.00
-
10.00
-
3
10
BARRA 2-3
θ - °
A 0.0005 m
E 2.00E+08 kg/m2
L 4.80 m
SOLUCION
BARRA 2-3
-
1.00
AE/L 20833
MATRIZ DE RIGIDEZ K - PARCIAL
F X1 ∆ X1
F Y1 ∆ Y1
F X2 ∆ X2
F Y2 ∆ Y2
F X3 ∆ X3
F Y3 ∆ Y3
senθ
cosθ
2.08E+04 0.00E+00
0.00E+00 0.00E+00
-2.08E+04 0.00E+00
0.00E+00 0.00E+00
K
4.17E+04 -2.08E+04 4.80E-05
-2.08E+04 2.08E+04 4.80E-05
K -1
DATOS
MATRIZ DE DESPLAZAMIENTOS
-
-
?
-
?
-
SOLUCION
-2.08E+04 0.00E+00
0.00E+00 0.00E+00
2.08E+04 0.00E+00
0.00E+00 0.00E+00
4.80E-05
9.60E-05
PROBLEMA. METODO DE LA CARGA UNITARIA FICTICIA
0°1
4.80 m
Nº BARRA L m
1 12 T 4.80 2 23 T 4.80
PROBLEMA. METODO DE LA CARGA UNITARIA FICTICIA
0°
2
4.80 m
A Pm2 kg
0.0005 20.00 1.00 - 0.0005 10.00 1.00 -
desplazamiento horizontal
desplazamiento vertical
μh μv
PROBLEMA. METODO DE LA CARGA UNITARIA FICTICIA
CERO
31
E μh PL/AE Kg/m2 m
2.00E+08 9.60E-042.00E+08 4.80E-04
∑= 1.44E-03 ∑=
δh= 1.44E-03 m
δv= 0.00E+00 m
PROBLEMA. METODO DE LA CARGA UNITARIA FICTICIA
μv PL/AE m 0.00E+000.00E+000.00E+00