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Ejemplo Barras - Metodo Rigideces
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MATRIZ DE FUERZAS F MATRIZ DE DESPLAZAMIENTOS
0 F X1 ? X1 -
1 2 F Y1 ? Y1 -
P F X2 10.00 X2 ?
F Y2 - Y2 -
4.80 m
BARRA 1-2
-
A 0.0005 m
E 2.00E+08 kg/m2
L 4.80 m
BARRA 1-2
sen -
cos 1.00
AE/L 20833
K
2.08E+04 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 2.08E+04 0.00E+00
0.00E+00 0.00E+00 0.00E+00 0.00E+00
K
2.08E+04
K -1
4.80E-05
F
0.00E+00 F X1
0.00E+00 F Y1
1.00E+01 F X2
0.00E+00 F Y2
= K -1 F
0.00E+00 u1
0.00E+00 v1
4.80E-04 u2
0.00E+00 v2
X = K
-10.00 X1
- Y1
10.00 X2
- Y2
EJEMPLO
SOLUCION
DATOS
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