Ejemplo Barras - Metodo Rigideces (2)

EJEMPLO

MATRIZ DE FUERZAS F MATRIZ DE DESPLAZAMIENTOS

F X1 ? ∆ X1 -

F Y1 ? ∆ Y1 -

F X2 10.00 ∆ X2 ?

F Y2 - ∆ Y2 -

F X3 10.00 ∆ X3 ?

F Y3 - ∆ Y3 -

0° 0°

1 3

2 10 10

4.80 m 4.80 m

BARRA 1-2 BARRA 2-3

θ - ° θ - °

A 0.0005 m A 0.0005 m

E 2.00E+08 kg/m2 E 2.00E+08 kg/m2

L 4.80 m L 4.80 m

BARRA 1-2 BARRA 2-3

senθ - senθ -

cosθ 1.00 cosθ 1.00

AE/L 20833 AE/L 20833

MATRIZ DE RIGIDEZ K - PARCIAL MATRIZ DE RIGIDEZ K - PARCIAL

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 2.08E+04 0.00E+00 -2.08E+04 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 2.08E+04 0.00E+00

0.00E+00 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

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-2.08E+04 0.00E+00 4.17E+04 0.00E+00 -2.08E+04 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 2.08E+04 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

K K K -1

4.17E+04 -2.08E+04 4.80E-05 4.80E-05

-2.08E+04 2.08E+04 4.80E-05 9.60E-05

4.17E+04 -2.08E+04

-2.08E+04 2.08E+04

K -1

4.80E-05 4.80E-05

4.80E-05 9.60E-05

F

0.00E+00 F X1

0.00E+00 F Y1

1.00E+01 F X2

0.00E+00 F Y2

1.00E+01 F X3

0.00E+00 F Y3

∆ = K -1 F

0.00E+00 u1

0.00E+00 v1

9.60E-04 u2

0.00E+00 v2

1.44E-03 u3

0.00E+00 v3

X = K ∆

-20.00 X1

- Y1

10.00 X2

- Y2

10.00 X3

- Y3

DATOS

SOLUCION