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CURSO: ANALISIS ESTRUCTURAL AVANZADO
TEMA: METODO POR ENSAMBLAJE
PROFESOR: ING. ERLYN GIORDANY SALAZAR HUAMAN
Solucionar la siguiente armadura:
6
5
4.00 ft
2
4
q1= 1
3
q2= 20.00
3.00 ft
10.00
MATRIZ DE RIGIDEZ DE LA BARRA 1
1
Angulo= 0 C = 1 C² / L = 0.3333333
S = 0 CS/ L = 0
L = 3 ft S² / L = 0
1 2 3 4
0.3333 0.0000 -0.3333 0.0000 1 2 4
0.0000 0.0000 0.0000 0.0000 2 1
K 1= -0.3333 0.0000 0.3333 0.0000 3 10
0.0000 0.0000 0.0000 0.0000 4
20
6
5
EA
MATRIZ DE RIGIDEZ DE LA BARRA 2
1
(0,0) (3,0)
(3 4)
2
1
λx = C = 0.6 C² / L = 0.072
λx = S = 0.8 CS/ L = 0.096
L = 5 ft S² / L = 0.128
5 6 1 2
0.0720 0.0960 -0.0720 -0.0960 5
0.0960 0.1280 -0.0960 -0.1280 6
K2= -0.0720 -0.0960 0.0720 0.0960 1
-0.0960 -0.1280 0.0960 0.1280 2
1 2 3 4 5 6 1 2
Q1 0.405 0.096 -0.333 0.000 -0.072 -0.096 1 D1 0.4053 0.0960
Q2 0.096 0.128 0.000 0.000 -0.096 -0.128 2 D2 0.0960 0.1280
Q3 =EA -0.333 0.000 0.333 0.000 0.000 0.000 3 D3 MT = EA -0.3333 0.0000
MATRIZ TOTAL DE LA ESTRUCTURA
EA
(3,4)
(0,0)
Q4 0.000 0.000 0.000 0.000 0.000 0.000 4 D4 0.0000 0.0000
Q5 -0.072 -0.096 0.000 0.000 0.072 0.096 5 D5 -0.0720 -0.0960
Q6 -0.096 -0.128 0.000 0.000 0.096 0.128 6 D6 -0.0960 -0.1280
-10.00 0.405 0.096 D1
-20.00 0.096 0.128 D2
D1 = 15.000 /EA
D2 = -167.5000 /EA
R3 -0.333 0.000 15.00 ´/ EA
R4 ´= EA 0.000 0.000 -167.50 ´/ EA
R5 -0.072 -0.096 0.00 ´/ EA
R6 -0.096 -0.128 0.00 ´/ EA
R3 = -5
R4 = 0
R5 = 15
´= EA
R6 = 20
DNX
DNY
EA ´- C ´- S C S DFX
L DFY
15 D1
q1 = EA -1 0 1 0 -167.50 D2
3 0 /EA D3
0 D4
q1 = -5.00 TN
q3= EA -0.6 -0.8 0.6 0.8 15.00 D1
5 -167.50 /EA D2
q =
BARRA 1
BARRA 2
FUERZAS DE LAS BARRAS
0.00 D5
0.00 D6
q3= 25.00
1 2 3 4 5 6
0.3333 0.0000 -0.3333 0.0000 0.0000 0.0000 1
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 2
K1 = EA -0.3333 0.0000 0.3333 0.0000 0.0000 0.0000 3
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 4
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 5
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 6
3
1 2 3 4 5 6
0.0720 0.0960 0.0000 0.0000 -0.0720 -0.0960 1
0.0960 0.1280 0.0000 0.0000 -0.0960 -0.1280 2
K2 = EA 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 3
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 4
-0.0720 -0.0960 0.0000 0.0000 0.0720 0.0960 5
-0.0960 -0.1280 0.0000 0.0000 0.0960 0.1280 6
3 4 5 6
-0.3333 0.0000 -0.0720 -0.0960 1
0.0000 0.0000 -0.0960 -0.1280 2
0.3333 0.0000 0.0000 0.0000 3
0.0000 0.0000 0.0000 0.0000 4
0.0000 0.0000 0.0720 0.0960 5
0.0000 0.0000 0.0960 0.1280 6
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje
Metodo por ensamblaje

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Metodo por ensamblaje

  • 1. CURSO: ANALISIS ESTRUCTURAL AVANZADO TEMA: METODO POR ENSAMBLAJE PROFESOR: ING. ERLYN GIORDANY SALAZAR HUAMAN Solucionar la siguiente armadura: 6 5 4.00 ft 2 4 q1= 1 3 q2= 20.00 3.00 ft 10.00 MATRIZ DE RIGIDEZ DE LA BARRA 1 1
  • 2. Angulo= 0 C = 1 C² / L = 0.3333333 S = 0 CS/ L = 0 L = 3 ft S² / L = 0 1 2 3 4 0.3333 0.0000 -0.3333 0.0000 1 2 4 0.0000 0.0000 0.0000 0.0000 2 1 K 1= -0.3333 0.0000 0.3333 0.0000 3 10 0.0000 0.0000 0.0000 0.0000 4 20 6 5 EA MATRIZ DE RIGIDEZ DE LA BARRA 2 1 (0,0) (3,0) (3 4)
  • 3. 2 1 λx = C = 0.6 C² / L = 0.072 λx = S = 0.8 CS/ L = 0.096 L = 5 ft S² / L = 0.128 5 6 1 2 0.0720 0.0960 -0.0720 -0.0960 5 0.0960 0.1280 -0.0960 -0.1280 6 K2= -0.0720 -0.0960 0.0720 0.0960 1 -0.0960 -0.1280 0.0960 0.1280 2 1 2 3 4 5 6 1 2 Q1 0.405 0.096 -0.333 0.000 -0.072 -0.096 1 D1 0.4053 0.0960 Q2 0.096 0.128 0.000 0.000 -0.096 -0.128 2 D2 0.0960 0.1280 Q3 =EA -0.333 0.000 0.333 0.000 0.000 0.000 3 D3 MT = EA -0.3333 0.0000 MATRIZ TOTAL DE LA ESTRUCTURA EA (3,4) (0,0)
  • 4. Q4 0.000 0.000 0.000 0.000 0.000 0.000 4 D4 0.0000 0.0000 Q5 -0.072 -0.096 0.000 0.000 0.072 0.096 5 D5 -0.0720 -0.0960 Q6 -0.096 -0.128 0.000 0.000 0.096 0.128 6 D6 -0.0960 -0.1280 -10.00 0.405 0.096 D1 -20.00 0.096 0.128 D2 D1 = 15.000 /EA D2 = -167.5000 /EA R3 -0.333 0.000 15.00 ´/ EA R4 ´= EA 0.000 0.000 -167.50 ´/ EA R5 -0.072 -0.096 0.00 ´/ EA R6 -0.096 -0.128 0.00 ´/ EA R3 = -5 R4 = 0 R5 = 15 ´= EA
  • 5. R6 = 20 DNX DNY EA ´- C ´- S C S DFX L DFY 15 D1 q1 = EA -1 0 1 0 -167.50 D2 3 0 /EA D3 0 D4 q1 = -5.00 TN q3= EA -0.6 -0.8 0.6 0.8 15.00 D1 5 -167.50 /EA D2 q = BARRA 1 BARRA 2 FUERZAS DE LAS BARRAS
  • 7.
  • 8.
  • 9.
  • 10.
  • 11. 1 2 3 4 5 6 0.3333 0.0000 -0.3333 0.0000 0.0000 0.0000 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 2 K1 = EA -0.3333 0.0000 0.3333 0.0000 0.0000 0.0000 3 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 4 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 5 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 6 3
  • 12. 1 2 3 4 5 6 0.0720 0.0960 0.0000 0.0000 -0.0720 -0.0960 1 0.0960 0.1280 0.0000 0.0000 -0.0960 -0.1280 2 K2 = EA 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 3 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 4 -0.0720 -0.0960 0.0000 0.0000 0.0720 0.0960 5 -0.0960 -0.1280 0.0000 0.0000 0.0960 0.1280 6 3 4 5 6 -0.3333 0.0000 -0.0720 -0.0960 1 0.0000 0.0000 -0.0960 -0.1280 2 0.3333 0.0000 0.0000 0.0000 3
  • 13. 0.0000 0.0000 0.0000 0.0000 4 0.0000 0.0000 0.0720 0.0960 5 0.0000 0.0000 0.0960 0.1280 6