Armadura
11
Ing. Juan Manuel Urteaga García Modulo de Elasticidad E= 29000.00 Ksi Universidad Privada Antenor Orrego Escuela Profesional de Ingeniería Civil Análisis Estructural 2 1 1 2 16' 12' 1 3 2 3 16' 12' 15 Kb
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Transcript of Armadura
Ing. Juan Manuel Urteaga García
Modulo de Elasticidad E= 29000.00 Ksi
Universidad Privada Antenor OrregoEscuela Profesional de Ingeniería Civil
Análisis Estructural
21
1
2
16'
12'1
3
2
3
16'
12'
15 Kb
1 12.00 144.00 3.002 16.00 192.00 3.003 20.00 240.00 3.00
Elemento D1 D2 D30 0 10 0 00 0 01 0 00 0 3/5
4/5 3/5 0
K1,1 = (1)(29000)(3)/(192)+(4/5)(29000)(3)/(192)(4/5) = 685.13 KlbK2,1 = (29000)(3)/(240)(4/5)(3/5)) = 174.00 KlbK3,1 = -(29000)(3)/(240)(4/5)(3/5)) = -174.00 Klb
Longitud (pies)
2
Area (in2)
Longitud (in)
3
Elemento
1
11
1
4/5
3/5
3/5
K3,1
K1,1
K2,1
EA2/L2(1)
EA3/L3 (4/5)
K1,2 = (29000)(3)/(240)(3/5)(4/5)) = 174.00 KlbK2,2 = (29000)(3)/(240)(3/5)(3/5)) = 130.50 KlbK3,2 = -(29000)(3)(240)(3/5)(3/5)) = -130.50 Klb
K1,3 = -(29000)(3)/(240)(3/5)(4/5)) = -174.00 KlbK2,3 = -(29000)(3)/(240)(3/5)(3/5)) = -130.50 KlbK3,3 = (29000)(3)/(240)(3/5)(3/5)+(29000)(3)/(144) = 734.67 Klb
K3,2
K1,2
K2,2
EA3/L3 (3/5)
K3,3
K1,3
K2,3
EA3/L3 (3/5)
EA1/L1 (1)
Solución del Sistema de Ecuaciones
{Q} = {K}{D}
0 685 174 -174 D1
-15 174 131 -131 D2
0 -174 -131 735 D3
{K}-1{Q} = {K}-1{K}{D}
{K}-1{Q} = {I}{D}
2.21E-03 -2.94E-03 0.00E+00 0 D1
-2.94E-03 1.32E-02 1.66E-03 -15 D2
0.00E+00 1.66E-03 1.66E-03 0 D3
Desplazamientos Globales
D1 = 4.41E-02 inD2 = -1.99E-01 inD3 = -2.48E-02 in
=
=
Desplazamientos Locales
d1,1 = D3 = -2.48E-02 0.00000 0.00000 1.00000 4.41E-02 -2.48E-02d2,1 = 0.00E+00 0.00000 0.00000 0.00000 -1.99E-01 0.00E+00
-2.48E-02
d1,2 = 0.00E+00 0.00000 0.00000 0.00000 4.41E-02 0.00E+00d2,2 = D1 = 4.41E-02 1.00000 0.00000 0.00000 -1.99E-01 4.41E-02
-2.48E-02
d1,3 = (3/5)D3 = -1.49E-02 0.00000 0.00000 0.60000 4.41E-02 -1.49E-02d2,3 = -8.39E-02 0.80000 0.60000 0.00000 -1.99E-01 -8.39E-02
-2.48E-02
Esfuerzos Locales
q1,1 = (29000)(3)(2.48E-02)/(144) - (29000)(3)(0)/(144) = -15.00 Klbq2,1 = -(29000)(3)(2.48E-02)/(144)+(29000)(3)(0)/(144) = 15.00 Klb
q1,2 = (29000)(3)(0)/(194) - (29000)(3)(4.41E-02)/(194) = -20.00 Klbq2,2 = -(29000)(3)(0)/((16)(12))+(29000)(3)(4.41E-02)/((16)(12)) = 20.00 Klb
q1,3 = (29000)(3)(-1.49E-02)/((20)(12)) - (29000)(3)(-8.39E-02)/((20)(12)) = 25.00 Klbq2,3 = -(29000)(3)(-1.49E-02)/((20)(12)) + (29000)(3)(-8.39E-02)/((20)(12)) = -25.00 Klb
(4/5)D1+(3/5)D2 =
Diagramas
25 Klb
15.Klb25 Klb
20 Klb 20 Klb
15.Klb
20 Klb
20 Klb
15 Klb
