Analisis Sismico
Páginas: 4 (873 palabras)
Publicado: 29 de enero de 2013
Tipo de suelo zona ll terreno de transición
Uso grupo B
Tipo 1
Coeficiente sísmico ( C ) sin reducir por ductiviidad análisis estatico =0.32
Reducción por ductivilidad Q=2 cs=C/Q = 0.32/2 =0.16
Método de análisis
Analisis estatico
Fi= C/Q Wihi/∑Wihi ∑Wi
A = 10 x 25 = 250
A1 = 10 x 6.16 = 61.6
A2 = 1.72 x 9.61 = 16.52
A3 = 5.22 x 9.23 = 48.18
A4 = 3.23 x 2.72 = 8.78
A5 =3.08x 2.58 = 7.94
143.02
A total = 106.98
Wi = peso del nivel 1,2
1 calculamos fuerza sísmica
2 revisamos la resistencia
Coeficiente sísmico
C=0.32
Cs = 0.16Cw= 0.16 [ ( 106.98)(670 kg/m²)+ (106.98)(740 kg/m²)]
Cw = 0.16 (71.6766 + 79.1652)
=0.16 (150.8418)
= 24.14 toneladas cortante
Calcular Fi =C/Q = Wi/∑ Wihi
∑w=csw Wihi/∑wihi
Fi = Wi/∑ Wihi xcs x∑w
Fi = 215.0298/690.201 = 0.16 x 150.8414
Fi = 215.0298/690.201 x 24.15
Fi = 0.32 x 24.15 = 7.72 toneladas F1 =7.72
=474.9912/690.201 x 24.15
=.68 x 24.15 =16.42 toneladas F2=16.42Revisión de la resistencia
Vr = fr (0.7 v*)
Fr = factor de reducción 0.6
V* = 3.51 kg/cm2
Vr= 0.6(0.7x3.5)=1.5 kg/cm
nivel | entrepiso | altura de entrepiso (m) | Wi(ton) | hi(m) | Wi hi |Fi(ton) | Vi(ton) | |
2 | 2 | 3 | 79.1652 | 6 | 475.1712 | 16.42 | 16.42 | |
1 | 1 | 3 | 71.6766 | 3 | 215.0298 | 7.72 | 24.15 | |
total | | | 150.8413 | | 690.201 | 24.15 | | |Planta baja sentido OX
muro | h/l | Vr(kg/cm²) | AT(cm²) | Vr(kg) |
A | 3/4.78 = 0.62 | 1.5 | 478X15=7170 | 10755 |
B | 3/2.65 = 1.13 | 1.5 | 265X15=3975 | 5962.5 |
C | 3/1.3 = 2.3 | 0.5 |130X15=1950 | 975 |
D | 3/0.7=4.2 | 0.15 | 70X15=1050 | 157.5 |
E | 3/1.1=2.72 | 0.35 | 110X15=160 | 577.5 |
F | 3/1=3 | 0.3 | 100X15=1500 | 450 |
G | 3/0.6=5 | 0.11 | 60X15=900 | 99 |
H |3/0.8=3.75 | 0.2 | 80X15=1200 | 240 |
I | 3/3.50=0.85 | 1.5 | 350X15=5250 | 7875 |
J | 3/3.23=0.92 | 1.5 | 323X15=4845 | 7267.5 |
K | 3/3.23=0.92 | 1.5 | 323X15=4845 | 7267.5 |
L | 3/1.1=2.72 | 0.43...
Uso grupo B
Tipo 1
Coeficiente sísmico ( C ) sin reducir por ductiviidad análisis estatico =0.32
Reducción por ductivilidad Q=2 cs=C/Q = 0.32/2 =0.16
Método de análisis
Analisis estatico
Fi= C/Q Wihi/∑Wihi ∑Wi
A = 10 x 25 = 250
A1 = 10 x 6.16 = 61.6
A2 = 1.72 x 9.61 = 16.52
A3 = 5.22 x 9.23 = 48.18
A4 = 3.23 x 2.72 = 8.78
A5 =3.08x 2.58 = 7.94
143.02
A total = 106.98
Wi = peso del nivel 1,2
1 calculamos fuerza sísmica
2 revisamos la resistencia
Coeficiente sísmico
C=0.32
Cs = 0.16Cw= 0.16 [ ( 106.98)(670 kg/m²)+ (106.98)(740 kg/m²)]
Cw = 0.16 (71.6766 + 79.1652)
=0.16 (150.8418)
= 24.14 toneladas cortante
Calcular Fi =C/Q = Wi/∑ Wihi
∑w=csw Wihi/∑wihi
Fi = Wi/∑ Wihi xcs x∑w
Fi = 215.0298/690.201 = 0.16 x 150.8414
Fi = 215.0298/690.201 x 24.15
Fi = 0.32 x 24.15 = 7.72 toneladas F1 =7.72
=474.9912/690.201 x 24.15
=.68 x 24.15 =16.42 toneladas F2=16.42Revisión de la resistencia
Vr = fr (0.7 v*)
Fr = factor de reducción 0.6
V* = 3.51 kg/cm2
Vr= 0.6(0.7x3.5)=1.5 kg/cm
nivel | entrepiso | altura de entrepiso (m) | Wi(ton) | hi(m) | Wi hi |Fi(ton) | Vi(ton) | |
2 | 2 | 3 | 79.1652 | 6 | 475.1712 | 16.42 | 16.42 | |
1 | 1 | 3 | 71.6766 | 3 | 215.0298 | 7.72 | 24.15 | |
total | | | 150.8413 | | 690.201 | 24.15 | | |Planta baja sentido OX
muro | h/l | Vr(kg/cm²) | AT(cm²) | Vr(kg) |
A | 3/4.78 = 0.62 | 1.5 | 478X15=7170 | 10755 |
B | 3/2.65 = 1.13 | 1.5 | 265X15=3975 | 5962.5 |
C | 3/1.3 = 2.3 | 0.5 |130X15=1950 | 975 |
D | 3/0.7=4.2 | 0.15 | 70X15=1050 | 157.5 |
E | 3/1.1=2.72 | 0.35 | 110X15=160 | 577.5 |
F | 3/1=3 | 0.3 | 100X15=1500 | 450 |
G | 3/0.6=5 | 0.11 | 60X15=900 | 99 |
H |3/0.8=3.75 | 0.2 | 80X15=1200 | 240 |
I | 3/3.50=0.85 | 1.5 | 350X15=5250 | 7875 |
J | 3/3.23=0.92 | 1.5 | 323X15=4845 | 7267.5 |
K | 3/3.23=0.92 | 1.5 | 323X15=4845 | 7267.5 |
L | 3/1.1=2.72 | 0.43...
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