Calculo de fuerzas sismicas estructuras

Páginas: 5 (1024 palabras) Publicado: 5 de febrero de 2012
GONZALEZ RIZO IVAN RAZIEL 6°A T/M 02-DIC-11

CARGA ENTREPISO = 900 KG
CARGA AZOTEA = 800 KG
CS = 0.06
A = 120 M2
CALCULO LOSA ENTREPISO
w1= w2= w3= w4= 120 m2 x 900 = 108,000 kg
CALCULO LOSA AZOTEA
W5= 120 m2 X 800 = 96,000 KG
POR TRIANGULOS IGUALES
CS4= 13 X 0.06/ 16 = 0.0487
CS3= 10 X 0.06/ 16 = 0.0375
CS2= 7 X 0.06/ 16 = 0.0262
CS1= 4 X 0.06/ 16 = 0.0150CALCULO FUERZAS SISMICAS
F1= CS1 x w1= 0.0150 x 108,000 = 1620 kg
F2= CS2 x w2= 0.0262 x 108,000 = 2830 kg
F3= CS3 x w3= 0.0375 x 108,000 = 4050 kg
F4= CS4 x w4= 0.0487 x 108,000 = 5260 kg
F5= CS5 x w5= 0.06 x 96,000 = 5760 kg
CORTANTE EN LA BASE DEL EDIFICIO
V=Cs x wT = 0.06 x 528,000 = 31,680 kg
COEFICIENTE DE CORRECCION
Ks = V/ Suma F= 31,680/19520 = 1.623
FUERZAS SISMICAS FINALES YCORTANTES SISMICAS
5760 X 1.623 = 9,348.48
5260 X 1.623 = 8,536.98
4050 X 1.623 = 6,573.15 Suma de Fuerzas= 31,680.96 kg
2830 X 1.623 = 4,593.09
1620 X 1.623 = 2,629.26

SE ANALIZARA EL NIVEL 5

1.- FUERZA TANGENCIAL
T2 = V5/n = 9,348.48/3 = 3,116.16 KG
T1 = T2/2 = 3,116.16/2 = 1,558.08 KG
2.- CALCULO DE MOMENTOS (M1 Y M2)
M1 = T1 x 1.5 = 1558.08 x 1.5 = 2,337.12 KG
M2= 2M1=2(2,337.12) = 4,674.24 KG

CORTANTE EN TRABES
V = 2M/L
V1= V2= 2X 2,337.12/ 8 = 584.28 KG
AXIAL EN COLUMNAS
N1= V1= 584.28 KG N2= V2 - V1 = 0 KG

ANALISIS NIVEL 1

DIAGRAMA DE CORTANTES
V1= 31,680.96 KG
V2 = 29,05I.70 KG
FUERZA TANGENCIAL SUPERIOR
T2 = V2/n = 29,051.70/3 = 9683.9 KG
T1 = T2/2 = 9,683.9/2 = 4841.95 KG
FUERZA TANGENCIAL INFERIOR
T2 = V1/n = 31,680.96/3= 10,560.32 KG
T1 = T2/2 = 10,560.32/2 = 5,280.16 KG

NUDO 1
M1 = T1 x 1.5 = 4,841.95 X 1.5 = 7,262.92 K-M
M2= T2 x 2.0 = 5,280.16X 2 = 10,560.32 K-M
M3= M1+ M2 = 17,823.24 K-M
NUDO 2
M3= M1+ M2 = 17,823.24 K-M
M4 = T1 x 1.5 = 9,683.9 X 1.5 = 14,525.85 K-M
M5= T2 x 2.0 = 10,560.32 X 2 = 21,120.64 K-M
NUDO 2

2M3= M4 + M5
2M3 = 2(17,823.249) = 35,646.48
CORTANTES
V1= 2M3/ L1 =4455.81 K
V2= 2M3/ L2 = 4455.81 K
AXIAL
N1= V1= 4,455.81 KG N2= V2 - V1 = 0 KG

SE ANALIZARA EL NIVEL 2

DIAGRAMA DE CORTANTES
V3= 24,458.61 KG
V2 = 29,05I.70 KG
FUERZA TANGENCIAL SUPERIOR
T2 = V3/n = 24,458.61/3 = 8,152.87 KG
T1 = T2/2 = 8,152.87/2 = 4,076.435 KG
FUERZA TANGENCIAL INFERIOR
T2 = V2/n = 29,051.70/3 = 9,683.9 KG
T1 = T2/2 = 9,683.9/2 = 4,841.95 KG
M1 =T1 x 1.5 = 4,076.435 X 1.5 = 6,114.652 K-M
M2= T2 x 1.5 = 4,841.95 X 1.5 = 7,262.925 K-M
M3= M1+ M2 = 13,377.577 K-M
M4 = T1 x 1.5 = 8,152.87 X 1.5 = 12,229.305 K-M
M5= T2 x 1.5 = 9,683.9 X 1.5 = 14,525.85 K-M
2M3 = 2(13,377.577) = 26,755.154 K-M
CORTANTE
V1= 2M3/ L1 = 3,344.39
V2= 2M3/ L2 = 3,344.39 K

SE ANALIZARA EL NIVEL 3

DIAGRAMA DE CORTANTES
V4= 17,885.46 KG
V3= 24,458.61KG
FUERZA TANGENCIAL SUPERIOR
T2 = V4/n = 17,885.46/3 = 5,961.82 KG
T1 = T2/2 = 5,961.82/2 = 2,980.91 KG
FUERZA TANGENCIAL INFERIOR
T2 = V3/n = 24,458.61/3 = 8,152.87 KG
T1 = T2/2 = 8,152.87/2 = 4,076.435 KG

M1 = T1 x 1.5 = 2,980.91 X 1.5 = 4,471.365 K-M
M2= T2 x 1.5 = 4,076.435 X 1.5 = 6,114.65 K-M
M3= M1+ M2 = 10,586.015 K-M
M4 = T1 x 1.5 = 5,961.82 X 1.5 = 8,942.73 K-M
M5= T2 x1.5 = 8,152.87 X 1.5 = 12,229.305 K-M
2M3 = 2(10586.015) = = 21,172.03 K-M
CORTANTES
V1= 2M3/ L1 = 2646.50 K
V2= 2M3/ L2 = 2646.50 K
SE ANALIZARA EL NIVEL 4

DIAGRAMA DE CORTANTES
V5= 9,348.48 KG
V4= 17,885.46 KG
FUERZA TANGENCIAL SUPERIOR
T2 = V5/n = 9,348.48/3 = 3,116.16 KG
T1 = T2/2 = 3,116.16/2 = 1,558.08 KG
FUERZA TANGENCIAL INFERIOR
T2 = V4/n = 17,885.46/3 = 5,961.82 KGT1 = T2/2 = 5,961.82/2 = 2,980.91 KG

M1 = T1 x 1.5 = 1,558.08 X 1.5 = 2,337.12 K-M
M2= T2 x 1.5 = 2,980.91 X 1.5 = 4,471.365 K-M
M3= M1+ M2 = 6,808.485 K-M
M4 = T1 x 1.5 = 3,116.16 X 1.5 = 4,674.24 K-M
M5= T2 x 1.5 = 5961.82 X 1.5 = 8,942.73 K-M
2M3 = 2(6808.485) = 13,616.97 K-M
CORTANTES
V1= 2M3/ L1 = 1,702.12 K
V2= 2M3/ L2 = 1,702.12 K

DIAGRAMA DE MOMENTOS

DIAGRAMA DE...
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