ARBE-CASTILLO-LUIS-MICHEL.xlsx
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Transcript of ARBE-CASTILLO-LUIS-MICHEL.xlsx
![Page 1: ARBE-CASTILLO-LUIS-MICHEL.xlsx](https://reader036.fdocument.pub/reader036/viewer/2022072001/563db792550346aa9a8c4ba6/html5/thumbnails/1.jpg)
DISEÑO DE ZAPATA AISLADA
DATOS DATOS
180 Tn 0.25 NPT +0.30
1.7 m 1.25 P P 0.30
2.1 Tn/m3 280 kg/cm2
65 Tn 0.00
210 kg/cm2 245 Tn HF= 2.00 DF= 1.70
500 kg/m2
3.5 kg/cm2
4200 kg/cm2
T NFC - 1.70
DIMENSIONAMIENTO DE LA COLUMNA
b.D= 4375 cm2 s
USARt2 t1 t1.t2 und
55 80 4400 cm2 OK0.55 0.8 0.44 m2
ESFUERZO NETO DEL TERRENO
30.30 Tn/m2 Pu= 387.08.09 m2
raiz2 de Asap 2.85 2.85 m2
T= 2.975 T= 3.00 m2
Lv1=Lv2 S= 2.725 S= 2.75 m2
del graficoLv1=(T-t1)/2= 1.10 OKLv2=(S-t2)/2= 1.10
REACCION NETA DEL TERRENO
46.91 Tn/m2
DIMENSIONAMIENTO DE LA ALTURA hz DE LA ZAPATApor punzonamiento
t1+d condicion del diseño
d/2 0.55 t2+d 2.75 βc=Dmayor/Dmenor= 1.45 < 2.00 ok0.80
3.00
PD= n=
DF= PS=
Ym= F´C=
PL= NTN ±
F´C= P=PD+PL=
S/CPISO=
σt=
FY=
σn=σt - yprom.hf - s/c=Azap=P/σn
como no existe excentricidad
Vu/Ø=VcVu/Ø=1/Ø{Pu-Wnu(0.80+d)(0.55+d)}………..(1)
𝑃𝑠/(𝑛.𝐹′𝑐)= 𝐿𝑣2𝐿𝑣1𝑡1
𝑡2
ℎ𝑧
𝑇=√𝐴𝑍+((𝑡1−𝑡2))/2𝑆=√𝐴𝑍−((𝑡1−𝑡2))/2
𝑊𝑛𝑢=𝑃𝑢/𝐴𝑠𝑎𝑝=
Vc=0.27(2+4/β)√(𝐹^′ 𝑐).𝑏𝑜.𝑑 ≤ 1.06√( ^ 𝐹 ′ 𝑐 .) .𝑏𝑜 𝑑Vc=1.06 √( ^ )𝐹 ′ 𝑐
Vc=1.06 √( ^ ).𝐹 ′ 𝑐 bo.d.............................(2)
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donde:bo=2*(t2+d)+2*(t1+d)=2.7+4d
igualando 1 y 2
formando una ecuacion de 2do grado de la ec.4
a= 507.73 DIAMETROb= 374.38467 3/4 " 0.75c= -366.36
d= 0.557
USARh= 0.6 m dprom= 50.59 cm
3/4 " 7.5 cm dprom= 0.51 m
VERIFICACION POR CORTANTEVdu=(Wnu*S)(Lv-d)= 62.94 Tn
83.92 Tn
107.72 tn > Vn ok
DISEÑO POR FLEXION
78.045 Tn-m
45.35 cm2REAJUSTANDO
3.88 cm As= 42.44 cm2a= 3.63 cm
VERIFICACION DE As min
25.04 cm2
14.89 15.00 USAR (para x)15.00 Ø 3/4 " @ 0.19
0.19
EN DIRECCION TRANSVERSAL46.30 cm2
16.24 17.00 USAR (para y)17.00 Ø 3/4 " @ 0.19
0.19
LONGITUD DE DESARROLLO DEL REFUERZO
Longitud disponible para cada barra y
Ld = 102.50 cm
Para barras en Traccion : x
Ab = 2.85 cm2Fc = 280.00 Kg/cm2Fy = 4200.00 Kg/cm2
db = 1.905 cmLd1 = 42.92 cmLd2 = 45.60 cm
Ø=
Vn=Vdu/Ø=
AST=AS*T/S=
Vc=1.06 √( ^ ).𝐹 ′ 𝑐 bo.d.............................(2)
Pu-Wnu(0.80+d)(0.55+d)=1.06 √(𝐹^′ 𝑐).bo.d......................(3)Pu-Wnu(0.80+d)(0.55+d)=1.06 √(𝐹^′ 𝑐).(2*(t2+d)+2*(t1+d)*d......................(4)
Vc=0.53*√(𝐹^′ 𝑐) ∗𝑏∗𝑑=
As = 𝑀𝑢/(Ø∗𝐹𝑦∗(𝑑− 𝑎/2))=a = (𝐴𝑠∗𝐹𝑦)/(0.85∗𝐹^′ 𝑐∗𝑏)=
n= 𝐴𝑠/𝐴Ø=s= (𝑠−2𝑟−Ø)/(𝑛−1)=
n= 𝐴𝑠/𝐴Ø=s= (𝑠−2𝑟−Ø)/(𝑛−1)=
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Ld3 = 30.00 cm Ld = 45.604 cm
Usar Ld = 36.483 cm < Ldisp = 102.500 cm conforme
Transferencia de fuerza en la interfase de columna y cimentacion
a.- Transferencia al Aplastamiento sobre la columna para Ø=0.65
f ' c = 280 kg / cm2
F'c = 280.00 Kg/cm2Pu = 387.0 TnPn = 595.38 Tn
P nb = 0.85*f ' c*Ac
Pnb = 1047.2 TnPn < Pnb conforme
b.- Resistencia al Aplastamiento en el concreto de la Cimentacion
Pn = 595.38Xo = 2.06 mtA2 = 6.19 mtA1 = 0.44 mt
(A2/A1)^0,5 = 3.75 usar 2.00
Ao = 0.88Pnb = 1570.8 Tn
Pn < Pnb conforme
Dowells entre columna y cimentacion A s min = 0.005 Acol
si Pn < Pnb usar Asmin = 22.00 cm2 0.0022m2
para zonas sismicas
EXCENTRICIDAD DE LA CARGA
DATOS
180 Tn COLUMNA
1.7 m 0.55 0.8
2.1 Tn/m3 e= 0
65 Tn e= 0.25
210 kg/cm2 e= 0.7
500 kg/m2 e= 0.9
3.5 kg/cm2
4200 kg/cm2 area= 68.0625
Resistencia al Aplastamiento de la columna Pnb
PD=
DF=
Ym=
PL=
F´C=
S/CPISO=
σt=
FY=
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1/4 " 0.75 1/4 " 0.323/8 " 0.75 3/8 " 0.711/2 " 0.75 1/2 " 1.275/8 " 0.75 5/8 " 1.983/4 " 0.75 3/4 " 2.851 " 0.75 1 " 5.1