fundaciones de lindero con tensor
TRANSCRIPT
![Page 1: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/1.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
NOTA:
PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO
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![Page 2: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/2.jpg)
DATOS: Ejemplo Libro Fratelli
250.00 3,500.00
2.60 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 78,000.00 Fmu = 1.5077
Pu (kg) = 117,600.60
bx (cm) = 50.00 by (cm) = 50.00 h1 (m) = 2.80
L (tensor) (m) = 4.00
h1 (m) ν1.15
1.20 1.20
1.30
36,000.00 Bx escogido = 120.00 cm
By = 300.00 cm By escogido = 300.00 cm
e x = (Bx-bx)/2 = 35.00 cm 3.27
Mu a-a = 2,401,012.25 kg*cm Mu b-b = 3,062,515.63 kg*cm
Mu eje momento máximo = 3,062,515.63 kg*cm
Adoptar d = 40.00 cm r = 10.00 cm
h= 50.00 cm
Verificación a Corte:
8.38 kg/cm2
v u 1-1 = 2.88 kg/cm2 v u 2-2 = 8.17 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
16.76 kg/cm2
box = 70.00 cm boy = 90.00 cm
97,020.50 kg
v u = 12.41 kg/cm2 < 16.76 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 3: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/3.jpg)
Acero en Fundación Perpendicular al Eje de Momento Menor:
Mu eje momento menor = 2,401,012.25 kg*cm
As menor = Mu menor / (0.90*0.90*Fy*(d-2))
As menor = 22.29 cm2 As menor / B perpendicular = 7.43 cm2/m
As min = 0.002*100*h = 10.00 cm2
As escogido = 10.00 cm2
Ø escogido = 0.50 Pulg Sep. = 12.67 cm
Sep. Escogida= 12.50 cm
Acero en Fundación Perpendicular al Eje de Momento Mayor:
Mu eje momento mayor = 3,062,515.63 kg*cm
As mayor = Mu mayor / (0.90*0.90*Fy*d)
As mayor = 27.01 cm2 As mayor / B perpendicular = 22.51 cm2/m
As min = 0.002*100*h = 10.00 cm2
As escogido = 22.51 cm2
Ø escogido = 0.75 Pulg Sep. = 12.66 cm
Sep. Escogida= 12.50 cm
Diseño del Tensor:
M volc. = P*ex = 2,730,000.00 kg*cm H = M volc. / h1 = 9,750.00 kg
28,389.68 kg O.K.
10.28 kg/cm2 0.015
8.80
n escogido = 9.00 Por Norma
777.66 kg/cm2 ≤ Fy / 2 = 1,750.00 kg/cm2
Acero Tensor = Ast = H / fs = 12.54 cm2
Ø escogido = 0.88 Pulg Nº cabillas = 3.23
N escogido = 4.00 AsT = 15.52 cm2
835.84 cm2 Tx*Ty = 28.91 cm
Tx escogido = 30.00 cm Ty escogido = 30.00 cm
0.0172 ≈ 0.015, asumido
0.15 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 117,600.60 kgMuc = 33,810.17 kg*m
bx (cm) = 50.00 by (cm) = 50.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =
![Page 4: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/4.jpg)
PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO
![Page 5: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/5.jpg)
PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO
![Page 6: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/6.jpg)
FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
![Page 7: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/7.jpg)
DATOS: EJE: A6 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg), (Pm + Pv) = 42,603.33 Fmu = 1.5000
Pu (kg) = 63,905.00
bx (cm) = 60.00 by (cm) = 60.00 h1 (m) = 1.50
L1(tensor) (m) = 2.59 L2(tensor) (m) = 3.87
h1 (m) ν1.15
1.20 1.15
1.30110.67
12,248.46 Bx escogido = 115.00 cm(Ver Anexo)
By = 106.51 cm By escogido = 115.00 cm
e = (Ver Anexo) 38.89 cm 4.83
Mu a-a = 840,489.67 kg*cm Mu b-b = 840,489.67 kg*cm
Mu eje momento máximo = 840,489.67 kg*cm
Adoptar d = 27.50 cm r = 7.50 cm
h= 35.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 5.68 kg/cm2 v u 2-2 = 5.68 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 73.75 cm boy = 73.75 cm
37,622.71 kg
v u = 10.91 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 mBx*By ≈
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 8: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/8.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu eje a-a = 840,489.67 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As menor = 8.98 cm2 As a-a / By= 7.81 cm2/m
As min = 0.002*100*h = 7.00 cm2
As escogido = 7.81 cm2
Ø escogido = 0.50 Pulg Sep. = 16.22 cm
Sep. Escogida= 15.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu eje b-b = 840,489.67 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 8.98 cm2 As b-b / Bx = 7.81 cm2/m
As min = 0.002*100*h = 7.00 cm2
As escogido = 7.81 cm2
Ø escogido = 0.50 Pulg Sep. = 16.22 cm
Sep. Escogida= 15.00 cm
Diseño de los Tensores, (Son Dos), (para H1 = H2) :
M volc. = P*e = 1,656,843.63 kg*cm H = M volc. / h1 = 11,045.62 kg
7,810.44 kg
15,506.35 kg O.K. NOTA: Si no cumple cambiar seccion pedestal,para disminuir excentricidad, o
9.42 kg/cm2 0.015 profundizar la fundación.
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensores = Ast1 = H1 / fs = 10.82 cm2 = Ast2 = H2 / fs
Ø escogido = 0.63 Pulg Nº cabillas = 5.46
N escogido = 6.00 As = 11.88 cm2
721.03 cm2 Tx*Ty = 26.85 cm Ac1 = Ac2
Tx escogido = 30.00 cm Ty escogido = 30.00 cm
0.0132 ≈ 0.015, asumido
0.09 cm
0.13 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 63,905.00 kgMuc = 19,053.70 kg*m
bx (cm) = 60.00 by (cm) = 60.00
H1 ≈ H2 ≈ H*cos (45o) =
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area conc. Tensor 1 ≥ H1*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L1 = Ɛs*L1 = fs*L1 / Es =
Alargamiento del Tensor = ∆L2 = Ɛs*L2 = fs*L2 / Es =
![Page 9: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/9.jpg)
FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
![Page 10: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/10.jpg)
DATOS: EJE: B1 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg), (Pm + Pv) = 91,586.67 Fmu = 1.5000
Pu (kg) = 137,380.00
bx (cm) = 60.00 by (cm) = 60.00 h1 (m) = 1.50
L1(tensor) (m) = 4.61 L2(tensor) (m) = 3.87
h1 (m) ν1.15
1.20 1.15
1.30162.27
26,331.17 Bx escogido = 165.00 cm(Ver Anexo)
By = 159.58 cm By escogido = 165.00 cm
e = (Ver Anexo) 65.76 cm 5.05
Mu a-a = 4,589,740.91 kg*cm Mu b-b = 4,589,740.91 kg*cm
Mu eje momento máximo = 4,589,740.91 kg*cm
Adoptar d = 47.50 cm r = 7.50 cm
h= 55.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 7.19 kg/cm2 v u 2-2 = 7.19 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 83.75 cm boy = 83.75 cm
101,986.36 kg
v u = 15.08 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 mBx*By ≈
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 11: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/11.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu eje a-a = 4,589,740.91 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As menor = 28.40 cm2 As a-a / By= 17.21 cm2/m
As min = 0.002*100*h = 11.00 cm2
As escogido = 17.21 cm2
Ø escogido = 0.63 Pulg Sep. = 11.50 cm
Sep. Escogida= 11.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu eje b-b = 4,589,740.91 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 28.40 cm2 As b-b / Bx = 17.21 cm2/m
As min = 0.002*100*h = 11.00 cm2
As escogido = 17.21 cm2
Ø escogido = 0.63 Pulg Sep. = 11.50 cm
Sep. Escogida= 11.00 cm
Diseño de los Tensores, (Son Dos), (para H1 = H2) :
M volc. = P*e = 6,022,739.20 kg*cm H = M volc. / h1 = 40,151.59 kg
28,391.46 kg
33,334.82 kg O.K. NOTA: Si no cumple cambiar seccion pedestal,para disminuir excentricidad, o
9.42 kg/cm2 0.015 profundizar la fundación.
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensores = Ast1 = H1 / fs = 39.31 cm2 = Ast2 = H2 / fs
Ø escogido = 0.75 Pulg Nº cabillas = 13.79
N escogido = 12.00 As = 34.20 cm2
2,621.00 cm2 Tx*Ty = 51.20 cm Ac1 = Ac2
Tx escogido = 50.00 cm Ty escogido = 50.00 cm
0.0137 ≈ 0.015, asumido
0.16 cm
0.13 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 137,380.00 kgMuc = 57,216.02 kg*m
bx (cm) = 60.00 by (cm) = 60.00
H1 ≈ H2 ≈ H*cos (45o) =
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area conc. Tensor 1 ≥ H1*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L1 = Ɛs*L1 = fs*L1 / Es =
Alargamiento del Tensor = ∆L2 = Ɛs*L2 = fs*L2 / Es =
![Page 12: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/12.jpg)
FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
h
Bx
h1
Ty
bx
P
H
R
ex
H
Muc
h1´
Bx
Bybx
by
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d/2
d/2
d/2
bx+d/2
H
H
H1
H2
![Page 13: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/13.jpg)
DATOS: EJE: D1 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg), (Pm + Pv) = 86,714.67 Fmu = 1.5000
Pu (kg) = 130,072.00
bx (cm) = 120.00 by (cm) = 60.00 h1 (m) = 1.50
L1(tensor) (m) = 4.61 L2(tensor) (m) = 3.87
h1 (m) ν1.15
1.20 1.15
1.30157.89
24,930.47 Bx escogido = 195.00 cm(Ver Anexo)
By = 127.85 cm By escogido = 135.00 cm
e = (Ver Anexo) 53.03 cm 4.94
Mu a-a = 1,876,038.46 kg*cm Mu b-b = 2,709,833.33 kg*cm
Mu eje momento máximo = 2,709,833.33 kg*cm
Adoptar d = 32.50 cm r = 7.50 cm
h= 40.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 7.60 kg/cm2 v u 2-2 = 7.60 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 136.25 cm boy = 76.25 cm
78,739.57 kg
v u = 13.41 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 mBx*By ≈
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 14: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/14.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu eje a-a = 1,876,038.46 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As menor = 16.97 cm2 As a-a / By= 12.57 cm2/m
As min = 0.002*100*h = 8.00 cm2
As escogido = 12.57 cm2
Ø escogido = 0.63 Pulg Sep. = 15.75 cm
Sep. Escogida= 15.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu eje b-b = 2,709,833.33 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 24.51 cm2 As b-b / Bx = 12.57 cm2/m
As min = 0.002*100*h = 8.00 cm2
As escogido = 12.57 cm2
Ø escogido = 0.63 Pulg Sep. = 15.75 cm
Sep. Escogida= 15.00 cm
Diseño de los Tensores, (Son Dos), (para H1 = H2) :
M volc. = P*e = 4,598,478.77 kg*cm H = M volc. / h1 = 30,656.53 kg
21,677.44 kg
31,561.56 kg O.K. NOTA: Si no cumple cambiar seccion pedestal,para disminuir excentricidad, o
9.42 kg/cm2 0.015 profundizar la fundación.
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensores = Ast1 = H1 / fs = 30.02 cm2 = Ast2 = H2 / fs
Ø escogido = 0.75 Pulg Nº cabillas = 10.53
N escogido = 12.00 As = 34.20 cm2
2,001.18 cm2 Tx*Ty = 44.73 cm Ac1 = Ac2
Tx escogido = 45.00 cm Ty escogido = 45.00 cm
0.0169 ≈ 0.015, asumido
0.16 cm
0.13 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 130,072.00 kgMuc = 50,583.27 kg*m
bx (cm) = 120.00 by (cm) = 60.00
H1 ≈ H2 ≈ H*cos (45o) =
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area conc. Tensor 1 ≥ H1*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L1 = Ɛs*L1 = fs*L1 / Es =
Alargamiento del Tensor = ∆L2 = Ɛs*L2 = fs*L2 / Es =
![Page 15: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/15.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
![Page 16: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/16.jpg)
DATOS: EJE: A7 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 58,544.67 Fmu = 1.5000
Pu (kg) = 87,817.00
bx (cm) = 60.00 by (cm) = 60.00 h1 (m) = 1.50
L (tensor) (m) = 2.59
h1 (m) ν1.15
1.20 1.15
1.30
16,831.59 Bx escogido = 110.00 cm(Ver Anexo)
By = 153.01 cm By escogido = 155.00 cm
e x = (Bx-bx)/2 = 25.00 cm 5.15
Mu a-a = 997,920.45 kg*cm Mu b-b = 639,151.96 kg*cm
Mu momento máximo = 997,920.45 kg*cm
Adoptar d = 22.50 cm r = 7.50 cm
h= 30.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 7.41 kg/cm2 v u 2-2 = 6.73 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 71.25 cm boy = 82.50 cm
57,541.38 kg
v u = 13.37 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 17: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/17.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu a-a = 997,920.45 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As a-a = 13.04 cm2 As a-a / By= 8.41 cm2/m
As min = 0.002*100*h = 6.00 cm2
As escogido = 8.41 cm2
Ø escogido = 0.50 Pulg Sep. = 15.06 cm
Sep. Escogida= 15.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu b-b = 639,151.96 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 8.35 cm2 As b-b / Bx = 7.59 cm2/m
As min = 0.002*100*h = 6.00 cm2
As escogido = 7.59 cm2
Ø escogido = 0.50 Pulg Sep. = 16.69 cm
Sep. Escogida= 15.00 cm
Diseño del Tensor:
M volc. = P*ex = 1,463,616.67 kg*cm H = M volc. / h1 = 9,757.44 kg
21,308.52 kg O.K.
9.42 kg/cm2 0.015
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensor = Ast = H / fs = 13.51 cm2
Ø escogido = 0.63 Pulg Nº cabillas = 6.83
N escogido = 8.00 AsT = 15.83 cm2
900.77 cm2 Tx*Ty = 30.01 cm
Tx escogido = 35.00 cm Ty escogido = 35.00 cm
0.0129 ≈ 0.015, asumido
0.09 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 87,817.00 kgMuc = 17,563.40 kg*m
bx (cm) = 60.00 by (cm) = 60.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =
![Page 18: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/18.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
NOTA:
PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
![Page 19: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/19.jpg)
DATOS: EJE: C8 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 109,469.33 Fmu = 1.5000
Pu (kg) = 164,204.00
bx (cm) = 50.00 by (cm) = 110.00 h1 (m) = 1.50
L (tensor) (m) = 2.59
h1 (m) ν1.15
1.20 1.15
1.30
31,472.43 Bx escogido = 125.00 cm(Ver Anexo)
By = 251.78 cm By escogido = 255.00 cm
e x = (Bx-bx)/2 = 37.50 cm 5.15
Mu a-a = 3,694,590.00 kg*cm Mu b-b = 1,692,347.60 kg*cm
Mu momento máximo = 3,694,590.00 kg*cm
Adoptar d = 37.50 cm r = 7.50 cm
h= 45.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 6.06 kg/cm2 v u 2-2 = 5.66 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 68.75 cm boy = 147.50 cm
111,964.59 kg
v u = 12.32 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 20: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/20.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu a-a = 3,694,590.00 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As a-a = 28.96 cm2 As a-a / By= 11.36 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 11.36 cm2
Ø escogido = 0.50 Pulg Sep. = 11.15 cm
Sep. Escogida= 11.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu b-b = 1,692,347.60 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 13.27 cm2 As b-b / Bx = 10.61 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 10.61 cm2
Ø escogido = 0.50 Pulg Sep. = 11.94 cm
Sep. Escogida= 11.00 cm
Diseño del Tensor:
M volc. = P*ex = 4,105,100.00 kg*cm H = M volc. / h1 = 27,367.33 kg
39,843.58 kg O.K.
9.42 kg/cm2 0.015
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensor = Ast = H / fs = 37.90 cm2
Ø escogido = 0.75 Pulg Nº cabillas = 13.30
N escogido = 14.00 AsT = 39.90 cm2
2,526.46 cm2 Tx*Ty = 50.26 cm
Tx escogido = 50.00 cm Ty escogido = 50.00 cm
0.0160 ≈ 0.015, asumido
0.09 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 164,204.00 kgMuc = 43,103.55 kg*m
bx (cm) = 50.00 by (cm) = 110.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =
![Page 21: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/21.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
![Page 22: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/22.jpg)
DATOS: EJE: B8 - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 52,898.00 Fmu = 1.5000
Pu (kg) = 79,347.00
bx (cm) = 60.00 by (cm) = 60.00 h1 (m) = 1.50
L (tensor) (m) = 3.87
h1 (m) ν1.15
1.20 1.15
1.30
15,208.18 Bx escogido = 105.00 cm(Ver Anexo)
By = 144.84 cm By escogido = 145.00 cm
e x = (Bx-bx)/2 = 22.50 cm 5.21
Mu a-a = 765,131.79 kg*cm Mu b-b = 494,208.69 kg*cm
Mu momento máximo = 765,131.79 kg*cm
Adoptar d = 22.50 cm r = 7.50 cm
h= 30.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 6.13 kg/cm2 v u 2-2 = 5.45 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 71.25 cm boy = 82.50 cm
48,712.41 kg
v u = 11.32 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 23: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/23.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu a-a = 765,131.79 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As a-a = 10.00 cm2 As a-a / By= 6.89 cm2/m
As min = 0.002*100*h = 6.00 cm2
As escogido = 6.89 cm2
Ø escogido = 0.50 Pulg Sep. = 18.38 cm
Sep. Escogida= 18.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu b-b = 494,208.69 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 6.46 cm2 As b-b / Bx = 6.15 cm2/m
As min = 0.002*100*h = 6.00 cm2
As escogido = 6.15 cm2
Ø escogido = 0.50 Pulg Sep. = 20.60 cm
Sep. Escogida= 18.00 cm
Diseño del Tensor:
M volc. = P*ex = 1,190,205.00 kg*cm H = M volc. / h1 = 7,934.70 kg
19,253.30 kg O.K.
9.42 kg/cm2 0.015
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensor = Ast = H / fs = 10.99 cm2
Ø escogido = 0.63 Pulg Nº cabillas = 5.55
N escogido = 6.00 AsT = 11.88 cm2
732.50 cm2 Tx*Ty = 27.06 cm
Tx escogido = 30.00 cm Ty escogido = 30.00 cm
0.0132 ≈ 0.015, asumido
0.13 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 79,347.00 kgMuc = 14,282.46 kg*m
bx (cm) = 60.00 by (cm) = 60.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =
![Page 24: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/24.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
![Page 25: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/25.jpg)
DATOS: EJE: D" - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 123,000.00 Fmu = 1.5000
Pu (kg) = 184,500.00
bx (cm) = 50.00 by (cm) = 110.00 h1 (m) = 1.50
L (tensor) (m) = 4.61
h1 (m) ν1.15
1.20 1.15
1.30
35,362.50 Bx escogido = 135.00 cm(Ver Anexo)
By = 261.94 cm By escogido = 270.00 cm
e x = (Bx-bx)/2 = 42.50 cm 5.06
Mu a-a = 4,937,083.33 kg*cm Mu b-b = 2,186,666.67 kg*cm
Mu momento máximo = 4,937,083.33 kg*cm
Adoptar d = 37.50 cm r = 7.50 cm
h= 45.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 7.54 kg/cm2 v u 2-2 = 6.75 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 68.75 cm boy = 147.50 cm
133,170.91 kg
v u = 14.66 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
![Page 26: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/26.jpg)
Acero en Fundación Perpendicular al Eje a-a:
Mu a-a = 4,937,083.33 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As a-a = 38.70 cm2 As a-a / By= 14.33 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 14.33 cm2
Ø escogido = 0.63 Pulg Sep. = 13.81 cm
Sep. Escogida= 13.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu b-b = 2,186,666.67 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 17.14 cm2 As b-b / Bx = 12.70 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 12.70 cm2
Ø escogido = 0.63 Pulg Sep. = 15.59 cm
Sep. Escogida= 13.00 cm
Diseño del Tensor:
M volc. = P*ex = 5,227,500.00 kg*cm H = M volc. / h1 = 34,850.00 kg
44,768.34 kg O.K.
9.42 kg/cm2 0.015
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensor = Ast = H / fs = 48.26 cm2
Ø escogido = 0.75 Pulg Nº cabillas = 16.93
N escogido = 18.00 AsT = 51.30 cm2
3,217.23 cm2 Tx*Ty = 56.72 cm
Tx escogido = 60.00 cm Ty escogido = 60.00 cm
0.0143 ≈ 0.015, asumido
0.16 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 184,500.00 kgMuc = 54,888.75 kg*m
bx (cm) = 50.00 by (cm) = 110.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =
![Page 27: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/27.jpg)
FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
h
Bx
h1
Ty
bx
P
H
R
H
Muc
h1´
ex
Bx
Byby
bx
Falla por Corte
bb
22
d
a
a
1
d
1Bx
By
bx
by
Falla por Punzonado
by+d
d/2
d/2
d/2
bx+d/2
![Page 28: Fundaciones de Lindero Con Tensor](https://reader036.vdocuments.pub/reader036/viewer/2022081717/544c9dfab1af9f7f538b467a/html5/thumbnails/28.jpg)
DATOS: EJE: D" - AUDITORIUM IGLESIA
210.00 4,200.00
4.00 Ø = 30.00
δ = 2/3*Ø = 20.00 tg δ = 0.3640
P (kg) = 135,513.33 Fmu = 1.5000
Pu (kg) = 203,270.00
bx (cm) = 110.00 by (cm) = 50.00 h1 (m) = 1.50
L (tensor) (m) = 4.61
h1 (m) ν1.15
1.20 1.15
1.30
38,960.08 Bx escogido = 190.00 cm(Ver Anexo)
By = 205.05 cm By escogido = 210.00 cm
e x = (Bx-bx)/2 = 40.00 cm 5.09
Mu a-a = 3,423,494.74 kg*cm Mu b-b = 3,097,447.62 kg*cm
Mu momento máximo = 3,423,494.74 kg*cm
Adoptar d = 37.50 cm r = 7.50 cm
h= 45.00 cm
Verificación a Corte:
7.68 kg/cm2
v u 1-1 = 6.79 kg/cm2 v u 2-2 = 6.79 kg/cm2
NOTA: Si no cumple adoptar otro " d "
Verificación a Punzonado:
15.36 kg/cm2
box = 128.75 cm boy = 87.50 cm
145,877.43 kg
v u = 13.27 kg/cm2 < 15.36 kg/cm2 O.K.
NOTA: Si no cumple adoptar otro " d "
F'c(kg/cm2)= FY(kg/cm2)=
Rs(kg/cm2)=
Tabla de Valores ν :
h1 ≤ 1.50 m
1.5 < h1 ≤ 3.00 m ν Escogido =
3.00 < h1 ≤ 5.00 m
Bx*By = ν * P / Rs = cm2
Rsu (kg/cm2) =
vu ≤ vc adm = 0.53*(F´c)^0.50 =
vu ≤ vc adm = 1.06*(F´c)^0.50 =
Vu = Pu - Rsu(box*boy) =
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Acero en Fundación Perpendicular al Eje a-a:
Mu a-a = 3,423,494.74 kg*cm
As a-a = Mu a-a / (0.90*0.90*Fy*d)
As a-a = 26.84 cm2 As a-a / By= 12.78 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 12.78 cm2
Ø escogido = 0.63 Pulg Sep. = 15.49 cm
Sep. Escogida= 15.00 cm
Acero en Fundación Perpendicular al Eje b-b:
Mu b-b = 3,097,447.62 kg*cm
As b-b = Mu b-b / (0.90*0.90*Fy*d)
As b-b = 24.28 cm2 As b-b / Bx = 12.78 cm2/m
As min = 0.002*100*h = 9.00 cm2
As escogido = 12.78 cm2
Ø escogido = 0.63 Pulg Sep. = 15.49 cm
Sep. Escogida= 15.00 cm
Diseño del Tensor:
M volc. = P*ex = 5,420,533.33 kg*cm H = M volc. / h1 = 36,136.89 kg
49,322.82 kg O.K.
9.42 kg/cm2 0.015
9.60
n escogido = 10.00 Por Norma
722.15 kg/cm2 ≤ Fy / 2 = 2,100.00 kg/cm2
Acero Tensor = Ast = H / fs = 50.04 cm2
Ø escogido = 0.75 Pulg Nº cabillas = 17.56
N escogido = 18.00 AsT = 51.30 cm2
3,336.03 cm2 Tx*Ty = 57.76 cm
Tx escogido = 60.00 cm Ty escogido = 60.00 cm
0.0143 ≈ 0.015, asumido
0.16 cm
Verificar Pedestal a Flexo-Compresión:
Para: Muc = Hu*h1´ = Pu*ex*h1´/h1 = Pu = 203,270.00 kgMuc = 56,915.60 kg*m
bx (cm) = 110.00 by (cm) = 50.00
H < P*tgδ P*tgδ =
fct < 0.65*(F´c)^0.50 = Se adopta ρ =
n = Es / Ec = 2.1*10^6 / 15100*(F´C)^0.50 =
fs = fct*(1+n*ρ) / ρ =
Estribos = máximo Ø 3/8" @ 30 cm
Area concreto Tensor ≥ H*(1/fct - n/fs) =
Nota: Recalculando ρ = As / Ac ≥ 0.01 ρ =
Alargamiento del Tensor = ∆L = Ɛs*L = fs*L / Es =