fundaciones de lindero con tensor

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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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) =

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 =

PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO

PARA RESOLVER, SIEMPRE COLOCAR LA FUNDACION EN EL SENTIDO SEÑALADO

FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

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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) =

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 =

FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

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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) =

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 =

FUNDACION DE LINDERO DOBLE CON TENSORES(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

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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) =

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 =

FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

h

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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) =

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 =

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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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) =

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 =

FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

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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) =

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 =

FUNDACION DE LINDERO CON TENSOR(Elaborada por el Ing. Gustavo Adolfo Dubuc Moreno. C.I.V.: 42.644)

h

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bx

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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) =

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 =

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

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Muc

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Bx

Byby

bx

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1Bx

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bx+d/2

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) =

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 =