suelos ii solucion

8/17/2019 Suelos II Solucion

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1Mecánica de suelos II

La zapata típica de una edificación tiene un área de 3.50 x 5.50 m y esta cimentada

a 1.70 m de profundidad, transmite una carga de 2.25 kg/cm2.cuyo perfil del suelo

es el siguiente

Considerar estratos de un metro obligatoriamente

a) Determinar y graficar los diagramas de los esfuerzos geos taticos, neutrales

y efectivos

b) Calcular el asentamiento total

Solución

=1.85

3 =

1.85

3

=2.15

3 =

2.15

3

=2.25

2 =

22.5

2

Arena mal graduada = . e=0.65

= . =.

= .

Arcilla inorgánica de media plasticidad

= . = . = . = .

OCR=1.33 a 3.10 m

Arcilla inorgánica de alta plasticidad= . = . = . = .

OCR=2. 70 a 6.25 m

Grava limosa

0.00

-2.70

-5.70

-8.60

-11.00


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N.F

= . /= . /

1.7 m

OCR=2.70m a 6.25 m

Cc=0.42 Cs=0.06

Gs=2.66 e=0.44

2.9 m

3 m

OCR=1.33 m a 3.10 m

Cs=0.05 n=0.45

Gs=2.45 Cc=0.25

1.9 m

0.8 m

2.45 m

1.50 m

0.50 m

6.45 m

5.50 m

3.50 m

1.50 m

2.50 m

0.50 m

0.9 m

1m

1m

1m

1m

1m

1metro

-11.00

-8.60

-5.70

-4.70

-2.70

0.00

8.15 m

7.20 m

6.20 m

5.20 m

4.20 m

3.20 m

3.50 m x 5.50 m

-0.80

-3.70

-6.70

-7.70

1.50 m2.50 m

4.50 m

Grava

= .

= .


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Calculamos:

Hc =N

e ∗D ⇒ =

0.115

0.65∗0.00093= 190 ≅ .

Calculando:

γ =Gs ∗γw + e

1 + e

Antes hallamos “e”

=1 −

⇒ =0.45

1 − 0.45= .

γ =Gs ∗γw + e

1 + e

=2.45∗1 + 0.81

1 + 0.81

= 1.80

2

γ =Gs ∗γw + e

1 + e =

2.66∗1 + 0.44

1 + 0.44 = 2.15

2

Hallamos los esfuerzos geos taticos, neutrales y efectivos

a. A una profundidad de 0.8 metros

= 0.8 ∗(1.85) = 1.482

= − ∗ = −1.9 ∗1 = −1.9 2= 1.48 − (−1.9) = 3.38 / 2

b. A una profundidad de 2.70 metros

= 1.48 + 1.9 ∗(2.15) = 5.572

= 0 = 02

= 5.57− 0 = 5.57 / 2

c. A una profundidad de 5.70 metros

= 5.57 + 3 ∗(1.80) = 10.972

= 3 ∗1 = 32

= 10.97 −3 = 7.97 / 2


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d. A una profundidad de 8.60 metros

= 10.97 + 2.90 ∗(2.15) = 17.2052

= 5.90 ∗1 = 5.90

2= 17.205 − 5.90 = 11.305

2

Dibujamos los diagramas de los esfuerzos geos taticos, neutrales y efectivos

11.305

7.97

5.57

3.381.48

3

5.90

0

-1.9

17.205

10.97

5.57

1.48


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N° Hi (m) ,

/

,

/

Zi(m) m n W0 Sobrecarga ∆

∆ + , ,

formula S(mm)

1 3.20 5.97 7.91 1.50 1.17 1.83 0.209 18.81 24.78 > 7.91 III 72.70

2 4.20 6.76 8.70 2.50 0.7 1.1 0.152 13.68 20.44 > 8.70 III 54.26

3 5.20 7.56 9.50 3.50 0.5 0.78 0.109 9.81 17.37 > 9.50 III 38.93

total 165.89

σ,

= 5.57+ 0.50(1.80− 1) = 5.97

σ,

= 5.57+ 1.50(1.80− 1) = 6.77

σ, = 5.57+ 2.50(1.80− 1) = 7.56

σ = 5.57 + 0.40(1.80− 1) = 5.89

3.10-2.70=0.40→ es lo que falta para llegar a 3.10 metros

σ,

= 1.33∗5.89 = 7.83

IPC= 7.83 −5.89 = .,

= 1.94+ 5.57 = 7.91,

= 1.94+ 6.77 = 8.70,

= 1.94+ 7.56 = 9.50

1.75 metros

2.75 metros


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n =2.75

1.50= 1.83 m =

1.75

1.50= 1.17

n =2.75

2.50= 1.1 m =

1.75

2.50= 0.7

n =2.75

3.50= 0.78 m =

1.75

3.50= 0.5

σ = w ∗w w =22.5ton

m2 dato

σ = 22.5 ∗0.209 ∗ = 18.81

σ = 22.5 ∗0.152 ∗ = 13.68

σ = 22.5 ∗0.109 ∗ = 9.81

1) ∆ + , = 18.81 + 5.97 = 24.78

2) ∆ + , = 13.68 + 6.76 = 20.44

3) ∆ + , = 9.81 + 7.56 = 17.37

=+

(

,

, ) + + (

, + ∆, )

= . = . = .

=0.05 ∗1

1 + 0.81log(

7.91

5.57) +

0.25 ∗1

1 + 0.81log(

24.78

7.91) = 72.70

=0.05 ∗1

1 + 0.81log(

8.70

6.76) +

0.25 ∗1

1 + 0.81log(

20.44

8.70) = 54.26

=0.05 ∗1

1 + 0.81

log(9.50

7.56

) +0.25 ∗1

1 + 0.81

log(17.37

9.50

) = 38.93


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N° Hi (m) ,

/

,

/

Zi(m) m n W0 Sobrecarga ∆

∆ + , ,

formula S(mm)

4 6.20 8.545 23.169 4.50 0.39 0.61 0.078 7.02 15.565 < 23.169 II 10.85

5 7.20 9.695 24.319 5.50 0.32 0.5 0.059 5.31 15.005 < 24.319 II 7.90

6 8.15 10.787 25.411 6.45 0.27 0.42 0.048 4.32 15.107 < 25.411 II 6.09

total 24.84

σ,

= 7.97+ 0.50(2.15− 1) = 8.545

σ,

= 7.97+ 0.50(2.15− 1) = 9.695

σ, = 7.97+ 0.50(2.15− 1) = 10.787

σ = 7.97 + 0.55(2.15− 1) = 8.602

6.25-5.70=0.55→ es lo que falta para llegar a 6.25 metros

σ,

= 2.70∗8.602 = 23.226

IPC= 23.226 −8.602 = .,

= 14.624+ 8.545 = 23.169,

= 14.624+ 9.695 = 24.319,

= 14.624+ 10.787 = 25.411

1.75 metros

2.75 metros


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n =2.75

4.50= 0.61 m =

1.75

4.50= 0.39

n = 2.755.50

= 0.5 m = 1.755.50

= 0.32

n =2.75

6.45= 0.42 m =

1.75

6.45= 0.27

σ = w ∗w w =22.5ton

m2 dato

σ = 22.5 ∗0.078 ∗ = 7.02

σ = 22.5 ∗0.059 ∗ = 5.31

σ = 22.5 ∗0.048 ∗ = 4.32

4) ∆ + , = 7.02 + 8.545 = 15.565

5) ∆ + , = 5.31 + 9.695 = 15.005

6) ∆ + , = 4.32 + 10.787 = 15.107

=

+

(

, + ∆, )

= . = . = .

=0.06 ∗1

1 + 0.44log(

15.565

8.545) = 10.85

=0.06 ∗1

1 + 0.44log(

15.005

9.695) = 7.90

=0.06 ∗1

1 + 0.44log(

15.107

10.787) = 6.09

Asentamiento total

= +

= . + .

= .


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0.209

0.152

0.109

0.078

0.059

0.048

suelos ii solucion

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