TALUD EJERCICIO

Determinar el Factor de Seguridad al vuelco del siguiente Talud

R
4m

P=7.5 ton

R

3

3m

Capa 1

g1= 1,68 Ton/m
f1=25
c = 15 kg/cm 2

3

5m

Capa 2

g2= 1,70 Ton/m
f1=5
2
c = 10 kg/cm

00°

45.
32m

Capa 3

Mecánica de Suelos II y Laboratorio

g3= 1,45 Ton/m
f1=7
2
c = 0 kg/cm

Universidad Autónoma del Beni U.A.B.

Docente: Luis Alberto Zambrana Monasterio

O
a
4m

f

g

e

R

b

P=7.5 tonH

R

A

Capa 1

3m

2'

11.18 cm
12.82 cm
8.38 cm

1'
3'

B

G

5.71 cm

4'

Capa 2

5m

6'

6.45 cm

7.21 cm

5'

2.40 cm
C
1

2m
2

F
2.40 cm

Capa 3

7'

8'

2.00 cm

E

D
Caracteristicasgeométricas.Del triángulo OAH: Cos(f+g+e)= 4 / 14
Del triángulo OBG: Cos(f+g)= 7 / 14
Del triángulo OCF: Cos(f)= 12 / 14
Del triángulo OCE: Cos(a)= 12 / 14
Longitudes de arco.Capa 1: Lc1=e*R
Capa 2: Lc2=g*RCapa 3: Lc3=f*R
Capa 3: Lc4=a*R

.... f+g+e=73.39 .... e=13.39 .... K=0.001087
... f+g=60 ... g=29 ... K=0.010667
f=31
..
..
. . K=0.013007
a=31
...
...
... K=0.013007

.... Lc1=13.39 * p/180 * 14
...Lc2=29 * p/180 * 14
... Lc3=31 * p/180 * 14
. . Lc4=31 * p/180 * 14

Mecánica de Suelos II y Laboratorio

.... Lc1= 3.27 m
... Lc2= 7.086 m
... Lc3= 7.54 m .
. . Lc4= 7.54 m . .

Universidad Autónomadel Beni U.A.B.

Q=0.117048
Q=0.245599
Q=0.261412
Q=0.261412

L total = 25.43 m

Docente: Luis Alberto Zambrana Monasterio

Cu
ña
Fi
gu
ra

Resistente

1
2

Área
(m 2 )

Brazo (m)
De Palanca

Momento(m * m 2 )

1 7.21 * 2 = 7.21
2

2.40

2

2

Q*R=0.261412*14 K*R=0.013007*14
2
Q*R=3.659
K*R=2.549

W, WL, q, N
2.40 * 7.21=17.30 WR = A*1m*g1
WR = 9.759*1m*1.68 = 16.39 ton
-1

3.659*2.549

9.328
M =26.628

A = 9.759
3

g1 =1.68 ton/m

L = M / A = 26.628 / 9.759 = 2.728
3
g2 =1.70 ton/m Deslizante

R

2

12.82

1'

2

K*R=0.2130
P= (5.43+5.1+3.27)/2
P= 6.90

2.7313

W DD1= A*1m*g2
WD1=10.28*1m*1.70 = 17.47 ton

91.016

0D1=sen (LD /R)
-1
0D1=sen (9.656 / 14) = 43.607
ND1= WD1* cos ( 0D1)

-1

A= 6.90(6.90-5.43)(6.90-5.1)(6.90-3.27)

11.18

2'

A= 8.141
P= (5.10+4.24+7.12)/2
P= 8.23

ND1=...