8/2/2019 PE 13 Slope Stability

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13 - Slope Stability

*01: Find the FS of a straight wedged element of slope.

*02: Swedish slip circle used to find the FS of a slope.

*03: Method of slices to find the FS of a slope.*04: Method of slices to find the FS of a slope.

*05: Field evaluation of the stability of a loaded river bank.

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*Slope-01:Factor of Safety of a straight line slope failure.(Revision: Oct.-08)

A slope cut to 1.5H:1V will be made in a shale rock stratum that has bedding planes that

have an apparent dip of 16 (see the figure below). If the acceptable factor of safety against

failure is at least 2 along the lower-most bedding plane, is this slope stable? Use a unit

weight of 20.1 kN/m3, and bedding strength parameters ofc = 22 kPa and = 30.

85.0 m

Original ground surface

1.5 11.3 m L

20.0 m 1

Rocks bedding planes

16

Proposed new ground surface

Solution:

( ) ( ) 3

The traingule of rock above the potential slip plane has a weight per unit width,

185.0 11.3 20.1 9, 650

2

The length of the slip plane is,

85.088.4

cos16

Therefore,

Resisting For

W

kN kN W m m

m m

L

mL m

F S

= =

= =

=( )

( )

( )2

cos tances

Driving Forces sin

22 88.4 9, 650 cos16 tan 30

9, 650 sin 6

2

1

2.7

cL W

W

O

kN kN m

m mF S

kN

m

K

+ = =

+

= =

>

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*Slope-02:Same as Slope-01 but with a raising WT.(Revision: Oct.-08)

In the previous problem the slope appeared to be stable with a factor of safety = 2.7. What

happens to that factor of safety if the water table rises to the level shown below? Use a unit

weight of 20.1 kN/m3, and bedding strength parameters are reduced by the effective

parameters ofc = 15 kPa and = 20.

85.0 m

Original ground surface

3.2 m

1.5 11.3 m

20.0 m Seepage 1

Rocks bedding planes

16

Proposed new ground surface

Solution:

T h e w e i g h t o f t h e r o c k t ri a n g l e p e r u n i t w i d t h i s s t il l 9 , 6 5 0

T h e le n g th o f t h e s lip p la n e i s s ti l l 8 8 .4 .

T h e p o r e w a t e r p r e s s u re is b a s e d o n a n e s ti m a t e o f i ts v a lu e a lo n g t h e le n g t h ,

a t w a t e r

k NW

m

L m

L

( )

( )

( )

( )

3

2

d e p t h a b o v e t h e p l a n e t h a t r a n g e f ro m 0 t o 3 . 2 m ; c o n s e r v a t i v e ly ,

9 .8 1 3 .2 3 1 .4

' c o s ta nR e s is t i n g F o r c e s

D r iv i n g F o r c e s s in

1 5 8 8 .4 9 , 6 5 0 c o s 1

w

w w

z

k Nu z m k P a

m

c L W uF S

W

k N k N m

m mF S

= = =

+ = = =

+

=

( )6 3 1 .4 ta n 2 0

9 , 6 5 0 s i n 1 6

T h e c o m p u t e d f a c to r o f s a fe t y o f 1 . 7 6 i s le s s th a n t h e m i n im u m a c c e p t a b le

v a lu e o f 2 , th e r e f o re t h i s d e s ig n i s N O T a c c e p t a b le .

1 .7 6 2

s in

k P a

k Nm

N G

N o t ic e t h a t a r i g W