Geoscience Reference
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=
((
F/f
)
2
+((
)
0.5
tan
)
/
w
)
1)/
F/f
with
f =
2
w
tan
/
(10.10)
is obtained for
F =
1. One may also determine the angle for
a specified safety, incorporating an overall safety factor
The critical angle
)
0
(see Chapter 8). Hence,
F
in (10.10) is replaced by
F
0
, with
F =
1 as the limit stability factor
.
For
example, consider a long slope (open mine pit) with a safety factor of
0
= 1.2.
Horizontal outflow of groundwater is expected (isotropy is assumed:
= 1). With
=
35
o
, and an average soil weight of
=
18 kN/m
3
,
an internal friction angle of
0
/f
)
2
application of (10.9) yields
f =
2
w
tan
/
=
0.778, and tan
)
=
((
F
1)
0.5
0
/
f =
((1.2/0.778)
2
+ 8/10
1)
0.5
+
'/
F
1.2/0.778 = 0.241. Thus,
)
F=
1.2
w
=
13.5
o
. For
F=
1
=
15.7
o
. Dynamic flow conditions can
aggravate the situation. Along coastlines where the sea tides and waves create
dynamic conditions of which 'horizontal outflow' is determining, it is observed
that slopes in the range of tan
0
= 1, one would find
)
<
10
o
are formed.
)
<
0.15 or
)
application 10.2
Consider a water-retaining embankment, shown in Fig 10.10. The soil is fully
saturated and cohesive (
=
0). The soil properties and structural dimensions are
shown in Table 10.2. The factor of safety is determined for three cases (1) stagnant
water with depth
H
, (2) water depth zero, and (3) the effect of a tension crack in the
crest of the embankment, when filled with water.
(1) stagnant water with a depth of 6 m
The sliding soil mass is divided in two zones: ABGF under the groundwater
level, and GCDEF above. The weight of these zones are
W
1
=
144 (18-10) = 1152
kN/m and
W
2
=42x18 = 756 kN/m. Applying (10.15)
yields
(Wd) =
(30 x 22.8
2
x 80 x
F = c
u
LR/
%
/180)/(1152 x 4.5 + 756 x 13) = 1.45
(2) water depth zero
The weight of zone ABGF becomes
W
1
= 144 x 18 = 2592 kN/m. Applying
(10.15) yields
F =
(30 x 22.8
2
x 80 x
/180)/(2592 x 4.5 + 756 x 13) = 1.01
(3) effect of a crack and hydraulic thrust
The depth of a tension crack is determined with (10.14b), which yields
z
c
=
2
c
u
w
z
c
2
/2
=
10 x 3.3
2
/2 =
54.5
kN/m, and the lever arm is
z
P
=
10 + 2 x 3.3/3 = 12.2 m.
Applying (10.15)
yields for case 1
/
=
2 x 30 / 18 = 3.3 m. The hydraulic thrust becomes
P =
F =
(30 x 22.8
2
x 80 x
/180)/(1152 x 4.5 + 756 x 13 + 54.5 x 12.2) = 1.39
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