Civil Engineering Reference
In-Depth Information
8.18
Zone of liquefaction:
The following table shows the liquefaction calculations.
The data indicate that the zone of liquefaction extends from a depth of 2.3 to 18 m.
Fill layer:
Since the zone of liquefation extends from a depth of 2.3 to 18 m, the thick-
ness of the liquefiable sand layer
H
2
is equal to 15.7 m. Entering Fig. 7.6 with
H
2
15.7 m
and extending the
a
max
0.2
g
curve, the minimum thickness of the surface layer
H
1
needed
to prevent surface damage is 3 m. Since the surface layer of unliquefiable soil is 2.3 m thick,
there will be liquefaction-induced ground damage. The minimum required fill layer to be
added at ground surface is equal to 0.7 m (3 m
2.3 m
0.7 m).
Settlement:
The following table shows the calculations for the liquefaction-induced set-
tlement of the ground surface. The calculated settlement is 54 to 66 cm using Figs. 7.1 and
7.2. The settlement calculations should include the 0.7-m fill layer, but its effect is negligi-
ble. The settlement calculations should also include the weight of the oil in the tank, which
could cause the oil tank to punch through or deform downward the upper clay layer, result-
ing in substantial additional settlement. As indicated in the next section, the factor of safety
for a bearing capacity failure is only 1.06, and thus the expected liquefaction-induced set-
tlement will be significantly greater than 66 cm.
Bearing capacity:
D
2
___
P
4
(
H
) (
oil
)
where
D
diameter of the tank,
H
height of oil in the tank, and
oil
unit weight of oil.
Therefore,
(20 m)
2
_______
P
4
(3 m) (9.4 kN/m
3
)
8860 kN
f
50 kPa
50 kN/m
2
For the circular tank:
P
DT
f
P
(20 m) (3 m) (50 kN/m
2
)
FS
R
__
______
______________________
8860 kN
1.06
CHAPTER 9
9.1
The area of the wedge is first determined from simple geometry and is equal to
41.4 m
2
(450 ft
2
). For a unit length of the slope, the total weight
W
of the wedge equals the
area times total unit weight, or 750 kN per meter of slope length (52,000 lb per foot of
slope length).
Static case:
Use Eq. (9.2
a
) with
F
h
0 and the following values.
c
14.5 kPa
14.5 kN/m
2
(300 lb/ft
2
)
0
Length of slip surface
L
29 m (95 ft)
Slope inclination
18°
