Civil Engineering Reference
In-Depth Information
Consider the lower soil:
At depth
6
: . ( . . ) .
(Note that at the interface of two cohesionless soil layers, the pressure values are the
same if the
φ
′
values are equal.)
=
6
m p
=
12 2
+
0 27 12 7 4
×
× =
25 9
kPa
a
Water pressure
At depth
=
2 m, the water pressure
=
0
At depth
=
6 m, the water pressure
=
9.81
×
4
=
39.2 kPa
The two pressure diagrams are shown in Fig.
7.8b and 7.8c
; the resultant pressure
diagram is the sum of these two drawings:
1
2
1
2
1
2
+
+ ×
=
= ×
12 2 2
.
×
(
12 2 4
.
×
)+ ×
13 7 4
.
×
39 2
.
×
4
167
kPa
Had no groundwater table been present and the soil remained saturated throughout,
the active thrust would have been:
1
2
=
= ×
0 27 22 5 6
.
×
.
×
2
109 4
.
kPa
( . .
i e a lower value
)
Example
7.3
illustrates the significant increase in lateral pressure that the presence of a water table
causes on a retaining wall. Except in the case of quay walls, a situation in which there is a water table
immediately behind a retaining wall should not be allowed to arise. Where such a possibility is likely an
adequate drainage system should be provided (see Section
7.10.1)
.
For both Example
7.2
and Example
7.3,
the area of the resulting active pressure diagram will give the
magnitude of the total active thrust, P
a
. If required, its point of application can be obtained by taking
moments of forces about some convenient point on the space diagram. If this approach is not practical
then the assumptions of Fig.
7.6
should generally be sufficiently accurate.
7.4 Rankine's theory: granular soils, passive earth pressure
7.4.1 Horizontal soil surface
In this case the vertical pressure due to the weight of the soil,
γ
h, is acting as a minor principal stress.
Figure
7.9a
shows the Mohr circle diagram representing these stress conditions and drawn in the usual
position, i.e. with the axis OX (the direction of the major principal plane) horizontal. Figure
7.9b
shows
the same diagram correctly orientated with the major principal stress, K
p
γ
h, horizontal and the major
principal plane vertical. The Mohr diagram, it will be seen, must be rotated through 90°.
In the Mohr diagram:
′
′
σ
σ
OB
OA
OC DC
OC DB
+
−
1
1
+
sin
sin
φ
φ
φ
1
=
=
=
′
=
tan
2
45
°+
−
2
3
hence
1
1
sin
sin
′
′
+
φ
φ
φ
K
p
=
′
=
tan
2
45
°+
−
2
As with active pressure, there is a network of shear planes inclined at (45°
−
φ
′
/2) to the direction of the
major principal stress, but this time the soil is being compressed as opposed to expanded.
