Environmental Engineering Reference
In-Depth Information
14.5.14 Example with Portion of Profile Removed by
Excavation and Backfilled with Nonexpansive Soil
Let us consider the possibility of excavating the upper por-
tion of the profile. This is considered under the category of
case I. If the excavated portion is backfilled with a non-
expansive soil, the total heave calculations are considered
under the category of case II. These two cases are shown
in Fig. 14.51. In case I, it is assumed that the active depth
is unaffected by the removal of the soil. The total heave
calculations in both cases are made for the condition where
the entire active depth is wetted.
The heave for any layer in the expansive soil in case I
can be computed by substituting the variables in Fig. 14.51
into Eq. 14.18:
Substituting Eq. 14.27 into Eq. 14.40 for
j
=
35 gives
l)
−
l)
C
s
H
1
H
1
=
(
1
−
0
.
430
+
log
(
1
−
(14.41)
+
e
0
0, there is no removal of soil and Eq. 14.41
reverts to Eq. 14.27. If
l
If
l
=
1, the entire active depth has
been excavated and there is no more tendency for heave to
occur. A comparison between Eq. 14.41 and Eq. 14.27 can
be written as
=
l)
1
l)
H
l
H
=
(
1
−
−
2
.
326 log
(
1
−
(14.42)
Let us now consider case II where the excavated portion
of the profile is backfilled with an inert or nonexpansive soil.
The total density of the backfill is assumed to be equal to
the density of the expansive soil. Similarly, the total heave,
H,
for case II can be calculated using the variables given
in Fig. 14.51:
log
ρg(
1
C
s
(
1
−
l)H
−
l)H(
2
i
−
1
)/
2
j
h
li
=
1
+
e
0
j
ρ
gH
(14.38)
where:
l)
1
j
l)
log
l
h
li
=
heave in an expansive soil layer when the active
depth is partially excavated and
j
C
s
H
1
2
i
−
1
H
b
=
(
1
−
+
(
1
−
+
e
0
2
j
l
=
the portion of the profile excavated (i.e.,
H
l
/H
).
i
=
1
(14.43)
The above equation can be rearranged to the following
form:
1,
the heave is zero. A comparison between Eq. 14.43 and
Eq. 14.27 can be written as
If
l
=
0, the maximum heave is computed, and if
l
=
log
2
i
l)
C
s
(
1
−
l)H
−
1
h
li
=
+
log
(
1
−
1
+
e
0
j
2
j
(14.39)
H
b
H
=−
2
.
326 log
(
1
−
l)
The total heave for case I is computed by summing the
individual magnitudes of heave in each soil layer:
1
j
l)
log
l
j
2
i
−
1
×
+
(
1
−
(14.44)
l)
1
j
l)
log
2
i
2
j
j
C
s
H
1
−
1
i
=
1
H
l
=
(
1
−
+
log
(
1
−
+
e
0
2
j
Dimensionless plots illustrating cases I and II are shown
in Fig. 14.52. These plots apply for all swelling indices
i
=
1
(14.40)
Figure 14.51
Stress distributions and definition of variables for partial excavation and backfill-
ing with nonexpansive soil within active depth.
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