Civil Engineering Reference
In-Depth Information
Solution:
2 5
10 000
.
ρ
c
=
m dp H
2
=
×
100 9 2 1000
× ×
.
=
230
mm
therefore, minimum settlement that must have occurred by the time the roadway is
constructed
=
230
−
25
=
205 mm. i.e.
205
230
U
=
=
90%
Assume that settlement commences at half the construction time for the embankment.
Then time to reach
U
=
6
2
90
%
= − =
12
9
months.
c t
H
0 187 9
9 2
.
×
v
T
=
=
=
0 020
.
z
2
.
2
Try 450 mm (0.45 m) diameter drains in a triangular pattern.
Select n
=
10. Then
R/r
=
10
and R
=
2 25
.
m
hence
2 25
0 525
.
.
a
=
=
4 3
.
m
Select a grid spacing of 3 m.
R
=
0 525 3
.
× =
1 575
.
m
1 575
0 225
.
.
n
=
=
7
c t
R
0 187 9
4 1 575
.
×
v
T
=
=
=
0 169
.
(
Note that no value for c w
as given so c must be used
v
)
r
h
4
2
×
.
2
1
100
100 66
U
100
(
100 16
)
(
)
=
−
−
−
=
71 4
. %,
which is not sufficient
Try a
=
2.25 m; R
=
1.18 m; n
=
5.25.
0 187 9
4 1 18
.
×
T
r
=
=
0 302
.
2
×
.
From graph, U
r
=
90%
1
100
100 16 100 90
Total consolidation percentage
=
100
−
(
−
)(
−
)
=
91
. %
6
The arrangement is satisfactory.
In practice no sand drain system could be designed as quickly as this. The object of
the example is simply to illustrate the method. The question of installation costs must
be considered and several schemes would have to be closely examined before a final
arrangement could be decided upon.
