Geoscience Reference
In-Depth Information
TABLE 6.2a DATA PROVIDED
H
[m]
n
C
r
C
c
C
weight [kN/m
3
]
d
16.0
12.8
14.0
20
17
18
1
2
1
1
5
0.40
0.42
0.35
0.001
0.12
0.04
0.006
0.76
0.14
0.0
0.005
0.002
3
TABLE 6.2b ELABORATION, = ' = 100 KPA, PRECONSOLIDATION STRESS 50 KPA
layer
1
2
3
e
0
0.667
0.724
0.538
a
0.0003
0.0303
0.0113
b
0.0016
0.1917
0.0395
c
0
0.0013
0.0006
u
kPa
0
0
10
10
60
S
total
m
0.001
0.143
0.160
total
0.305
S
creep
m
0.0
0.010
0.025
total
0.035
A
kPa
8
16
33
33
123
'
A
kPa
8
16
23
23
63
'
B
kPa
58
66
73
73
113
'
C
kPa
108
116
123
123
163
1
0.0015
0.1500
0.1363
0.0382
0.0260
remark
1
2
2
2
3
4
5
Use the following remarks (Table 6.2b)
1. The total stress at each layer interface (top and bottom); for the first layer in the
middle.
2. See Fig 6.1. See also Table 5.2b.
3. Formula (6.2) is applied:
l
= 1
- v
D
/
v
A
. = 1
- (..)
-a
(..)
-b
(..)
-c
4. The total settlement
S
(reloading, virginal loading and creep) in each layer is
approximated by
H
(
1
bottom
)/2, except for the first layer.
5. The settlement due to the creep component only after 27 years (10
4
days) is
shown separately (last column of Table 6.3). For example, for layer 2 (top) one
finds, applying (6.2)
1
top
+
top
creep
top
total
top
S
creep
= H
= H
H
=
l
l
l
elastic
plastic
'
A
)
-a
(
'
B
)
-b
(
t
D
/t
C
)
-c
)
- H
(1
-
(
'
A
)
-a
(
'
B
)
-b
)
H
(1
-
(
'
B
/
'
C
/
'
B
/
'
C
/
'
A
)
-a
(
'
B
)
-b
(1
-
(
t
D
/t
C
)
-c
)
= H
(
'
B
/
'
C
/
= 1(66/16)
-0.0303
(116/66)
-0.1917
(1 - (10
4
)
-0.0013
) = 0.0102 m
It appears that, in this case, the creep in total is about 12% of the primary
(elasto-plastic) settlement. What would be the result when layer 2 and 3 are
switched?
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