Geoscience Reference
In-Depth Information
Note from
B/A
that for
G
<
the phase shift is negative (
>
0). Finally, the
amplitude and phase shift become
(
A
2
+B
2
)
= a
2
exp(
U = a
(1
2
;
exp(
G
)
cos(
G
) +
;
2
G
))
(16.10a)
0
0
=
atan(
B/A
) = atan[
;
sin(
G
)
/(exp(
G
)
;
cos(
G
))]
(16.10b)
The graphical representation of these formulas is given in Fig 16.4b, for various
values of
. The cases for
< 1, shown in Fig 16.4b, reveals that at certain depth
G
> 1.45 or
z
> 2.06
(
c
v
/
2
) the amplitude
U
becomes larger than the partial loading
a
= 0 (see Fig 16.4a).
In case of incompressible water, the pore pressure is then 7% larger than the total
loading
0
. A maximum is found at
z
= 3.25
(
c
v
/
2
) of 1.07
a
0
, for
) the local
maximum of
U
arrives before the maximum of the loading, in fact a negative phase
shift or a negative retardation! For
z
= 0 this negative phase shift is at maximum
0
! The phase shift
shows that in the area 0 <
z
<
(2
c
v
/
2
/4, with a local amplitude equal to zero. The negative phase shift is small, about
0.03
). The maximum pore
pressure response and the maximum loading will hardly show any phase shift.
, when
U
reaches its maximum at
z
= 3.25
(
c
v
/
2
1.0
1.0
0.75
0.8
U/a
0.5
0
0.25
0.6
0.0
a
0
0.4
0.0
0.25
0.2
0.5
G
0.75
0.0
1.0
1 2 3 4 5
Figure 16.4b The pore-pressure amplitude and phase shift
A set of isochrones for the induced pore pressure for the case of a fully drained
surface (
= 0) including the effect of pore fluid compressibility, i.e. for
a
= 0.9
and
a
= 1.0, is presented in Fig 16.5a.
The value
0.01 m
2
/s and for clay:
(
c
v
/
2
) has dimension of length. For sand:
c
v
10
-5
m
2
/s. For tidal loading
1.45x10
-4
1/s, the length
c
v
) becomes for
sand 8.3 m, and clay 0.26 m. For large waves with a period of
T
= 10 s (Note
2
(
c
v
/
2
2
=
2
) becomes for sand 0.13 m, and for clay 0.003 m.
Hence, consolidation takes place near the seabed surface, and the assumption of a
/
T
=0.63 rad), the length
(
c
v
/
2
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