Geoscience Reference
In-Depth Information
2
U
=
S
1
i
(1- i
)/(1+
)
(16.25)
The total harmonic response is therefore
2
)
u
=
U
exp(
i
2
t
)
= S
1
exp(
i
2
t
)
i
(1-
i
)/(1+
(16.26)
The real part of this expression yields
2
)
Re(
u
)
= S
1
cos(
2
t +
/4
-
) /
(1+
(16.27)
U
U
U
A
1
A
1
A
1
S
1
U
S
1
U
S
1
U
S
1
S
1
S
1
V
V
V
U
U
U
U
U
U
A
1
A
1
A
1
L
L
L
Figure 16.10
S
1
u
1
t
!HI#H2
Figure 16.11 Harmonic average response to cyclic total loading, showing negative
retardation
2
Solution (16.27) presents a response with a decay of
/
(1+
), and a delay of
(
/4 +
)/
2
, which is a negative delay for
>
/4, see Fig 16.11. The condition
>
L
1
L
2
/3
c
v
< 1. For a sandy seabed, under cyclic loading with a
period in the order of
T
= 10 s or
/4 yields
=
2
/
T
= 0.63, and length scales
L
1
and
L
2
in
the order of decimetres (vertical consolidation of thin layers), the condition of
negative delay, i.e.
2
= 2
< 1, is met for a seabed consolidation coefficient
c
v
larger
L
1
L
2
/3
= 2x10
-3
m
2
/s, which is also likely in many cases.
For a thicker soft clayey seabed, the length scales
L
1
and
L
2
are normally in the
order of decametres and the corresponding consolidation coefficient
c
v
is usually in
the order of 10
-5
m
2
/s, or smaller. In this case the condition of negative delay, i.e.
than the order of
2
is in the order of 10
-4
rad/s.
The observed transient excess pore pressure response in coastal zones due to
tidal water level changes shows a damped response with, indeed, a negative delay
< 1, is met for slower waves, like tidal waves, where
2
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