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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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