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Then in February 1995, after five months of uneventful performance, suddenly a
strong leakage was observed. Next, a pump broke down, a pronounced sand inflow
was noticed, and after a temporary repair the grout-column wall locally collapsed
(Fig 15.11). The subsequent mud and water inflow could not be stopped. In the
end, the building pit had to be put under water in order to repair the grout-column
wall. Total damage was about 5 MEuro. It is worthwhile to mention, that, if the
ongoing erosion process would have been modest another two months, nothing
would have been noticed and no damage would have happened, as by then the
construction of the metro station basement would have been completed.
application 15.1
Consider a cylindrical vertical borehole with a radius r 0 = 0.25 m in a soft
cohesive soil with undrained strength c u = 20 kPa, shear modulus G = 400 kPa,
saturated weight
= 14 kN/m 3 . The soil has a low permeability. The hole is filled
with water. At a depth of 5 m the situation is considered. Here the pore pressure is
u = 50 kPa. The total stress was 70 kPa, so a cavity pressure of q = - 20 kPa is
representative and the cavity is contracting.
Applying (15.13) gives a plastic zone r p = r 0 e - ( q + c )/2 c = r 0 , implying that there
is just no plastic zone. (15.14) gives the local displacement w 0 =
cr 0 e - ( q + c ) /c / 2 G
= -20x0.25 xe - (-20+20)/20 /(2x400) = - 0.00625 m = - 0.6 cm. Rather small.
Next, the water is taken out. Hence the representative cavity pressure is q = - 70
kPa, which gives r p = r 0 e - ( q + c )/2 c = 0.25 e - (-70+20)/40 = 0.88 m; in a ring wide 0.88 -
0.25 = 0.63 m around the cavity the soil is in a plastic state. The corresponding
displacement is w 0 =
cr 0 e - ( q + c ) /c / 2 G = -20x0.25 e - (-70+20)/20 /(2x400) = - 0.076
m = -7.6 cm. Quite large!
A critical situation occurs when, with (15.17), q = - c (1 + ln I r ) = -20(1+
ln(400/20) = -79.9 kPa. At a depth of 5 m the cavity will not collapse, but at a
depth 79.9/14 = 5.7 m it will. Here, the soil weight will play a role.
application 15.2
Consider a tunnel lining with radius R at a depth H under the ground surface. The
groundwater table is at the surface, i.e. D = 0 in Fig 15.8. Let K 0 = 4 (a highly
over-consolidated soil), R/H = 1/3,
s = 20 kN/m 3 ,
w = 10 kN/m 3 , and
t = 0 (the
tunnel lining is rigid and relatively weightless). The components of the traction t at
the lining surface, i.e. the normal pressure induced by the local soil stress, is
described by its components (see Fig 15.11)
t x = sin 2
x + cos 2
x + cos
sin
z and t z = sin
cos
z
Application of the approximate soil stress according to (15.23) and (15.24) to this
situation yields
x /
s H = ( K 0
d D + (
w + K 0 (
s
w )( H + R cos
) /
s H = 3/2+½ cos
z /
s H = (
d D +
s ( H
R )) /
s H =
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