Geoscience Reference
In-Depth Information
Tensile load upon impact
Initially it has been assumed that all of the fall energy can be absorbed by the geotextile
alone, i.e. it will survive the impact irrespective of the fill material. If this is not the case,
and in many cases this is not possible, the influence of the filling of the geotextile container
will have to be taken into account as well. Without the influence of the fill material:
EE
val
geo
With the formulae (6.5) and (6.8) and
⋅
TJ
ε
J
The tensile load can be calculated on the basis of the fall velocity
v
(See Appen-
∞
dix G):
AJ
S
10 8 1990
18 3
⋅
⋅
=
.
kN
/
m
N
Tv
79
270
ρ
v
=⋅
∞
.
in
[kg/m
3
] and the geotextile tensile stiffness load to be resisted (270 kN/m). In this example
this is (much) greater than the allowable tensile load of the geotextile (140 kN/m).
Since the fall velocity has a large influence on the result, it is better to determine
the fall velocity more accurately using the formulae given in Appendix G where the
fall velocity just before impact on the bed is calculated as 6.7 m/s, making the maxi-
mum tensile load
Note: for the dimensions to be correct the calculation must be made with
ρ
223 kN/m. This is still too high for the geotextile, but the energy
is also dissipated by other mechanisms, as described below.
=
Fall energy
It can also be assumed that the kinetic energy of the falling geotextile container is not
only absorbed by the geotextile but in part also by the fill material (sand). For this
situation formula (6.8) applies:
1
⋅
2
2
E
A v
0
1990 10 86
4
kJ/m
E
E
+
A
fill
fall
ρ
geo
2
Energy dissipation by the sand in the container
When the sand in the geotextile container is unsaturated, the effective stress in the sand will
increase during the fall (the previously cited 'Bezuijen effect'). The extent of this increase
depends on the level of saturation and the permeability of the material. In this example
it is assumed that 80% of the increased pressure in the water (100 KPa) results in an
increase in effective stress of:
σ
i
′
=
0.8
×
100
=
80 kN/m
2
. Substituting in formula (6.7):
J/m
E
′
A
(
A
A
A
A
(
A
A
2
)
390
.k
5
=
⋅
⋅
A
−
fill
σ
j
ϕ
ϕ
)
280
t
()
30
j
