Geoscience Reference
In-Depth Information
Dimensions and shape
For the lower tube elements the theoretical maximum radius can be determined from
the known circumference:
12 3
2
S
=
.
R
196
96
m
=
=
2
π
2
π
⋅
⋅
Based on a degree of filling of 70%, it follows from Table 5.3 (not based on the
more accurate Timoshenko method, see 5.5.2) that:
70%
=
0.63 m
70%
=
4.27 m
70%
=
1.92 m
Filling time and filling length
Based on the sand capacity of 381 m
3
/hour, the length of tube that will be filled within
an hour can be estimated. The diameter of the tube is assumed to be 100% filling
π ⋅
12.1 m
2
. For a degree of filling of 70% this is 8.4 m
2
. Over 1 hour the tube
will be filled up to 381/8.4
R
2
=
45 m. This is a simplified relationship. In actual fact the
tube will be filled quicker where the sand is introduced, while further on it will take
longer before the tube is full.
Note: In general the total tube cross-section will not be available during filling
(8.4 m
2
), but only a proportion (maybe only ¼) because the cross-section will be partly
filled with sand.
=
Required tensile strength
The geotextile tubes have a circumference of 12.3 m and lie entirely under water.
From Figure 5.9 a tensile strength of 20 kN/m can be determined. The initial degree
of filling is then around 80%, but it is assumed that this degree of filling is necessary
to reach 70% when the filling stops. Due to strength-reduction material factors the
required tensile strength will be raised by a factor of 3.5 to arrive at a design strength
(see 5.4)
70 kN/m. Use is made of a
woven polypropylene geotextile with a maximum tensile strength of 80 kN/m. This
provides an overall safety factor of 80/70
⇒
required design tensile strength
=
20
⋅
3.5
=
=
1.14, so the requirement for the tensile
strength is fulfilled.
Sand density
Table 2.3 presents the formula for the required pore size (
O
90
) of the geotextile. For
dynamic load (wave attack) the requirement is:
12
/
O
D
C
u
⋅
10
90
12
/
⎛
⎛
292
100
⎞
complies
15 100
256
=
256
⇒
⋅
100
⋅
⎝
μ
