It is defined by the EI each line should represent in that
plane: E * t * h3/12 = 700 *0.005 * 13/12=0.292 kNm2.
A line represents a height for the bending stiffness of 1
m because they are separated in a grid of 1 m and thus
each line is a beam with a height of 1 m (as defined in
Figure A4.7) and a width (thickness of skirt) of 0.005 m.
The heavy skirt has 10 times this width and thus has an
EI of 2.92 kNm².
Normal to the skirt plane the light skirt should have an EI
of E * 1 * t3/12 = 700 * 0.0053/12 = 7.29 * 10-6 kNm2.
The EI of the heavy skirt should then be 700 * 0.053/12 =
7.29 * 10-3 kNm2 .
The lower stiffness in normal direction is not integrated
into the model as it would still need to be applied to the
lines. When a line has a weak and a strong direction, it will
just turn around its axis until the force acts on its weak
direction. This creates unstable models and thus is not
applied.
Because of the hinged connections between skirt and
boom, it is possible for the skirt to have lateral movements
relative to the boom. These movements have been
found realistic from observations in the model. The difference
in skirt movements between reality and the
model due to the high estimate of the stiffness in normal
direction is therefore assumed to remain small.
The axial stiffness of the skirt is also concentrated in the
lines in the same way. So each line has an axial stiffness
of E * A = 700 * 1 * 0.005 = 3.5 kN in the light skirt and 3.5
* 10 = 35 kN in the heavy skirt.
Based on small model tests beforehand, 3 ballast masses
have been chosen for comparison. A mass of 10 kg/m,
another of 50 kg/m and one of 100 kg/m, the mass is applied
to the bottom 6D Buoys in the skirt as they are introduced
in chapter A4.3.
In this way they are concentrated masses at the points
where the skirt axial stiffness is also concentrated. Their
volume is linked to the density of steel of 7,850 kg/m³
through the formula:
V = m/p = 50/1,850 = 6.37 * 10-3 m3.
For the height they are assumed as cubic and thus the
height (and length of the sides) would be:
h = 3√V = 3√(6.37 * 10-3) = 0.185 m.
With these values, a mass moment of inertia of
(2*h2*m)/12 = (2*0.1852*0.05)/12 = 0.29 * 10-3 t * m2 is
calculated and used. As an example the numbers were
filled in for the ballast mass of 50 kg/m, the values for the
other masses are different.