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ship stability
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Kul-24.3300 Ship Buoyancy and Stability Model Answer 4.10.2012 Assignment 2. Markus Tompuri 1(3)
Task 1. Equilibrium position requires that the total moment around an arbitrary point zero. In this case the external moment Mext is acting on the ship and thus this moment has to be compensated by the static restoring moment Mst. Consequently:
stextstext MMMM 0
When the heeling angle is small ( < 10) the static restoring moment Mst can be approximated as:
sin0GMM st
sin0GMM ext
where:
g
sin0GMgM ext
m 1.299
)sin(6sm 9.81mkg 1001m 16500
Nm 000 000 22
sin
233
0
g
MGM ext
KGKMKGMBKBGM 00000
m 12.001
m 1.299 - m 13.3
00
GMKMKG
Task 2. Metacentric height:
KGMBKBGM 0000
where 0KB is the height of the center of buoyancy from the base line K:
Kul-24.3300 Ship Buoyancy and Stability Model Answer 4.10.2012 Assignment 2. Markus Tompuri 2(3)
The buoyancy of the semisubmersible platform consists of the following 6 components:
length width height volume v.c.g. vol*vcg
pontoon PS 60 10 10 6000 5 30000
pontoon SB 60 10 10 6000 5 30000
column 1 10 10 10 1000 15 15000
column 2 10 10 10 1000 15 15000
column 3 10 10 10 1000 15 15000
column 4 10 10 10 1000 15 15000
sum
16000 7.5 120000 The submerged height of the columns is limited to the draft of T = 20m. Thus we obtain:
m5.7KB0
00MB is the metacentric radius, it is calculated on the basis of the inertial moment of the waterplane
area and the volume of displacement:
TIMB 00
For a single column the inertia moment of the waterplane area is:
44
33
, m333.833m12
1010
12
lbI columnT
By applying the Steiner’s rule we obtain:
2,,4 dAII columnwcolumnTT
where the waterplane area of a single column is 2
, m100columnwA and the distance from the x-axis is
d = 25 m Resulting in:
442 m3.253333m25100333.8334 TI
Thus the metacentric radius is:
m833.15m16000
m3.2533333
4
00
TIMB
KG is the height of the center of gravity G from the base line level K. This has been given in the assignment. Thus the metacentric height is:
Kul-24.3300 Ship Buoyancy and Stability Model Answer 4.10.2012 Assignment 2. Markus Tompuri 3(3)
m333.2
m21m833.15m5.7
0000
KGMBKBGM
For the other direction (around the y-axis) we get IL = 93 333 m4. And consequently GML = -7.67 m, meaning that the platform is unstable and will capsize! For offshore structures the stability must be checked for all directions (0 .. 360 deg), i.e. the so-called Azimuth directions. Task 3. Let’s start from the moment equation:
stextstext MMMM 0
In this case there is no external heeling moment, thus Mext = 0. The restoring moment can be expressed by using the righting lever arm. Thus for a ship with vertical sides:
sin
2
tansin
2
000 MBGMGZM st
Consequently, by combining the previous equations we obtain:
0sin2
tansin
2
000
MBGM
02
tan2
000
MBGM
00
02 2tan
MB
GM
where:
m 5.32
m 3 - m 5
200000
TKMKBKMMB
After substituting the input values we get:
086.0
m 3.5
m 0.152tan 2
3.16