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