NTNUFaculty of Engineering Science and TechnologyDepartment of Marine Technology

EXERCISE 1Suggested Solution

TMR 4195 DESIGN OF OFFSHORE STRUCTURES

Exercise: Overturning stability for floating buoys

a) Weight and ballast of buoy A and the drought of buoy B

Buoy A

Buoyancy of the cylindrical part:

Bcyl=4302123102[MN ]=106 MN

For simplicity we say that g = 10 m/s density = 1.0 tonnes / m. Further, we intend to work with MN as force units, hence the factor 10 [MN/m].

Buoyancy of the conical part:

Bcone=4 30

2153

202103

102[MN ]=25 MN

This yields a total buoyancy equal to: Btot = 131 MN

Weight:

Topside 25.0 MNSmall cylinder 0.4 *15 6.0 MNLarge cylinder 0.75 *17 12.8 MNSum 43.8 MNBallast weight 87.2 MN

Buoy B:

From Table 2

Total weight:

W B=20402510MN=95 MN

Buoyancy of conical section

Bcone=1434022030215102=48.4 MN

Bcyl=W totBcone=9548.4=46.5 MN

This yields:

Bcyl=4[3025z bottom]=46.5

z bottom=11.5m

b) Initial metacentric heights

Buoy A:

Center of gravity CG:

zCG=[2518615

212.817

28528.53

2] /131=16.5

BM= I=

D4

64

=

64204

131102=0.6 m

Note the very small contribution from the water plane stiffness!!!

h1

h2

Conical volumes

r1

r2

At last, the initial GM value is found by:

GM=BMGB=BMCGCB =0.616.512.5=4.6m

where CB is given in the exercise text as -12.5 m

Buoy B:

Center of gravity:

CG=[20740810511.52

]/95=4.0m

Center of buoyancy yields:

CB=[2.248.446.5511.52

]/95=5.2m

This gives the following GM value:

GM=BMGB= ICGCB

GM=64

404

95009.3=3.9m

Note the large contribution form the water plane stiffness!!!

c) Heeling angle due to acting ice forces

Buoy A:

The ice action yields a set-off of 62 m. The vertical component of the mooring system is approx 2MN.

05

10

15

20

25

0 20 40 60 80 100Displacment [m]

Moo

ring

forc

e [M

N] Horizontal force

Vertical force

Figure: Finding the buoy set off, and the vertical mooring force

The total vertical force is 3MN (1 MN from ice action and the other from the mooring line). This yields another set down of z, calculated from :

1024202 10

3dnex

2 dnew /23

=3

this gives dnew = 17.87 m, which results in the additional set-down:

z=20d new

2=1.06m

The new BM value is now

BM new=I new

=64

17.844

1313102=0.4m

Hence the GM is reduced by 0.2 m to 4.4 m

We can now find the overturning moment:

M O=F ICE , Horizzline=1010=100 MNm

Restoring moment

M R=GM sinF ICE ,vert10F mooring ,vert0.430

This yields a heeling angle of = 6.15. The heeling angle is within the accepted range.

Buoy B:

The ice action of 13 MN yields a set-off of 13 MN yields a set -off of 70 m with a vertical mooring force of 2.6 MN. The net vertical force is now 2.6 1.0 = 1.6 MN. A short estimation of draft change yields:

z= 1.6g AW

= 1.6106

1044

102=0.13m

Note that since we assume a constant waterline surface, it will only be a coarse estimation. However, it does shows that the change in draft is small, hence we neglect the change in buoyancy in the following.

We now find the acting moments. First, the over turning moment :

M O=1310140=170 MNm (13 horizontal ice force, 1 vertical ice force)

Resisting momentM R=GM sin2.60.430=170

=arcsin [M O31.2GGM

]=arcsin [ 17031.2952.61.03.9

]

=0.384 [rad ]=22.0o

The heeling angle is not within the accepted range!