model answer 1

Kul-24.3300 Ship Stability 30.9.2013

Assignment 1.

Markus Tompuri 1(1)

1.

The buoyancy force is:

gmg

where m is the mass of the water that the ship displaces (i.e. displacement), g is acceleration due to

gravity and is the volume of the water that the ship displaces (i.e. volume of displacement). is the density of the water.

When it is assumed that the lightweight and deadweight of the ship do not change, also the

displacement must be constant, so that:

21

gg 2211

12

12

Thus the change in the volume of displacement is:

1

1

2

11

2

111

2

112 11

md

Since the sides of the ship are vertical near the waterline, it can be assumed that the water plane

area is constant, and thus the change of draft is:

m 17.0

m 0004mkg 1002

kg 10300001

mkg 1025

mkg 1002

1

23

3

3

3

1

1

2

1

ww A

m

A

ddT

The sign of the change is significant. In this case the draft decreases (on a return voyage the draft

would increase).


Assignment 1.

Markus Tompuri 1(1)

2.

The lifting process consists of 4 different phases:

I: lifting from the bottom

II: lifting while the whole object is submerged

III: lifting while the object is partly submerged

IV: lifting in the air (whole object above water level)

Let's calculate the volume of the object V1:

AhV 11

where:

2rA (area of the bottom of the object)

where

m5.02

D

r (radius of the bottom)

32211 m 571.1m 0.5m2 rhV

Then we can calculate the mass of the object:

bVm

where V is the volume of the concrete shell:

21 VVV

where V2 is the volume of the void inside the concrete shell:

3

22

22 m 0.1962

m 0.5m1

2

d

hV

Thus:

kg 12.3436mkg 2500m 0.1961.571m 33321 bVVm


Assignment 1.

Markus Tompuri 1(1)

Phase I

Lets define the forces acting on the object when the positive direction is upwards.

Weight of the object W:

kN .7133sm 9.81kg 12.3436 2 mgW

Lifting force of the buoyancy :

kN49.15sm 9.81mkg 10051.571 231 gV w

suction force of the sea bottom :

AgHpApp wh 00

where p0 is the atmospheric air pressure (= 101325 Pa) and ph is the hydrostatic pressure at the

bottom. A is the area of the bottom of the object:

22

1 m 785.0 rA

Additionally there is a force due to the porous water pressure ppo which is acting on the opposite

direction to the suction force. It was said to be 50% of the hydrostatic pressure, i.e. 0.5ph. Thus we

get:

kN -118.24

m 785.010msm 9.81mkg 10052

1Pa 101325

2

1

2

1

2

1

223

0

000

AgHp

AppApppAppp

w

hhhpoh

So the force P, that is needed to lift the object from the bottom is obtained from the equation:

0 WP

kN 136,5kN 24.118kN 15.49kN 33.71 WP

Phase II

The suction force is not acting anymore, thus we obtain:

kN 18.2kN 49.15kN 71.33 WP


Assignment 1.

Markus Tompuri 1(1)

Phase III

Lets mark the height of the object above the sea level with h. The part below the water level provides buoyancy, which can be written as a function of h:

ghrVh w 21

The needed force P is:

0 hWP

Phase IV

No more buoyancy, and thus:

0WP kN 33.7 WP

All four phases are presented in the figure below as a function of z.

0

20

40

60

80

100

120

140

160

180

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13

P (

kN

)

z (m)


Assignment 1.

Markus Tompuri 1(1)

3.

a) The initial stability of a ship is presented by the initial metacentric height 0GM . It is

determined as:

KGMBKBGM 0000

Thus the factors affecting the initial stability are:

0KB height of the center of buoyancy from the baseline K. 0KB depends on the shape

of the hull and the draft T.

00MB is the metacentric radius, it depends on the inertia moment of the waterplane

area IT and the volume of displacement , namely:

T

IMB 00

KG is the height of the center of gravity G from the baseline level K. So this factor depends on the weight distribution of the ship (lightship and deadweight).

b) For a submarine the location of the center of buoyancy does not change (when the submarine is submerged) when the submarine heels. If the center of gravity is above the center of

buoyancy the pair of gravity and buoyancy forces acts as increasing the heeling moment and

thus the vessel will capsize.

B

G

B

G

Epvakaa tilanne.Voimapari

kallistaa alusta lis.

Vakaa tilanne.Voimapari

tasapainottaa alusta.

W

W

DD

stable condition: the

forces cause a

restoring moment

unstable condition:

the forces cause a

heeling moment


Assignment 1.

Markus Tompuri 1(1)

For surface ships the location of the center of buoyancy changes as the ship heels. In transverse direction the change of this

point B is initially larger than the change of the center of gravity G. Thus this pair of forces is restoring.

D

W

B

G

model answer 1

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