Hydrostatics Solutions

Solutions of Exercises onO¤shore Hydrostatics

(Lecture Code OT4620)

J.M.J. Journée and W.W. Massie

November 5, 2003

In this ”quiz book”, the solutions of a number of exercises on hydromechanical problems ino¤shore activities are given. As far as possible, these exercises are given here in the sameorder as the underlying theory has been treated in the textbook of the lecture OT4620:

O¤shore Hydromechanics(First Edition)

byJ.M.J. Journée and W.W. Massie

1

1. Miscellaneous on Static Loads

One has a large open tank with three 100 mm diameter drain holes in its bottom. Thereis atmospheric pressure all around the tank. Before the tank is …lled with water (½ = 1000kg/m3) each hole is closed o¤ in a di¤erent way:

A vertical steel pipe - just as long as the height of the tank sides - is used to close o¤ onehole. The pipe has a diameter of 200 mm with a seal along its lower edge. The pipeweighs 600 N.

The second opening is closed o¤ by a steel (½ = 7850 kg/m3) disk, 200 mm in diameter,with a seal along its perimeter. The plate weighs 70 N and has a nylon cord leadingto the water surface.

The third opening has a plywood (½ = 750 kg/m3) cover which weighs 7 N. It is otherwiseidentical to the steel cover.

Questions:

a) With the pipe and the two covers in place, can the tank be …lled?

b) Assuming that the tank has been …lled to a depth of 5 m, which of the three drainswill be easiest to open? Explain also why!

Solutions:

a) The tank can be …lled. The pressures on the pipe and covers act only on the sidestill the water depth is higher than the top of the covers. The seals prevent waterpressure from working on the bottom of the covers so that the wooden cover doesnot ‡oat!

b) Force to remove cover:

Force to lift pipe:F = weight of pipe = 600 N

Force to lift steel cover:

F1 = ½gh ¢ A = 1000 ¢ 9:81 ¢ 5 ¢ ¼4

¢ 0:22 = 1541 N

F2 = weight of cover = 70 N

F = F1 + F2 = 1541 + 70 = 1611 N

Force to lift wood cover:

F = F1 + F2 = 1541 + 7 = 1548 N

So, pipe is easiest to lift!

2

2. Miscellaneous on Static Floating Stability

Questions:

a) What is the relation between a ship’s mass and its volume of displacement.

b) Give a de…nition (in words) of:- the center of gravity of a structure, G,- the center of buoyancy of a structure, B,- the initial metacenter, M , and- the metacenter, NÁ:

c) Prove for a rectangular pontoon that:

BM =ITr and BNÁ =

ITr ¢

µ1 +

1

2tan2 Á

¶

What is the generally used name of the second expression?

d) Explain why B =M = NÁ for a fully submerged structure such as a submarine.

e) Show that for a rectangular barge - with length L, breadth B and uniform draft T -one can write:

BM =B2

12 ¢ Tf) Prove that for a wall-sided ship, the vertical separation of BÁ and G at angle of heel

Á is given by:

BÁZ =µBG+

1

2BM ¢ tan2 Á

¶¢ cos Á

where BG and BM relate to the upright condition.Show that - when this ship has a free ‡uid surface in a wall-sided tank - the previousexpression becomes:

BÁG =µBGS +

1

2

³BM ¡GSGF

´¢ tan2 Á

¶¢ cos Á

where the subscripts S and F relate to the solid and ‡uid centers of gravity.

g) Show that the dynamical stability of a wall-sided vessel up to an angle Á is given bythe expression:

PÁ =µBG ¢ (cos Á ¡ 1) + 1

2BM ¢ tanÁ ¢ sin Á

¶¢ ½gr

h) A wall-sided vessel with a negative GM lolls to an angle Á0. Express this angle Á interms of GM and BM.

i) A laden rectangular barge with a breadth of 6.00 m has a metacentric height of GM =¡0:20 m when ‡oating at a uniform draft of 3.00 m. Calculate the angle of loll Á0.

3

j) A small mass is added to a wall-sided ship on the center line at such a position as toleave trim unchanged. Show that the metacentric height is reduced if the mass isadded at a height above keel greater than T ¡GM , where T is the draft.

k) A rectangular homogeneous block (30 m long, 7 m wide and 3 m deep) is half as denseas the water in which it is ‡oating. Calculate the metacentric height, GM , and thestability lever arm, GZ, at 15 degrees and 30 degrees heel.

l) In one way or another, you are involved in the design process of converting an existingship with a length of about 200 meters into a stone-dumping vessel. You have beeninformed that the location of the center of buoyancy, B, at the fully laden draft andzero trim has been determined (via hydrostatic calculations) at 3.25 meters in frontof the amidships section. However, accurate mass calculations show you that thecenter of gravity, G, of the ship in this loading condition will be of about 4.75 metersin front of the amidships section.Assuming correct hydrostatic calculations, what will be your comment on this infor-mation with respect to the actual under-water geometry of the vessel?

m) One has a glass tube nearly …lled with water and closed at each end.We have also an hourglass (in Dutch: ”zandloper”) ‡oating in the water in the tube.The hourglass has a maximum diameter a bit less than the internal diameter of thetube. The total weight of the hourglass (including the sand in it) is only slightly lessthe weight of the same total volume of water - it just barely ‡oats.If we abruptly invert the glass tube then:- the air bubble comes to the top immediately,- the hourglass remains at the bottom, and- the sand moves slowly from the upper hourglass chamber to the lower one.Some time later, the hourglass slowly rises to the top of the tube.Explain why it initially stays at the bottom!

Solutions:

a) m = ½r

b)

c) Scribanti formula; for derivation see textbook.

d) There is no water plane area, so IT = 0 which yields BM = BNÁ = 0

e) BM = ITr =

112LB

3

LBT = B2

12¢T

f)

g)

h) Á = § arctanÃr

2jGMjBM

!

i) Á = 32:3 degrees.

4

j)

k) GM = 1:972 m, GZ(15) = 0:536 m and GZ(30) = 1:047 m.

l) Because of a stability lever arm - between the centers of buoyancy and gravity - theship will trim until this arm becomes zero.

m) Initially, the hourglass has a negative GM and it heels against the glass tube. Asa consequence, it remains at the bottom due to friction. After a certain time theGM becomes positive and the hourglass gets an upright position without frictionanymore. Then it rises slowly to the top of the tube.

5

3. Float-On Float-O¤ Pontoon

Figure 0.1: Float-On Float-O¤ Pontoon

In a harbor with fresh water (½ = 1.000 ton/m3) an empty rectangular Float-On Float-O¤Pontoon, as given in …gure 0.1 has an amidships draft of 1.55 meter in an upright even keelcondition. A rough inclining experiment has been carried out, by fully …lling a port sidetank aft (tank VI), which is bounded by half the length of the pontoon and the longitudinalmiddle line plane of the pontoon, with fresh water (½wb = 1.000 ton/m3). The measuredangle of heel Á1 was 1.46 degrees. In the calculations, the volumes of plating, frames andother structure parts may be ignored.

a) Determine the position of the centre of gravity G0 and the initial metacentric heightG0M0 of the empty pontoon.

b) Determine the trim angle µ1 during the inclination experiment.

c) Determine the drafts at the four corners of the pontoon during the inclination experi-ment.

d)Determine the angle of heel Á2 in case of an 80 per cent …lled tank, during the inclinationexperiment.

During operation, the pontoon will be sunk down in a protected bay by loading ballastseawater (½ = ½wb = 1.025 ton/m3) in all 16 tanks. This ballast water is supposed to havean equal height h in all 16 tanks.

e) Determine the initial metacentric height GM at an even keel draft of 7.50 meter,supposing that water will just not cover the deck.

6

f) Determine the GM -value at an even keel draft of 7.50 meter, supposing that water hadjust covered the deck.

g) Determine the GM -value, when the pontoon has been sunk down until an even keeldraft of 11.50 meter.

Now, the pontoon picks up a drill-rig, with the following speci…cations:¡ upright even keel condition¡ mass = 4920 ton¡ KG = 20.00 meter¡ water plane dimensions: 40 x 40 meter¡ no free surfaces of liquids in any tank

h) Determine the initial metacentric height of the rig.

i) Determine the initial metacentric height of pontoon+rig, supposing that they just hiteach other when de-ballasting the pontoon during loading of the rig (centre of rigabove centre of pontoon).

j) Determine the initial metacentric height of pontoon+rig when all water ballast hasbeen removed from the pontoon.

k) Determine the angle of heel when, due to an inaccurate loading of the pontoon, thecentre of the rig is amidships but 1.0 meter outside the middle line plane of thepontoon.

Solutions:

a) G0: amidships at middle line plane with KG0 = 4.02 m and G0M0 = 45.15 m.

b) µ1 = 0:380

c) T1i = 1.28, 2.00, 2.04 and 2.76 m, respectively.

d) Á2 = 1:160.

e) GM = 9.93 m.

f) GM = 1.55 m.

g) GM = 1.62 m.

h) GM = 25.94 m.

i) GM = 5.51 m.

j) GM = 10.86 m.

k) Á = 2:610.

7

Detailed Solutions

Volume of displacement r0 of empty pontoon:

r0 = L ¢B ¢ T0= 108:00 ¢ 30:00 ¢ 1:55 = 5022 m3

Mass of displacement ¢0 of empty pontoon:

¢0 = ½ ¢ r0

= 1:000 ¢ 5022 = 5022 ton

Solution of Part 3-aDetermine the position of the centre of gravity G0 and the initial metacentric height G0M0

of the empty pontoon. The empty pontoon lies at an even keel condition. So, the centreof buoyancy of the pontoon and also the centre of gravity of the pontoon are situatedamidships at half the length of the pontoon.Mass of water ballast ¢wb in a fully …lled tank:

¢wb = ½wb ¢ l ¢ b ¢ h= 1:000 ¢ 27:00 ¢ 7:50 ¢ 7:50 = 1519 ton

Centre of gravity of water ballast in a fully …lled tank with respect to half the length ofthe pontoon, the middle line plane and the base plane:

xwb = ¡13:50 m

ywb = +3:75 m

zwb = +3:75 m

The calculation will be carried out in two steps:

1. Suppose the water ballast concentrated on a vertical line through the centre of thewater plane in a horizontal plane through the centre of gravity of the water ballast(parallel sinkage).

2. Shift the centre of gravity of the water ballast to the right position by adding aheeling moment (heel).

Step 1: Suppose a parallel sinkage.The new draft T1 becomes:

¢1 = ½ ¢ L ¢B ¢ T1 = 1:000 ¢ 108:00 ¢ 30:00 ¢ T1 = 3240 ¢ T1= ¢0 +¢wb = 5022 + 1519 = 6541

So : T1 =6541

3240= 2:02 m

Centre of buoyancy above keel KB1:

KB1 =T12=2:02

2= 1:01 m

8

Metacenter above centre of buoyancy B1M1:

B1M1 =ITr1

=112LB3

LBT1=

B2

12 ¢ T1=

30:002

12 ¢ 2:02 = 37:13 m

Step 2: Add a heeling moment.A transverse shift of the centre of gravity of the water ballast over a distance ywb will resultin a angle of heel Á1 = 1:46

0.Replace this shift by a heeling moment MH :

MH = ¢wb ¢ g ¢ ywb ¢ cosÁ1The righting stability moment of the pontoonMS is:

MS = (¢0 + ¢wb) ¢ g ¢G1M1 ¢ sinÁ1Because of the equilibrium MH = MS, it follows for the initial metacentric height G1M1:

G1M1 =¢wb ¢ ywb

(¢0 + ¢wb) ¢ tanÁ1=

1519 ¢ 3:75(5022 + 1519) ¢ tan(1:460) = 34:18 m

This formula for G1M1 can be used, because Á1 is very small:

M1NÁ = B1M1 ¢ 12

¢ tan2 Á1= B1M1 ¢ 0:000325 ¼ 0:00

Herewith is the position of the centre of gravity of the pontoon including water ballastKG1 known:

KG1 = KB1 + B1M1 ¡G1M1

= 1:01 + 37:13 ¡ 34:18 = 3:96 m

For the empty pontoon, the position of the centre of gravity KG0 follows from the …rstmoment of masses of the pontoon including water ballast with respect to the base plane:

¢1 ¢KG1 = ¢0 ¢KG0 + ¢wb ¢ zwb(pontoon + wb) (pontoon) (wb)

So:

KG0 =¢1 ¢KG1 ¡ ¢wb ¢ zwb

¢0

=(5022 + 1519) ¢ 3:96 ¡ 1519 ¢ 3:75

5022= 4:02 m


KB0 =T02=1:55

2= 0:78 m

9


B0M0 =ITr0

=112LB

3

LBT0=

B2

12 ¢ T0=

30:002

12 ¢ 1:55 = 48:39 m

Herewith is the initial metacentric height of the empty pontoon G0M0 known:

G0M0 = KB0+ B0M0 ¡KG0= 0:78 + 48:39¡ 4:02 = 45:15 m

Solution of Part 3-bDetermine the trim angle µ1 during the inclination experiment.Centre of buoyancy above keel KB1:

KB1 = 1:01 m

Longitudinal metacenter above centre of buoyancy BML:

B1M1L =ILr1

=112BL

3

LBT1=

L2

12 ¢ T1=

108:002

12 ¢ 2:02 = 481:19 m

Centre of gravity above keel KG1:

KG1 = 3:96 m

Herewith is the initial longitudinal metacentric height of the pontoon including waterballast G1M1L known:

G1M1L = KB1+ B1M1L ¡KG1= 1:01 + 481:19¡ 3:96 = 478:24 m

A longitudinal shift of the centre of gravity of the water ballast over a distance xwb willresult in a trim angle µ1.Replace this shift by a trimming moment MHL:

MHL = ¢wb ¢ g ¢ xwb ¢ cos µ1

The longitudinal righting stability moment of the pontoon MSL is:

MSL = ¢0 + ¢wb ¢ g ¢G1M1L ¢ sin Á1

Because of the equilibrium MHL = MSL, it follows for the trim angle µ1:

tan µ1 =¢wb ¢ xwb

(¢0 + ¢wb) ¢G1M1L

=1519 ¢ 13:50

(5022 + 1519) ¢ 478:24 = 0:380

10

Solution of Part 3-cDetermine the drafts at the four angular points of the pontoon during the inclinationexperiment.Half the heel displacement is:

B

2¢ tanÁ1 =

30:00

2¢ tan(1:460) = 0:38 m

Half the trim displacement is:

L

2¢ tan µ1 =

108:00

2¢ tan(0:380) = 0:36 m

Drafts at angular points of pontoon:Starboard aft: 2.02 + 0.38 + 0.36 = 2.76 mStarboard forward: 2.02 + 0.38 ¡ 0.36 = 2.04 mPort side aft: 2.02 ¡ 0.38 + 0.36 = 2.00 mPort side forward: 2.02 ¡ 0.38 ¡ 0.36 = 1.28 m

Solution of Part 3-dDetermine the angle of heel Á2 in case of an 80 per cent …lled tank, during the inclinationexperiment.Mass of water ballast ¢wb in the 80 per cent …lled tank:

¢wb = 0:80 ¢ 1519 = 1215 ton

Centre of gravity of water ballast in the 80 per cent …lled tank with respect to half thelength of the pontoon, the middle line plane and the base plane:

xwb = ¡13:50 m

ywb = +3:75 m

zwb = 0:80 ¢ 3:75 = +3:00 m

Suppose a parallel sinkage.The new draft T2 becomes:

¢0 + ¢wb = ¢2 = ½ ¢ L ¢B ¢ T2= 1:000 ¢ 108:00 ¢ 30:00 ¢ T2 = 5022 + 1215

So : T2 = 1:92 m


KB2 =T22=1:92

2= 0:96 m

The position of the centre of gravity KG2 follows from the …rst moment of masses of thepontoon and the water ballast with respect to the base plane:

¢2 ¢KG2 = ¢0 ¢KG0 + ¢wb ¢ zwbSo:

KG2 =¢0 ¢KG0 + ¢wb ¢ zwb

¢2

=5022 ¢ 4:02 ¡ 1215 ¢ 3:00

5022 + 1215= 3:82 m

11


B2M2 =ITr2

=112LB3

LBT2=

B2

12 ¢ T2=

30:002

12 ¢ 1:92 = 39:06 m

Herewith is the initial metacentric height of the pontoon with the 80 per cent …lled tankwith frozen water ballast G2M2 known:

G2M2 = KB2+ B2M2 ¡KG2= 0:96 + 39:06¡ 3:82 = 36:20 m

Now we let the water ballast unfreeze and the reduction G2G02 of the metacentric heightbecomes:

G2G02 =½wb ¢ it½ ¢ r2

=1:000 ¢ 1

12 ¢ 27:00 ¢ 7:5031:000 ¢ (5022 + 1215) = 0:15 m

Herewith is the reduced initial metacentric height of the pontoon with the 80 per cent …lledtank G02M2 known:

G02M2 = G2M2 ¡G2G02= 36:20 ¡ 0:15 = 36:05 m

Add a heeling moment.A transverse shift of the centre of gravity of the water ballast over a distance ywb will resultin a angle of heel Á2.Replace this shift by a heeling moment MH :

MH = ¢wb ¢ g ¢ ywb ¢ cosÁ2The righting stability moment of the pontoonMS is:

MS = ¢2 ¢ g ¢G02M2 ¢ sin Á2Because of the equilibrium MH = MS, it follows for the angle of heel Á2:

tanÁ2 =¢wb ¢ ywb¢2 ¢G02M2

=1215 ¢ 3:75

(5022 + 1215) ¢ 36:05 or: Á2 = 1:160

Note 1:In this exercise, the e¤ect of the free surface in the tank is very small.In case of frozen cargo, the angle of heel Á2 would be:

tanÁ2 =¢wb ¢ ywb¢2 ¢G2M2

=1215 ¢ 3:75

(5022 + 1215) ¢ 36:20 or: Á2 = 1:150

12

Note 2:The small reduction of the metacentric height has been obtained in the design of thepontoon by a subdivision in the transverse direction in 4 compartments. As a result ofthis, the transverse moment of inertia (second moment of areas) of the free surface of thewater ballast it has been reduced considerably.With a tank over the full breadth, the reduction G2G02 of the metacentric height would be:

G2G02 =½wb ¢ it½ ¢ r2

=1:000 ¢ 1

12 ¢ 27:00 ¢ 30:0031:000 ¢ (5022 + 1215) = 9:78 m

Solution of Part 3-eDetermine the initial metacentric height GM at an even keel draft of 7.50 meter, supposingthat water will just not enter the deck.Volume of displacement r of pontoon:

r = L ¢B ¢ T= 108:00 ¢ 30:00 ¢ 7:50 = 24300 m3

Total mass of displacement ¢ of pontoon:

¢ = ½ ¢ r= 1:025 ¢ 24300 = 24908 ton

Mass of water ballast ¢wb in pontoon:

¢wb = ¢ ¡ ¢0

= 24980 ¡ 5022 = 19886 ton

Height h of water level in ballast tanks:

h =¢wb

½ ¢ L ¢B=

19886

1:025 ¢ 108:00 ¢ 30:00 = 5:98 m

Centre of gravity of total water ballast with respect to half the length of the pontoon, themiddle line plane and the base plane:

xwb = 0:00 m

ywb = 0:00 m

zwb =h

2=5:98

2= 2:99 m

For the ballasted pontoon, the position of the centre of gravity KG follows from the …rstmoment of masses of the empty pontoon and the water ballast with respect to the baseplane:

¢ ¢KG = ¢0 ¢KG0 + ¢wb ¢ zwb

13

So:

KG =¢0 ¢KG0 + ¢wb ¢ zwb

¢

=5022 ¢ 4:02 + 19886 ¢ 2:99

24908= 3:20 m

Centre of buoyancy above keel KB:

KB =T

2=7:50

2= 3:75 m

Metacenter above centre of buoyancy BM :

BM =ITr =

112LB

3

LBT=

B2

12 ¢ T=

30:002

12 ¢ 7:50 = 10:00 m

The reduction of the metacentric height GG0 due to the free surface of the water ballastbecomes:

GG0 =

P½wb ¢ it½r

=16 ¢ 1:025 ¢ 1

12¢ 27:00 ¢ 7:503

24908= 0:62 m

Herewith is the reduced initial metacentric height of the pontoon G0M known:

G0M = KB +BM ¡KG¡GG0= 3:75 + 10:00¡ 3:20¡ 0:62 = 9:93 m

Solution of Part 3-fDetermine the GM -value at an even keel draft of 7.50 meter, supposing that water hadjust entered the deck.With respect to Exercise 5, here BM changes only:

BM =ITr

The transverse moment of inertia (second moment of areas) of the water plane can befound by subtraction of moments of inertia of rectangles:

IT =1

12¢n9:00 ¢

³30:003 ¡ 16:003

´+18:00 ¢

³30:003 ¡ 23:003

´o

= 39428 m4

or by using Steiner’s rule:

IT = 2 ¢½ 112

¢³9:00 ¢ 7:003 +18:00 ¢ 3:503

´+ 9:00 ¢ 7:00 ¢ 11:502 +18:00 ¢ 3:50 ¢ 13:252

¾

= 39428 m4

14

With this, the metacenter above centre of buoyancy BM becomes:

BM =ITr =

39428

24300= 1:62 m


G0M = KB + BM ¡KG¡GG0= 3:75 + 1:62¡ 3:20¡ 0:62 = 1:55 m

Solution of Part 3-gDetermine the GM -value, when the pontoon has been sink down until an even keel draftof 11.50 meter.Volume of displacement r of pontoon:

r = 24300 + 2 ¢ (9:00 ¢ 7:00 + 18:00 ¢ 3:50) ¢ (11:50¡ 7:50) = 25308 m3


¢ = ½ ¢ r= 1:025 ¢ 25308 = 25941 ton


¢wb = ¢ ¡ ¢0

= 25941 ¡ 5022 = 20919 ton


h =¢wb

½ ¢ L ¢B=

20919

1:025 ¢ 108:00 ¢ 30:00 = 6:30 m


xwb = 0:00 m

ywb = 0:00 m

zwb =h

2=6:30

2= 3:15 m

For the ballasted pontoon, the position of the centre of gravity KG follows from the …rstmoment of masses of the empty pontoon and the water ballast with respect to the baseplane:

¢ ¢KG = ¢0 ¢KG0 + ¢wb ¢ zwbSo:

KG =¢0 ¢KG0 + ¢wb ¢ zwb

¢

=5022 ¢ 4:02 + 20919 ¢ 3:15

25941= 3:32 m

15

Centre of buoyancy above keel KB follows from the …rst moment of volumes of the indi-vidual parts of the under water geometry of the pontoon with respect to the base plane:

r ¢KB = 108:00 ¢ 30:00 ¢ 7:50 ¢ 3:75+2 ¢ (9:00 ¢ 7:00 + 18:00 ¢ 3:50) ¢ (11:50¡ 7:50) ¢ 9:50

= 100701 m4

So:

KB =100701

25308= 3:98 m


BM =ITr =

39428

25308= 1:56 m


GG0 =

P½wb ¢ it½r

=16 ¢ 1:025 ¢ 1

12¢ 27:00 ¢ 7:503

25941= 0:60 m


G0M = KB + BM ¡KG¡GG0= 3:98 + 1:56¡ 3:32¡ 0:60 = 1:62 m

Solution of Part 3-hDetermine the initial metacentric height of the rig.

rrig =¢rig½

=4920

1:025= 4800 m3

Trig =rrig

Lrig ¢Brig=

4800

40:00 ¢ 40:00 = 3:00 m

KBrig =Trig2

=3:00

2= 1:50 m

BM rig =ITrigrrig

=112Lrig ¢B3rig

Lrig ¢Brig ¢ Trig=

B2rig12 ¢ Trig

=40:002

12 ¢ 3:00 = 44:44 m

GM rig = KBrig + BM rig ¡KGrig = 1:50 + 44:44 ¡ 20:00 = 25:94 m

Solution of Part 3-iDetermine the initial metacentric height of pontoon+rig, supposing that they just hit eachother when in‡ating the pontoon during loading the rig (centre of rig above centre ofpontoon).Draft T at even keel of pontoon:

T = 7:50 + 3:00 = 10:50 m

16

Volume of displacement r of pontoon:

r = 24300 + 2 ¢ (9:00 ¢ 7:00 + 18:00 ¢ 3:50) ¢ (10:50¡ 7:50) = 25056 m3


¢ = ½ ¢ r= 1:025 ¢ 25056 = 25683 ton


¢wb = ¢ ¡ ¢0

= 25683 ¡ 5022 = 20661 ton


h =¢wb

½ ¢ L ¢B=

20661

1:025 ¢ 108:00 ¢ 30:00 = 6:22 m


xwb = 0:00 m

ywb = 0:00 m

zwb =h

2=6:22

2= 3:11 m

For the ballasted pontoon + rig, the position of the centre of gravity KG follows from the…rst moment of masses of the empty pontoon, the water ballast and the rig with respectto the base plane of the pontoon:

(¢ + ¢rig) ¢KG = ¢0 ¢KG0 + ¢wb ¢ zwb + ¢rig ¢ zrig

With zrig = 7:50 + 20:00 = 27:50 meter, it is found:

KG =¢0 ¢KG0 + ¢wb ¢ zwb + ¢rig ¢ zrig

¢+¢rig

=5022 ¢ 4:02 + 20661 ¢ 3:11 + 4920 ¢ 27:50

25683 + 4920= 7:18 m

The centre of buoyancy above keel KB follows from the …rst moment of volumes of theindividual parts of the under water geometry of the pontoon and the rig with respect tothe base plane of the pontoon:

(r+rrig) ¢KB = 108:00 ¢ 30:00 ¢ 7:50 ¢ 3:75+2 ¢ (9:00 ¢ 7:00 + 18:00 ¢ 3:50) ¢ 3:00 ¢ 9:00+4800 ¢ (7:50 + 1:50)

= 141129 m4

17

So:

KB =141129

25056 + 4800= 4:73 m


BM =IT

r +rrig

=39428 + 40:00 ¢ 40:003

25056 + 4800= 8:47 m


GG0 =

P½wb ¢ it½r

=16 ¢ 1:025 ¢ 1

12¢ 27:00 ¢ 7:503

25056 + 4800= 0:51 m


G0M = KB + BM ¡KG¡GG0= 4:73 + 8:47¡ 7:18¡ 0:51 = 5:51 m

Solution of Part 3-jDetermine the initial metacentric height of pontoon+rig when all water ballast has beenremoved from the pontoon.

¢ = ¢0 + ¢rig = 5022 + 4920 = 9942 ton

r =¢

½=

9942

1:025= 9700 m3

T =rL ¢B =

9700

108:00 ¢ 30:00 = 3:00 m

KB =T

2=3:00

2= 1:50 m

BM =ITr =

B2

12 ¢ T =30:002

12 ¢ 3:00 = 25:00 m

KG =5022 ¢ 4:02 + 4920 ¢ 27:50

5022 + 4920= 15:64 m

GG0 = 0:00 m

GM = KB +BM ¡KG = 1:50 + 25:00¡ 15:64 = 10:86 m

Solution of Part 3-kDetermine the angle of heel when, due to an inaccurate loading of the pontoon, the centreof the rig is amidships 1.0 meter outside the middle line plane of the pontoon.

MH = 4920 ¢ g ¢ 1:00 ¢ cos ÁMS = (5022 + 4920) ¢ g ¢ 10:86 ¢ sin ÁtanÁ =

4920 ¢ 1:00(5022 + 4920) ¢ 10:86 = 0:0456 or: Á = 2:610

18

4. Metacentric Height

Figure 0.2: Unloading a Ship

A laden wall-sided ship with its own derrick on board, as given in …gure 0.2, is ‡oatingwithout heel at an even keel condition in fresh water (½ = 1.000 ton/m3) with a volume ofdisplacement r of 12000 m3. The load, which is a mass p of 250 ton, is placed in one ofthe holds on the tank top of the double bottom. The centre of gravity of this mass lies inthe middle line plane, 3.00 meter above the base plane of the ship. The suspension pointof the cargo in the derrick lies 25.00 meter above the base plane of the ship. When thederrick is turned outboard fully, this suspension point lies 15.00 meter from the middle lineplane of the ship.As soon as the mass has been hoisted from the tank top of the double bottom, the shipheels 2.0 degrees. After hoisting the mass further and turning it outboard fully, the angleof heel becomes 17.0 degrees.

Determine the initial metacentric height G0M of the ship, before the unloading operations.The in‡uence of the mass of the turning derrick may be ignored.

Solution: G0M = 0.44 m.

Detailed Solution

Mass free of tank top: Á = 20.This equilibrium will be achieved when the righting stability momentMS equals the heelingmoment MH , which is zero:

MS = ½gr ¢GZ= ½gr ¢GNÁ ¢ sinÁ= ½gr ¢

³GM +MNÁ

´¢ sin Á

19

= ½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ

= MH = 0

This equilibrium at Á = 20 gives the following equation with the unknowns BM and GM :

GM +1

2BM ¢ tan2 Á = 0

or:GM + 0:000610 ¢BM = 0

Mass turned outboard fully: Á = 170.This equilibrium will be achieved when the righting stability momentMS equals the heelingmoment MH :

MS = ½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ

= MH = 250 ¢ g ¢ 15:00 ¢ cosÁ

This equilibrium at Á = 170 gives a second equation with the unknowns BM and GM :

1:000 ¢ g ¢ 12000 ¢µGM +

1

2BM ¢ tan2(170)

¶¢ sin(170) = 250 ¢ g ¢ 15:00 ¢ cos(170)

or:GM + 0:046736 ¢BM = 1:0221

The solution of these two equations with the unknowns BM and GM gives:

BM = 22:16 m

GM = ¡0:01 m

The position of the original centre of gravity G0 follows from the …rst moment of massesof the ship with hoisted cargo with respect to the base plane:

12000 ¢KG = 12000 ¢KG0 + 250 ¢ (25:00 ¡ 3:00)

From this follows the vertical shift of the centre of gravity:

G0G = KG¡KG0 =250 ¢ (25:00¡ 3:00)

12000= 0:46 m

The metacentric height before the unloading operations of the cargo becomes:

G0M = G0G+GM

= 0:46 ¡ 0:01 = 0:45 m

20

5. Unloading a Pontoon

Figure 0.3: Unloading a Pontoon

An empty rectangular pontoon, subdivided in 8 watertight compartments of equal dimen-sions, has the following principal dimensions:¡ length L = 40.00 meter¡ breadth B = 6.00 meter¡ depth D = 2.50 meter¡ mass ¢0 = 200 ton

The pontoon is ‡oating at an even keel condition in fresh water (½ = 1.000 ton/m3). Thevertical position of the centre of gravity above the base plane KG0 is 1.20 meter. Thepontoon is laden with two masses of p = 50 ton each, of which the centres of gravity arepositioned at half the length of the pontoon, yp = 1.75 meter from the middle line planeof the pontoon and zp = 1.25 meter above the deck of the pontoon, see …gure 0.3.

a) Determine the initial metacentric height of pontoon+masses.

b) Determine the angle of heel of the pontoon after unloading one of the masses. It maynot be assumed that this angle is small.

c) To obtain an upright position again, two tanks will be equally …lled with water ballast(½wb = 1.000 ton/m3). Determine the mass of the water ballast and the reducedinitial metacentric height.

Solutions:

a) GM = 0.975 m.

b) Á = 11:30.

c) ¢wb = 58.33 ton and G0M = 1.35 m.

21

Detailed Solutions

Solution of Part 5-a

¢ = ¢0+ 2 ¢ p = 200 + 2 ¢ 50 = 300 ton

r =¢

½=

300

1:000= 300 m3

The pontoon is laden with the centre of gravity of the two masses together at the centreof the water plane, so it will sink down parallel to the water plane. The under water hullform is a rectangular pontoon at an even keel condition without heel.So:

T =rL ¢B =

300

40:00 ¢ 6:00 = 1:25 m

KB =T

2=1:25

2= 0:625 m

BM =ITr =

B2

12 ¢ T =6:002

12 ¢ 1:25 = 2:40 m

The vertical position of the centre of gravity of the laden pontoon follows from the …rstmoment of masses with respect to the base plane:

¢ ¢KG = ¢0 ¢KG0 + 2 ¢ p ¢ (D + zp)

So:

KG =¢0 ¢KG0 + 2 ¢ p ¢ (D + zp)

¢

=200 ¢ 1:20 + 2 ¢ 50 ¢ (2:50 + 1:25)

300= 2:05 m

Herewith, the initial metacentric height is known:

GM = KB + BM ¡KG= 0:625 + 2:40 ¡ 2:05 = 0:975 m

Solution of Part 5-bThe pontoon is laden with one mass p on a distance c from the middle line plane.Expecting a relative large angle of heel, the calculations will be carried out as follows:

1. Place the mass p above the centre of the water plane in a horizontal plane throughthe centre of the mass and, as a result of this, let the pontoon sink deeper parallelto the water plane. Determine in this situation the righting stability moment MS inrelation to the angle of heel Á.The righting stability moment MS of the pontoon is given by:

½gr ¢GNÁ ¢ sinÁ = ½gr ¢³GM +MNÁ

´¢ sinÁ

= ½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ

22

2. Replace the shift of the mass p in a horizontal direction over a distance c to the actualplace by a heeling moment MH , which depends on the angle of heel Á too.

MH = p ¢ g ¢ c ¢ cos Á

3. Finally, the equilibrium MS = MH should be ful…lled.

½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ = p ¢ g ¢ c ¢ cos Á

or: µGM +

1

2BM ¢ tan2 Á

¶¢ tan Á = p ¢ c

½r

Detailed further Working-out of Part 5-bDisplacements of laden pontoon:

¢ = ¢0 + p = 200 + 50 = 250 ton

r =¢

½=

250

1:000= 250 m3

Due to the parallel sinkage, the under water hull form remains a rectangular pontoon atan even keel condition without heel, so:

T =rL ¢B =

250

40:00 ¢ 6:00 = 1:04 m

KB =T

2=1:04

2= 0:52 m

BM =ITr =

B2

12 ¢ T =6:002

12 ¢ 1:02 = 2:89 m


r ¢KG = r0 ¢KG0 + p ¢ (D+ zp)

So:

KG =r0 ¢KG0 + p ¢ (D + zp)

r=

200 ¢ 1:20 + 50 ¢ (2:50 + 1:25)250

= 1:71 m


GM = KB + BM ¡KG= 0:52 + 2:89 ¡ 1:71 = 1:70 m

As pointed out before, an equilibrium will be achieved when the righting stability momentMS equals the heeling moment MH :

µGM +

1

2BM ¢ tan2 Á

¶¢ tanÁ = p ¢ c

½r

23

Inserting the calculated data in this equation gives:µ1:70 +

1

2¢ 2:89 ¢ tan2 Á

¶¢ tan Á = 87:5

1:000 ¢ 250or:

1:445 ¢ tan3 Á +1:700 ¢ tanÁ = 0:350

This third degree equation in tan Á can be solved iteratively by the Regula-Falsi method,by calculating the left hand side LHS of the equation as a function of Á until a value equalto the value of the right hand side RHS = 0:350 has been found:

Á = 10:00 ! LHS = 0:308Á = 15:00 ! LHS = 0:483Á = 11:20 ! LHS = 0:348Á = 11:30 ! LHS = 0:351 RHS = 0:350 So: Á ¼ 11:30.

Solution of Part 5-cTwo tanks will be …lled up to an equal height hwb with water ballast (½wb = 1.000 ton/m3).The distance ywb of the centre of the water ballast to the middle line plane is 1.50 meter.The pontoon is in an upright position, when the heeling moment MH caused by the massp is equal to the moment of the water ballast Mwb:

MH = p ¢ g ¢ yp = 50:0 ¢ g ¢ 1:75= Mwb = ¢wb ¢ g ¢ ywb = ¢wb ¢ g ¢ 1:50

This equilibrium gives the mass of the water ballast:

¢wb =87:5

1:50= 58:33 ton

and herewith it is found for the laden and ballasted pontoon:

rwb =¢wb½wb

=58:33

1:000= 58:33 m3

hwb =rwb

2 ¢ lt ¢ bt=

58:33

2 ¢ 10:00 ¢ 3:00 = 0:97 m

zwb =hwb2

=0:97

2= 0:485 m

¢ = ¢0 + p+ ¢wb = 200 + 50 + 58:33 = 308:33 ton

r =¢

½=308:33

1:000= 308:33 m3

The under water hull form is a rectangular pontoon at an even keel condition without heel.So:

T =rL ¢B =

308:33

40:00 ¢ 6:00 = 1:285 m

KB =T

2=1:285

2= 0:64 m

BM =ITr =

B2

12 ¢ T =6:002

12 ¢ 1:285 = 2:335 m

GG0 =

P½wb ¢ it½r =

2 ¢ 1:000 ¢ 112

¢ 10:00 ¢ 3:003308:33

= 0:145 m

24

The vertical position of the centre of gravity of the laden and ballasted pontoon followsfrom the …rst moment of masses with respect to the base plane:

¢ ¢KG = ¢0 ¢KG0 + p ¢ (D + zp) + ¢wb ¢ zwb

So:

KG =¢0 ¢KG0 + p ¢ (D + zp) + ¢wb ¢ zwb

¢

=200 ¢ 1:20 + 50 ¢ (2:50 + 1:25) + 58:33 ¢ 0:485

308:33= 1:48 m

Herewith, the reduced initial metacentric height is known:

G0M = KB +BM ¡KG¡GG0= 0:64 + 2:335¡ 1:48¡ 0:145 = 1:35 m

25

6. Lift Operation by a Pontoon

Figure 0.4: Lift Operation by a Pontoon

A rectangular pontoon has the following principal dimensions:¡ length L = 60.00 meter¡ breadth B = 12.00 meter¡ depth D = 6.00 meter

One of the double bottom tanks is partly …lled with fuel with a density ½f = 0.900 ton/m3.The length of this tank lf is 20.00 meter and the breadth bf is 6.00 meter. The pontoon is‡oating at an even keel condition with a draft T0 = 2.50 meter in fresh water (½ = 1.000ton/m3). The vertical position of the centre of gravity of the pontoon, including fuel, abovethe base plane KG0 is 4.00 meter. A sketch of the pontoon in this situation is given in…gure 0.4.Then, a mass of p = 100 ton will be hoisted from the quay. When the derrick is turnedoutboard fully, the suspension point of the cargo in the derrick lies 13.00 meter above thebase plane and 8.00 meter from the middle line plane of the pontoon.

Determine the maximum angle of heel of the pontoon during hoisting this load. Thein‡uence of the mass of the turning derrick may be ignored.

Solution: Á = 16:50.

Detailed Solution

Expecting a relative large angle of heel, the calculations will be carried out as follows:

1. Place the mass p above the centre of the water plane in a horizontal plane through thesuspension point at the end of the derrick and, as a result of this, let the pontoon sinkdeeper parallel to the water plane. Determine in this situation the righting stability

26

moment MS in relation to the angle of heel Á.The righting stability moment of the pontoon (with ”frozen fuel”) is given by:

½gr ¢GNÁ ¢ sinÁ = ½gr ¢³GM +MNÁ

´¢ sinÁ

= ½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ

Due to the free surface of the fuel in one of the double bottom tanks (unfreezing the”frozen fuel”) this righting stability moment will be reduced by:

½gr ¢GG00 ¢ sin Á = ½gr ¢GG0 ¢µ1 +

1

2tan2 Á

¶¢ sinÁ

Then the righting stability moment of the pontoon with ”liquid fuel” becomes:

MS = ½gr ¢µGM +

1

2BM ¢ tan2 Á ¡ GG0 ¢

µ1 +

1

2tan2 Á

¶¶¢ sinÁ

2. Replace the shift of the mass p in a horizontal direction over a distance c to the actualplace by a heeling moment MH , which depends on the angle of heel Á too.



½gr ¢µGM +

1


µ1 +

1

2tan2 Á

¶¶¢ sin Á = p ¢ g ¢ c ¢ cosÁ

or: µGM +

1


µ1 +

1

2tan2 Á

¶¶¢ tanÁ = p ¢ c

½rThe solution of this third degree equation in tanÁ gives the angle of heel Á.

Detailed further Working-outDisplacements of empty and laden pontoon:

r0 = L ¢B ¢T0 = 60:00 ¢ 12:00 ¢ 2:50 = 1800 m3

¢0 = ½ ¢ r0 = 1:000 ¢ 1800 = 1800 ton

¢ = ¢0 + p = 1800 + 100 = 1900 ton

r =¢

½=

1900

1:000= 1900 m3

The under water hull form is a rectangular pontoon at an even keel condition without heel,so:

T =rL ¢B =

1900

60:00 ¢ 12:00 = 2:64 m

KB =T

2=2:64

2= 1:32 m

BM =ITr =

B2

12 ¢ T =12:002

12 ¢ 2:64 = 4:55 m

27


¢ ¢KG = ¢0 ¢KG0 + p ¢ zp

So:

KG =¢0 ¢KG0 + p ¢ zp

¢

=1800 ¢ 4:00 + 100 ¢ 13:00

1900= 4:47 m


GM = KB + BM ¡KG= 1:32 + 4:55 ¡ 4:47 = 1:40 m

The reduction of the initial metacentric height due to the free surface of the fuel in thedouble bottom tanks is:

GG0 =

P½f ¢ it½r =

½f ¢ 112

¢ lf ¢ b3f¢

=0:900 ¢ 1

12 ¢ 20:00 ¢ 6:0031900

= 0:17 m

As pointed out before, an equilibrium will be achieved when the righting stability momentMS equals the heeling moment MH :

µGM +

1


µ1 +

1

2tan2 Á

¶¶¢ tan Á = p ¢ c

½r


1

2¢ 4:55 ¢ tan2 Á ¡ 0:17 ¢

µ1 +

1

2tan2 Á

¶¶¢ tanÁ =

100 ¢ 8:001:000 ¢ 1900

or:2:190 ¢ tan3 Á +1:230 ¢ tanÁ = 0:421


Á = 10:00 ! LHS = 0:229Á = 15:00 ! LHS = 0:372Á = 20:00 ! LHS = 0:553Á = 16:30 ! LHS = 0:414Á = 16:50 ! LHS = 0:421 RHS = 0:421 So: Á ¼ 16:50.

28

7. Deckload on a Drill Platform

Figure 0.5: Loading a Drill Platform

A ‡oating drill platform, as given in …gure 0.5 consists of 2 sets of 3 vertical cylindricalcolumns with a diameter of 6.00 meters. The 3 columns of each set are connected toeach other at the bottom by a horizontal cylinder with a diameter of 6.00 meters. Theplatform has a draft of 25.00 meters in an upright even keel condition in seawater (½ =1.025 ton/m3). The centre of gravity G0 is situated at 26.00 meters above the base plane.A load with a mass p of 100 ton will be placed at deck above column A.

a) Determine the drafts at the columns A, C, D and F.

To avoid heel and trim, the columns C and E will be partially …lled with ballast water.

b) Determine the required amount of ballast water.

c) Determine the resulting draft of the platform.

d) Determine the initial metacentric height, including the free surface correction.

Note: The moment of inertia (second moment of areas) of a circular area with a radius Ris IT = ¼=4 ¢R4.

Solutions:

a) TA = 38.10 m, TC = 29.18 m, TD = 13.05 m and TF = 21.97 m.

b) hC = 3.45 m and hE = 6.90 m.

c) T = 27.30 m.

d) G0M = 1.44 m.

29

Detailed Solutions

Empty drilling platform:

r0 = 6 ¢³¼ ¢ 3:002 ¢ 25:00

´+ 4 ¢

³¼ ¢ 3:002 ¢ (40:00¡ 6:00)

´= 8086:5 m3

¢0 = ½ ¢ r0 = 1:025 ¢ 8086:5 = 8288:6 ton

Solution of Part 7-aPut the load p at the centre of the water plane and 26.00 meter above the base plane:

¢ = ¢0 + p = 8288:6 + 100:0 = 8388:6 ton

r =¢

½=8388:6

1:025= 8184:0 m3

¢T0 =1001:025

6¢ (¼¢3:002) = 0:575 m

T = T0 + ¢T0 = 25:000 + 0:575 = 25:575 m

The remaining stability terms in this loading condition are:

KB =6¢ (¼ ¢ 3:002 ¢ 25:575) ¢25:575

2+ 4¢ (¼¢3:002 ¢ 34:00) ¢3:00

8184:0= 8:19 m

BMT =ITr =

6 ¢³¼4 ¢ 3:004 + ¼ ¢ 3:002 ¢ 30:002

´

8184:0= 18:70 m

BML =ILr =

6 ¢ ¼4

¢ 3:004 +4 ¢ ¼ ¢ 3:002 ¢ 40:0028184:0

= 22:16 m

KG =8288:6 ¢ 26:00 + 100 ¢ 45:00

8388:6= 26:23 m

GMT = KB + BMT ¡KG= 8:19 + 18:70¡ 26:23 = 0:66 m

GML = KB + BML ¡KG= 8:19 + 22:16¡ 26:23 = 4:12 m

Moment of equilibriumThe righting stability moment MS is given by:

½gr ¢GNÁ ¢ sin Á = ½gr ¢³GM +MNÁ

´¢ sinÁ

= ½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ

Replace the shift of the mass p in a horizontal transverse direction over a distance c to theactual place by a heeling moment MH, which depends on the angle of heel Á too.


The equilibrium MS =MH should be ful…lled, so:

½gr ¢µGM +

1

2BM ¢ tan2 Á

¶¢ sinÁ = p ¢ g ¢ c ¢ cos Á

30

or: µGM +

1

2BM ¢ tan2 Á

¶¢ tanÁ = p ¢ c

¢

The transverse moment of equilibrium gives:

µGMT +

1

2BMT ¢ tan2 Á

¶¢ tanÁ = p ¢ cy

¢

or: µ0:66 +

1

2¢ 18:70 ¢ tan2 Á

¶¢ tan Á = 100 ¢ 30:00

8184:0

which gives a third degree equation in tan Á:

9:35 ¢ tan3 Á +0:66 ¢ tan Á = 0:3576


Á = 10:000 ! LHS = 0:1676Á = 15:000 ! LHS = 0:3567Á = 15:020 ! LHS = 0:3577 RHS = 0:3576 So: Á ¼ 15:020.

So, the solution of this third degree equation in tan Á gives: Á ¼ 15:00

The longitudinal moment of equilibrium gives:µGML +

1

2BML ¢ tan2 µ

¶¢ tan µ = p ¢ cx

¢

or: µ4:12 +

1

2¢ 22:16 ¢ tan2 µ

¶¢ tan µ = 100 ¢ 40:00

8184:0

which gives a third degree equation in tan µ:

11:08 ¢ tan3 µ +4:14 ¢ tan µ = 0:4768

This third degree equation in tan µ can be solved by the Regula-Falsi method, as shownbefore for tanÁ:

µ = 5:000 ! LHS = 0:3696µ = 10:00 ! LHS = 0:7907µ = 6:270 ! LHS = 0:4696µ = 6:360 ! LHS = 0:4768 RHS = 0:4768 So: µ ¼ 6:360.

So, the solution of this third degree equation in tan µ gives: µ ¼ 6:40

The drafts at the corners of the platform can be found by a linear superposition of thee¤ects of heel and trim:

TA = T + 30:00 ¢ tanÁ + 40:00 ¢ tan µ = 25:575 + 8:067 + 4:458 = 38:10 m

TC = T + 30:00 ¢ tanÁ ¡ 40:00 ¢ tan µ = 25:575 + 8:067¡ 4:458 = 29:18 m

TD = T ¡ 30:00 ¢ tanÁ ¡ 40:00 ¢ tan µ = 25:575 ¡ 8:067 ¡ 4:458 = 13:05 m

TF = T ¡ 30:00 ¢ tanÁ + 40:00 ¢ tan µ = 25:575 ¡ 8:067 + 4:458 = 21:97 m

Solution of Part 7-b

31

To bring the platform again in an upright position two columns have to be …lled partlywith water ballast:¡ column C has to be …lled with 100 ton of water ballast:

hC =100

½ ¢ ¼ ¢ 3:002 = 3:45 m

¡ column E has to be …lled with 200 ton of water ballast:

hE =200

½ ¢ ¼ ¢ 3:002 = 6:90 m

Solution of Part 7-cThe increase of the draft due to 300 ton water ballast becomes:

¢T =300

1:025 ¢ 6¢ (¼¢3:002) = 1:725 m

so the new draft of the platform in an upright position will be:

T1 = T + ¢T = 25:575 + 1:725 = 27:30 m

Solution of Part 7-dFinal displacement:

¢1 = 8388:6 + 300 = 8688:6 ton

r1 =¢1½=8688:6

1:025= 8476:7 m3

The stability terms in this loading condition are:

KB1 =6¢ (¼¢3:002 ¢ 27:30) ¢27:30

2+ 4¢ (¼¢3:002 ¢ 34:00) ¢3:00

8476:7= 8:82 m

B1M1 =ITr1

=6 ¢

³¼4 ¢ 3:004 + ¼ ¢ 3:002 ¢ 30:002

´

8476:7= 18:06 m

KG1 =8388:6 ¢ 26:23 + 100 ¢ 3:45

2+ 200 ¢ 6:90

2

8388:6= 25:42 m

G1M1 = KB1 +B1M1 ¡KG1= 8:82 + 18:06¡ 25:42 = 1:46 m

The reduction of the metacentric height becomes:

G1G01 =P½ ¢ it½r1

=2 ¢ ¼

4¢ 3:004

8476:7= 0:015 m

With this the reduced initial metacentric height becomes:

G01M1 = G1M1 ¡G1G01= 1:46¡ 0:015 = 1:44 m

32

8. Loading a Semi-Submersible

Figure 0.6: Loading a Semi-Submersible

A ‡oating structure, as given in …gure 0.6 consists of 2 sets of 2 vertical cylindrical columnswith a diameter of 8.00 meters. The 2 columns of each set are connected to each other atthe bottom by a horizontal cylinder with a diameter of 10.00 meters. The structure has adraft of 17.00 meters at an upright even keel condition in fresh water (½ = 1.000 ton/m3).The centre of gravity G is situated at 16.60 meters above the base plane of the structure.With a crane on the ‡oating structure, a mass p of 100 ton will be laden from a supplyship. The top of the crane (suspension point) is in the middle line plane of the structure,50.00 meters forward of half the length of the structure and 60.00 meters above the baseplane of the structure.

Determine the drafts at the four columns A, B, C and D during hoisting the load. It maynot be assumed that the angle of inclination is small.

Solutions: TA = TD = 12.48 m and TB = TC = 22.52 m.

Detailed Solutions

Expecting a relative large angle of trim, the calculations will be carried out as follows:

1. Place the mass p above the centre of the water plane in a horizontal plane through thesuspension point at the end of the derrick and, as a result of this, let the pontoon sinkdeeper parallel to the water plane. Determine in this situation the righting stabilitymoment MS in relation to the angle of trim µ.The righting stability moment of the semi-submersible is given by:

MS = ½gr ¢GNµ ¢ sin µ = ½gr ¢³GM +MNµ

´¢ sin µ

= ½gr ¢µGM +

1

2BM ¢ tan2 µ

¶¢ sin µ

2. Replace the shift of the mass p in a horizontal direction over a distance c to the actualplace by a trimming moment MH , which depends on the angle of trim µ too.

MH = p ¢ g ¢ c ¢ cos µ

33


½gr ¢µGM +

1

2BM ¢ tan2 µ

¶¢ sin µ = p ¢ g ¢ c ¢ cos µ

or: µGM +

1

2BM ¢ tan2 µ

¶¢ tan µ = p ¢ c

½rThe solution of this third degree equation in tan µ gives the angle of trim µ.

Detailed further Working-outDisplacements of empty and laden semi-submersible:

r0 = 2 ¢ 80:00 ¢ ¼4

¢ 10:002 +4 ¢ (17:00¡ 10:00) ¢ ¼4

¢ 8:002 = 13974 m3

¢0 = ½ ¢ r0 = 1:000 ¢ 13974 = 13974 ton

¢ = ¢0 + p = 13974 + 100 = 14074 ton

r =¢

½=14074

1:000= 14074 m3

Awl = 4 ¢ ¼4

¢ 8:002 = 201:1 m2

¢T0 =p

½Awl=

100

1:000 ¢ 201:1 = 0:50 m

T = T0 +¢T0 = 17:00 + 0:50 = 17:50 m

From the under water hull form can be calculated:

KB =2 ¢ 80:00 ¢ ¼4 ¢ 10:002 ¢ 5:00 + 4 ¢ (17:50¡ 10:00) ¢ ¼4 ¢ 8:002 ¢ 13:75

14074= 5:94 m

BM =ITr =

4 ¢³¼4

¢ 4:004 + ¼4

¢ 8:002 ¢ 30:002´

14074= 12:91 m

The vertical position of the centre of gravity of the laden semi-submersible follows fromthe …rst moment of masses with respect to the base plane:

¢ ¢KG = ¢0 ¢KG0 + p ¢ zpSo:

KG =¢0 ¢KG0 + p ¢ zp

¢

=13974 ¢ 16:60 + 100 ¢ 60:00

14074= 16:91 m


GM = KB + BM ¡KG= 5:94 + 12:91 ¡ 16:91 = 1:94 m

As pointed out before, an equilibrium will be achieved when the righting stability momentMS equals the trimming moment MH :

µGM +

1

2BM ¢ tan2 µ

¶¢ tan µ = p ¢ c

½r

34


1

2¢ 12:91 ¢ tan2 µ

¶¢ tan µ = 100 ¢ 50:00

1:000 ¢ 14074or:

6:455 ¢ tan3 µ +1:940 ¢ tan µ = 0:3553

This third degree equation in tan µ can be solved iteratively by the Regula-Falsi method,by calculating the left hand side LHS of the equation as a function of µ until a value equalto the value of the right hand side RHS = 0:3553 has been found:

µ = 10:00 ! LHS = 0:3775µ = 5:000 ! LHS = 0:1741µ = 9:450 ! LHS = 0:3527µ = 9:510 ! LHS = 0:3553 RHS = 0:3553 So: µ ¼ 9:50.

The drafts at the 4 columns are:

TA = TD = 17:50 ¡ 30:00 ¢ tan(9:50) = 12:48 m

TB = TC = 17:50 + 30:00 ¢ tan(9:50) = 22:52 m

35

9. Loading a Drill Platform

Figure 0.7: Loading a Drill Platform

A ‡oating structure with three cylindrical ‡oaters has a draft of 4.90 meters at an uprighteven keel condition in fresh water (½ = 1.000 ton/m3). Figure 0.7 shows the dimensionsand details of the structure, just before a loading operation.With a crane on the ‡oating structure, a mass p of 47.10 ton will be laden from a supplyship. The position of the top of the crane (suspension point) is given in …gure 0.7.

a) Determine the angle of inclination during hoisting the load.

b) Determine the drafts at the centres of the ‡oaters A, B and C.

To avoid a heeling angle, ballast water has been pumped in ‡oater A.

c) Determine the height of the ballast water in ‡oater A.

d) Determine the reduced initial metacentric height in this condition.

Note: The moment of inertia (second moment of areas) of a circular area with a radius Ris IT = ¼=4 ¢R4.

Solutions:

a) Á = 4:330.

b) TA = 3.56 m, TB = 5.87 m and TC = 5.87 m.

c) hA = 0.75 m.

d) G0M = 12.21 m.

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Detailed Solutions

Empty drilling platform:

r0 = 3 ¢ ¼4

¢ 10:002 ¢ 4:9 = 1154:5 m3

¢0 = ½ ¢ r0 = 1:000 ¢ 1154:5 = 1154:5 ton

Solution of Part 9-aPut the load p at the centre of the water plane and 50.40 meter above the base plane:

¢ = ¢0 + p = 1154:5 + 47:1 = 1201:6 ton

r =¢

½=1201:6

1:000= 1201:6 m3

T0 =1201:6

3 ¢ ¼4

¢ 10:002 = 5:10 m


KB =T

2=5:10

2= 2:55 m

BM =ITr =

3 ¢ ¼4 ¢ 5:004 + ¼4 ¢ 10:002 ¢ 20:202 +2 ¢ ¼4 ¢ 10:002 ¢ 10:102

1201:6= 41:23 m

KG =1154:5 ¢ 30:00 + 47:1 ¢ 50:40

1201:6= 30:80 m

GM = KB + BM ¡KG= 2:55 + 41:23 ¡ 30:80 = 12:98 m

When expecting a small heel angle, the righting stability moment MS is given by:

MS = ½gr ¢GM ¢ sin Á

Replace the shift of the mass p in a horizontal direction over a distance c to the actualplace by a heeling moment MH :


The equilibrium MS =MH should be ful…lled, so:

½gr ¢GM ¢ sinÁ = p ¢ g ¢ c ¢ cos Á

or:

tanÁ =p ¢ c

¢ ¢GM =47:1 ¢ 25:101201:6 ¢ 12:98 = 0:0758

So the angle of heel becomes:Á = 4:330

Solution of Part 9-bThe drafts at the ‡oaters are:

TA = 5:10¡ 20:20 ¢ tan(4:330) = 3:56 m

TB = 5:10 + 10:10 ¢ tan(4:330) = 5:87 m

TC = TB = 5:87 m

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Solution of Part 9-cHeel will be avoided when the moment due to the water ballast equals the moment due tothe load MH :

¢wb ¢ 20:20 = MH = 1182:21 so: ¢wb = 58:5 ton

Then the height of the water ballast in the ‡oater A becomes:

hwb =58:5

¼4

¢ 10:002 = 0:75 m

Solution of Part 9-dThe displacement becomes:

¢1 = ¢+ ¢wb = 1201:6 + 58:5 = 1260:1 ton

r1 =¢

½=1260:1

1:000= 1260:1 m3

T1 =1260:1

3 ¢ ¼4

¢ 10:002 = 5:35 m


KB1 =T12=5:35

2= 2:67 m

BM1 =ITr1

=3 ¢ ¼

4¢ 5:004 + ¼

4¢ 10:002 ¢ 20:202 + 2 ¢ ¼

4¢ 10:002 ¢ 10:102

1260:1= 39:32 m

KG =1201:6 ¢ 30:80 + 58:5 ¢ 0:375

1260:1= 29:39 m

GM = KB +BM ¡KG= 2:67 + 39:32¡ 29:39 = 12:60 m

The reduction of the metacentric height becomes:

G1G01 =P½ ¢ it½r1

=¼4

¢ 5:0041260:1

= 0:039 m

With this the reduced initial metacentric height becomes:

G01M1 = G01M1 ¡G1G01= 12:60 ¡ 0:39 = 12:21 m

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10. Buckling of a Drill String

Questions:

a) Explain why a drill string in a deep oil well …lled with mud will buckle, even thoughthe entire string is under tension when hanging in air.

b) What is the e¤ect (on buckling) of adding an additional downward force on the top ofthe drill string?

Solutions:

a) When the well is …lled with mud an upward vertical pressure force acts on the bottomof the string. This places the lower segment in compression; an unconstrained columnin compression. If its length is long enough, it will buckle.

b) If we assume that the end of the string is on the well bottom (but that the mud pressurestill acts on it) then adding weight on top will only increase the compression at thelower end and aggravate the situation.

Hydrostatics Solutions

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