Static stability of floating structures

Delfi University of Technology

STATIC STABIlITY OF

FlOATING STRUCTURES

(Lecture Notes OT-4621)

J.M.J. Journée

Delft University Of Technology

Repoit No. 1110-K August 1997

'11.J IJeift

Paeulty of Mechanical Engineering and Marine Technology

2

i

Introduction'

i

2

Definitions

₃

3 Equations of Equilibrium

7

4'

Righting Stability Moment

₁₁

5

Static Stbility Curve

₁₇

6 Loading A Flòating Structure

21

7 Free Surface Correction

25

8 Exercises

29

8.1 No. 1: Scribanti Formula 29

8.2 No. 2: Float-On Float-Off Pontoon 30 8.3 No. 3: Unloading. A Ship with A Crane 32

8.4 No. 4: Unloading A Pontoon. . 34

8.5 No. 5: Lift Operation by A Crane Pontoon 36 8.6 No. 6: Loading. A Drilling Platform 38 8.7 .N:o. 7: Löadi:ng A Semi-Siibmersib1e with A Crane 40 8.8 No. 8: Loading A Drilling Platform with A Crane 42

.9

Solutions

₄₅

9.1 No. 1: Scribanti Formula ., 45

9.2 No., 2: Float-On Float-Off Pontoon 47 9.3 No. 3: Unloading A Ship with A Crane , 59

94

No. 4: Unloading APontoon . . . ₆₁

9.5 No 5: 'Lift Operation by A. Crane Pontoon 65

9.6 No 6; Loading A .Dri1li.n Platform

...

69

9.7 No. 7: Loading A Semi-Submersible with A Crane. 73 9.8 No. 8: Loading A Drilling Platform with' A Crane 7.5

Chapter 1

introduction

These lecture notes on static stability of floating structures in still water are based on the lecture notes "iEvenwicht en Stabiliteit" on this subject by Journée (19.88), used in the past for the Lecture X2-MT1 Offshore Hydromechanics.. When evalúating' and translating: these Dutch lecture notes for the new OT-4621 Lectúre on Offshòre Hydromechanics, for some explainatory texts fruitful use has been made of the chapters 3 and 4 of volume of the book "Basic Ship Theory" by Rawson and Tüp,per (.1983). Finally, a number of detailed solved exercises on typical stability problems with wall-sided floating structures have been added to these lecture notes.

Being on a floating structure, one takes for granted that it will float the right way LUE. This it will only do, if it is correctly designed Furthermore, in service it will experience many forces (e.g. from wind and waves) trying to turn it over. It must be capable of resisting these by' what is termed its stability. Too much stability is undesirable, because in a seaway it can cause unwanted excessi:ve motions and it can be costly. Thus, as with so many 'other design features, stability is a compromise. Because a floating structure Will meet varied conditions during its life,. its ability to survive should ideally be expressed in statistical terns and stability standards should be set accordingly. No floating structure can be guaranteed u.nder all conditions.

The static stability of a floating structure encompasses the uprighting properties of the structure when it is brought out of its state of equilibrium by a disturbance to its force and/or moment of equilibrium. As a result of these loads, the structure will translate linearly andjor rotate about its centre of gravity.

The static as well as the dynamic properties of the structure play a role in this. In these lecture nôtes however, only the static properties of the structure will be considered. lt is supposed here that disturbances of the state of equilibrium will be brought abou.t so slowly that ali dynamic effécts can .be ignored.

Chapter 2

Definitiòns

The body axes and the notations, as used here in static stability calculations, are presented in figure 2.1 below.

Figure 2.1: Body Axes and Notations

Pianes have been defined here as generally used in ship hydrostatics.

Most ships have only one plane of symmetry, called the middle line plane xz, which becomes the principal plane of reference.

The design water plane or load water plane, called here simply water plane, is

a

plane perpendicular to the middle line, plane, chosen as a plane of rèférence' at the. still water surface.

The base plane is a plane perpendicular to the middle line plane through the keel or the bottom öf the ship. It may or may not be parallel to the water plane.

Planes perpendicular to both the middle line plane and the water plane are called

trans-verse pianes.

A transverse plane at half the. length of the ship, so amidships, is called the midship sectioñ of the ship.

So-called hydrostatic fórces and moments, caused by the surrounding water, will act on

a structure floating in still water. The buoyancy (Dutch:: "waterverplaatsing" or drijende kracht") ofa structure immersed in a fluid is the vertical upthrust it experiences due to the displacement of t:he flúid., The structure, in fact, experiences all of the hy-drostatic pressures which were present before it displaced the fluid. This buoyancy is the resultant of all the forces due to hydrostatic pressures on elements of 'the undèr water

por-tion. This is,, according to Archimedes' principle (law of flotation), equal to the weight of the fluid displaced. The 'centre of the volume of fluid displaced by a floating structure is known as the centre of buoyancy B (Dutch: "dru'lçkingspunt"), see figure 2.2' (a)', The first moment of the under water volume about the centre of buoyancy i's zero.

z Óf

1G

The projections of the centre of' buoyancy B of a ship in the plan and in the section are

known' as the longitudinal centre of buoyancy (LCR) and the vertical centre of buoyancy (VCB). Should the ship not be symmetrical below the water'plane, the centreof buoyancy will not be situated in the middle line plane. Its projections in plan may then be referred to as the transverse centre of buoyancy (TCB).

Defining it formally, the centre of gravity G (Dutch: "massazwaartepunt") of a structure is' that point through which, for static considerations, the wholé weight Of the structure may be assumed to act, see figure 2.2 (b). The first moment of mass or weight about the centre of gravity is zero.

For the sake of simplicity and understanding, hereafter the disturbances and the hydrostatic

forces and moments are considered in the plaùe of the dawing only. The structure is considered to be a floaling cylinder' with a constant but arbitrarily shaped cross section. The centre of gravity Ç and' the 'centre of buoyancy B of the structure are supposed to be positioned in the' piane of drawing.

Rotations in the plane of drawing are defined here as heel. For trim, the same principle holds 'as will' be explained here for 'heel. For combinations o'f heel and trim; a superposition can be used. This will be shown' in some of the exercises, given at the end of these 'lecture

notes..

(buoyancy force)

pgV

Figure 2.2: Definition of Centres and Forces

In this figure:

p mass density of water g' acceleration of gravity

V volume of displacement of the floating structure (nabla) m mass' of the floating structure

G

'(gravity force)

Often in literature on 'Ship and offshore hydromechanics,. the mass of 'displacement pV 'is given by the symbol '(delta).

Stability 5

A floating structure is said to be in a state of eqùilibrium when the resultant of all the

forces acting on it is zero and t:he resulting moment of these forces is also zero.

Three di erent states of equilibrium or natures of stability can be distinghùised:

If the structure, subject to a small disturbance from a position of equilibrium, tends

to return to this state it is said to be in a state of stable equilibrium or to possess positive stability. This principle is given in figure 2.3.

Figure 2.3: Stable Equilibrium or Positive Stability

If, following the disturbance, the structure remains in its néw position, then it is said to be in a state'of neutral equilibrium or ta possess neutral stability, This.

principle is given in figure 2.4.

Figure 2.4: NeutraI Equiliibrium or Neutral Stability

If, following the disturbance, the excursion from the equilibrium position tends to increase, the :structure is said to be in a state of unstable equilibrium or topossess

negative stability. This principie is given in figure 2.5.

Chapter 3

Euations of Equilibrium

If a structure is floating freely in rest in a fluid, the follów.ing equilibrium conditions in the plane of drawing in figure: 2.2 will have been fulfilled:

- horizontal equilibrium: the sum of the' horizontal forces equals zero vertical equilibrium: the sum of the vertical forces equals zero'

- rotational equilibrium: the sum of the' moments about 'G (or any other point) equals

zero.

'Horizontal Equilibrium

A translation in a horizontal direction leads to no resultant hydrostatic force, so that the structure is in neutral equilibrium for this' type of disturbance. This: equilibrium is not of further interest here

Vertical Equilibrium

For a floating structure, a vertical downward movement (sinking deeper) results in an augmented buoyancy force which will tend to move t'he structure upwards. It tends to return the structure to its state of equilibrium 'and the structure is stable for this 'type of disturbance.

Archimedes' principle holds 'for the vertical equilibrium:

pgV =gm

if an additional mass pis placed on this structure, its previous equilibrium will 'be disturbed. The structure will sin'k deeper and heel until a new state of equilibrium has been reached. The new vertical equilibrium is given by:

pg(V+LW)' = g(m+p)

in which:

L V = increase of t'he volume of displacement of t'he floating structure

If t'he mass, p has been 'placed n the structure in such ,a manner that it only sinks deeper: parallel to the Water plane, the change of draught T 'follows from:

¿W = ¿.TAWL

p, or:

p, PAWL

'Here, AWL is the area of the water plane, whereby it has been' supposed that thisarea is constant over the draught interval T.

Rotational Equilibrium

Rotation about G, heel,, results in a moment acting on the. floating structure, about which no 'generalization is possible. The structure may display a stable, a neutral or an unstable equilibrium'.

if anexternal heeling moment (Dutch: "kenterend 'moment"). acts on't'he structure as given

in figure 3.1,, it 'follows from the rotational equilibrium:

MH = pg'Vy = gmy

Figure 3.1: 'Heeling Moment.

From this follòws too that if 'nó external moment acts on the structure, the lever arm' y should' be zero, so that:

MH = O results'in:

y = O

This means that for any 'floating structure 'in rest, 'the centre of buoyancy B and t'he centre of gravity G will be situated on the same vertical line. If this is not so, the' structure will heel' Or trim until they are 'situated on one vertical line.

This means too that the longitudinal position of the centre of gravity can be found easily from' the longitudinal position of the centre of buoyancy, which can be derived from the under water' geometry of the structure.

Shifting Masses and Volumes

If on a structure a mass p is shifted over a distance. c; as' shown in figure 3.2, the overall centre of gravity 0o will be shifted to G parallel to this. displacement over a distance equal

to:

This can be seen easily by writing down the first moment of masses ith respect to a

vertical 'line' through Go:,

L'

pV.GoG1 = pVO + p.c

new '=

old + change

p. C

=

Stability :9

C

Figure 3.. 2: Shifting Masses

Here, the t tal mass pV includés the mass p.

In case of a freely suspendèd mass p,for instance cargohanging on a hoisting rope of a crane on the structure, this mass p shouJd be considered to be concentrated at the suspension. point of the crane hoisting rope., his is the case immediately after hoisting the cargo. The vertical distance of the cargo above deck or beneath its suspension point is of no

importance at all At each angle of heel, the cargo acts a vertical force acting downwards at the suspension point of the cargo iii 'the' crane The centre of gravity cf the structure including cargo does 'not shift as the inclination of the structure changes.

Likewise can be pòir ted out that in case of heeling of the floating structure, the centre of buoyancy shifts parallel to a line through the centres of the volúmes' of the emerged and the immersed water displacement wedges The volume 'of the emerged wedge, in fact, has been. shifted to the immersed wedge, see figure 3.3.

Figure 3.3: Shifting Buoyancy

difference in angle of heel intersect each other on á line with respect to which the first moment of volumes of the two wedges is zero This is a line through the centre of the water plane.

This means that the structure heels and/or trims about a line through the centre of the water plane, the centre of floatation. In case of a heeling ship with symmetric water planes, this is half the breadTh of the water plane.

Chapter 4

Righting Stability Moment

A structure, as given in figure 4.1, is flöating,at rest,

Figure 4,1: Equi1ibrii.im at Zero. AngIe of Heel

if one adds an (external) heeling moment MH to this structure, the structure will heel with an angle . As a result of this heeling, the shäpe of the under water part of the structure will change. The centre of 'buoyancy shifts from B to B on a line parallel to the' line through 'the centresof the emerged and immersed wedes , see figure 4,2.

An equilibrium will be achieved when the righting stability moment M equals the (exter-nat) heeling morn ent MH:

MS=MH

in which the righting stability möment is deflnedt by:

M

pgV.GZ

For the righting stability lever arm can be written:

.GZ=GNsin

With this, the righting stability moment becomes: M

pgV.GN.sin

The so-called metacentre N (Dutch: "metacentrurn") is the point of intersection of the lines through the vertical buoyant forces at an angle of heel zero and at an angle of heel .

emerged wedge

gm,

pgV

Figure 4.2:: Equilibrium at Angle of Heel

The position of this metacentre N, is dependent on the slape of the structure at and near t.o its water piane, an area also called the "area between water and wind". Together with the angle of heel , the shape of the emerged and immersed wedges control the shift of the

centre of buoyancy from B to B and thereby also the position of N.

A floating structure is said to be wall-sided if, for the angles of heel to be considered, those portions of the outer bottom covered or uncovered by the moving water pláne are verfical with the structure upright.

In case of of a wall-sided structure (seefigure 4.3), vçrticai lines are boundaries of the wedges in the cross sections and thereby the emerged and immersed wedges being described by right angle triangles, the position of the metacentre N can be cakulated easily:

BN =

.

(i

-- tan2 Scribanti formula

in which 'T is the transverse moment of inertia (second moment of areas) of the unheeled water piane about the axis of inclination for both half water planes,

In the solution of Exercise No. 1 at theend of these lecture notes, the validity of this formula has been shown for a rectangular barge.

The Scribanti formula shows that the metacentre N is almost fixed for small angles of heel. Until angles of heel of atout 10 degrees this point is called initiàl metacentre M (Dutch "aanvangsmetacentrum") At small angles of inclination, tan2 in the Scribanti formula is very small, when compared to 1.0, and can be ignored.

S, for small angles of heel;

immersedwedge

and

GZ = GMsin

Until angles of heel of about 10 degrees, this is a fairly good approximation forships too,

BM=

'T

Stability 13

or:

gm

_pgV

Figure 4.3: WaIl-Sided Structure with Large Angles of Heel

which are wall-sided at the "area between water and wind" over a rélatively large part. of the length.

At larger angles of heel, the metacer tre depends on the angle of heel and N ,.. so the full

Scri.banti formula,, has to be used:

= .

(i

+

tan2 and GN, .

in here, the term (IT'/V) i represents the effect of the horizontal part of the shift of Ze

to z and the term (IT/V) .

tan2 represents the effect of the vertical, part of the shift of z to z. This is made clear in the solution of Exercise No. I, given at the end of these lecture notes.

The Scribanti formula can be written also as:

BN, = BM+MN

$M.(i+tn2q)

BM =

and

MN = -M.tan2,

Keep :fl mind that at angles of heel, at which the water plane changes strongly, this formula cannot be used anymore. Fûr instance, this is the case when the bilge comes out of the water or the deck enters the water This possibility aIways has to be checked, when carrying out stability calculations.

It has been shown above that the metacentre is defined as the intersectión of lines through the vertical buoyant force at ₌ and at =

- initial metacentre M: 00 and 2 = very small

- metacentre N :

= 0 and

2 = larger

For symmetric under water forms like ships, these two metacentres are situated' in the middle line plane, a. fact that requires a third definition:

- pro-metacentre: and

2 = + iq

in which ¿q is a very small increase of a larger angle of heel qf.

This pro-metacentre, which is not on the middle.line plane of the ship, will not be discussed here.

The stability lever arrn

= GN sin

will be determined' by the hydrostatic properties of the submerged structureand the position of the centre of gravity of this structure. This is reason why the following expression 'for GN, has been introduced:

GN=KB+BNKG

In here, K is the keelpoint of the structure (see figure 4.4).

Figure 4.4: Metacentre Points

The magnitude of KB follows from the under water vOlume of the structure and the magnitude of'k follows from the mass distribution of the structure.

It has been mentioned already that, if within the range of considered angles of 'heel the shape of the water plane does not change very much (wall-sided structure), then.:

Figure 4.5: Submerged Flóating Structure

For fully submerged floating structures can be written:

GN =GM

= KBKG

Stability 15 in which 'T V Using this, the expression: for BN can be written as:

GN = KB ± BM

(1-l-.tan2)

-For small angles of heel, this expression becomes:

GM=KB+BMKG

leading to:

GN, =

M + BMtan

Fully submerged structures, like tunnel segments. during installation or submarines (see figure 4.5), have no water plane. The definitions of BN and BM show that these válues are then zero, which means that the metacentre coincides with the centre of buoyancy.

Chapter 5

Static Stability Curve

If a floating structure is brought under a certain angle of heel , the righting stability moment is given by:

Ms

pgVGZ

= pgV GN, 'sin

pgv.{GM+MN}.sinçb

Figure 5.1: Stability Lever Arm GZ In this:

= GNsin4

{M±MN}.sin

= stabilityleverarm

This value determines the magnitude of the stability moment.

For practical applications i is very convenient to present the stability in the form of'righting moments or lever arms about the centre of gravity G, as the floating structure is heeled

at a constant displacement This is expressed as a function of the angle of heel Such a function will in general appear as in figure 5.2 and: is known as the static stability curve 'or 'the GZ curve.

Because the stability lever arm is strongly dependent on the angle of heel ,. a graphic

of , as'given in' figure 5.2, is very suitable for judging the stabillit3. For an arbitrarily floating structure form, this curve will not be symmetrical with respect to ç5 O

It is obvious that for symmetric forms like ships, the curve of static stability will be symmetric with respect to O. In that case, the right half part of this curve will be

presented only, see figure 5.3.

1

i

Figure 5.2: Static Stability Curve

10 20 30 40 50 60 70

(deg)

-Figure 5.3: Ship Static Stability Curve

The heel angle at point A in this figure, at which the second derivative of the curve changes sign, is roughly the angle at which the increase of stability due to side wall effects (Scribanti formula) starts to be counteracted by the fact that the deck enters the water or the bige comes above the water.

Stability 19

Figure 5.4 shows the static stability curve in case of a negative initial metacentric height GM, while ? becomes positive at some reasonable angle of heel , the so-called angle

of loll'.

G

Figure 5.4: Static Stability Curve with A Negative GM

If the floating structure is momentarily at 'some angle of heel less than ., the moment acting. on the structure due to GZ tends to increase t:he heeL if the angle is greater than ,, the moment tends to reduce the heel. Thus the angle. is a position of stable

equilibrium. Unfortunately, since the curve' is symmetrical about the origin, 'as _i is decreased, the floating structure' eventually passes: through the. upright concition and will then' suddenly lurch ovét towards the angle : on the opposite side and overshöot this val:ue before reaching a 'steady state. This causes an unpleasant rolling motion, whiçh is dften the only direct indication that the heel to one side is due to a negative GM rather than to a positive heeling moment acting on the structure.

'Some important characteristics of the static stability curve can be mentioned here: 1. Slope' at. The Origin

For small angles of heel, the righting lever arm is' proportional to the angle f incli-nation, the metacentre is effectively a fixed point. 'it f011ows, that the tangent to the 'GZ curve at the origin represents the metacentric height GM

This can 'be shown easily for the case of a wail-sided structure:

{GNcs.sin}

=

{(M-i-

LBM.tan2qf).sinq}

This derivative becomes GM for zero heel. This means t'hat the initial metacentric height GM can be of great importance for the course of the curve, especially at smaller angles' of heel, and for the area under the curve (see item 5' below).

Maximum GZ Value

The maxi urn

vaiue is proportional to the largest steady heeling moment that the floating structure can sustain without capsizing. Its value and the angle at which it occurs are both important. The shape of the above water part of the floating structure is of great importance on the maximum attainable value of the stabiility lever arm.

Range of Stability

At some angle of heel, often greater than 90 'degrees, the value reduces to zero and becomes negative for larger inclinations. This angle is known

as the angle of

vanishing 'stability (Dutch: "kenterhoek"). The range of angles for Which GZ is positive is known as the range of stability (Dutch: "stabiliteitsomvang"). This range is of great importance on the maximum attainable area under the stability curve and thereby also on the maximum potential energy that the structure can absorb via :a roll motion. The shape of the above water part has a large influence on the angle of vanishing stability; compare curves I and H in figures '5.1 and 5.2. For anglés within the range of stability, a floating structure will return to the upright state when the heeling moment is removed.

Angle of Deck Edge Immersion

For most floating structures, there is a point of inflexion in the curve, corresponding roughly to the angle at which the deck 'edge becomes immersed. This point is not so much of interest in its own right as in' .the fact that it provides guidance to the designer upon the possible effect on stability of'certain design changes. So, the shape

of the above water prt of the structure can have a large influence on the static

stability.

More .or less the same can be said when the bilge or the 'bottom chine emerges.

Keep in mind that for wall-sided structures, when the deck enters the water or the bottom chine comes above the water level, the immersed and emerged wedges are no longer triangular; calculations become much more cumbersome.

Area Under The Static Stability Curve.

When judging the stability properties of 'a structure flòating upright, the work that has to be done to reach a heel angle is an important parameter:

P*

_Aslçb

=

pgv.j GN.sinq.'d

This means that the area under the static stability curve is an important magnitude for the jûdgement of the stability properties. It represents the ability of the floating structure to absorb roll energy imparted to it by winds, waves or any other external

Chapter 6

Loading A Floating Structure

When a load has been put at an arbitrarily place on a floating structure,. this structure will sink deeper parallel to the original water plane and it will heel and/or trim as well. This parallel sinkage is caused by an increase of the total mass of the structure. Heel or trim after parallel sinkage will be caused by the generatiOn of a rnoment because of a shift of both the centre of gravity and the centre of buoyancy, displacing them from their vertical coimear positions.

Suppose a floating structure with a volume of displacement V0, a centre of gravity G0 and a centre of buoyancy B0, as given in figure 6.1

Then, a mass p will be placed on this structure, which results in a new volume of

displáce-ment V = Vo +

V0, a néw centre of gravity G and a new centre.of buoyancy .B.7 No moment will be generated when the centres of gravity and buoyancy shift horizontally over an equal distance. With the first moment of masses., it can be seen that this is true when the mass p has been placed above, at or undér the centre of the, by parallel sinkage caused, added displacement V0 of the structure, caused by parallel sinkage.

mass. p

CoG of edded displacement

Figure 6.1: Parallel Sinkage

The first moments of the masses of the structure and the masses of displacementare defined

here with respect to a line through G0 and B0:

{pVo+p

}.a

{pVo}.O

_{+ {}

}.c

{ pVo + p ¿w0}. b = {pVo} . o + {p .

Because p p V0, it follòws: a = b.

This means. that the new centre of buoyancy B and and the new centre of gravity G are situated on one vertical line., So no additional heeling moment remains.

The equilibrium after loading the mass at an arbitrarily place on the structure, can be. determined in two steps:

A. Place the mass p in the required horizontal plane vertically above or under the in-creased volume due to parallel sinkage of the structure over a distance T0.

When the water plane does not change very much over this increase of draught, one

can write:

PAWL

Then, the centre of the added displacement lies vertically above the. centre of the original water plane at a distance T0/2 from this original water piane.

Figure 6.2: Load Above Water Plane Centre

The centre of buoyancy B0 shift to B because of the added volume of displacement

V0 ₌

p/p

AWL.

The centre of gravity G0 shifts to G because of the increase of the mass of the structure with p = p. V0.

These shifts can be calculated easily with the first, moments ofmasses and volumes, especiailly when the shape of the water plane does not change very much during the draught increase ¿T0: horizontal part of horizontal part of vertical part f vertical part of

G0G=

B0B=

G0G=

B0B=

p a p.

pVo+p

a p

p.v0+p

(+ b)

. pVo ±p

(+zT/2)

. p V0 .+ p

Stability 23

In case of symmetric hull forms with respect to a plane through G0, B0 and M0, the centre of the water plane lies in this piane too. Then, because a = O,. both the horizontal part of G0,G and the horizontal part of B0B are zero.

The metacenters are given by:

B0M0 = and

BM=

'T

V0 + LW0

B. Replace the resulting moment, caused by a movement of the nassp horizontally over a distance c to its desired position, by a heeling moment MH..

Figure. 6.3: Shifting of A Load

This heeling moment MH is equal to the righting stability moment Mg:

MH=MS

g.p.c.cos.q = g.pV.GN.sinq

The heel angle q follows from this equilibrium of moments:

i pV . GN:. sin q

= arccos

P.0

If the static stability curve lever arms GZ sin 47 are knon as functions of ç5,

then ç5 can be solved in an iterative manner.

In these expressions, the mass Of displacement pV includes the mass p:

V = V0+LW0 =

p

For wall-sided structures, one can express the angle of heel qS as:

arccos pV

{M +

M.

tan2 . sin çS

or using tan :

BMtan'q+M.tançb =

2 pV

This third degree equation iu tan can be sólved iteratively by the Regula-Falsi method, by calculating the Left Hand Side (LHS)' of the equation as a function of until a value equal to the value of the Right Hand Side (RHS) has been found. For sma11' angles of heel, at which the shape of the water plane does not change very much, the expression for becomes:

= arctan

I

pV.GM

can be solved directly in this situation.

Much of the data used in stability calculations depends only on the geometry of the struc-ture. The total mass of the structure m follows from the volume of displacement V. The longitudinal position of the centre of gravity G follows simply from the longitudinal posi-tion of the centre of buoyancy B,, which can be derived from the under Water geometry of the structure.

The vertical position of the centre of gravity must be known ibefôre the stability can be assessed finally for a given lòading condition. Since the vertical position of the centre of gravity KG may be sometimes ten tinïes greater as the initial metacentric height GM, it must be known very accurately if the metacentric height is to be assessed with reasonable accuracy. KG can be calculated for a variety of loading conditions, provided ft is accurately known for one precisely specified loading condition.

'By measuring the draughts fore and aft in the upright condition ( o 00) and the

knowii geometry of the structure, the volume of displacement V can be found.

With this displacement and the measured angle of heel ( _{= q). after shifting a known} mass p on the structure over a distance c, the vertical position of the centre of gravity can be found easily by writing the previous equation for small values of in another form:

P.0

GM=

pV. tan(i

-The known underwater geometry of the structure and this "measured" CM-value gives the vertical position of the centre of gravity G.

An experiment to determine the metacentric hei'gh't in this way is called an' inclining experiment (Dutch: "hellingproef"). The purposes of this experiment are to determine the displacement and the positión of the centre of gravity of 'the structure in a precisely known condition. It is usually carried out when the structure is as. nearly complete as 'possible during initial building. An additional inclining may 'be carried out followIng an extensive modernization or a major refit.

Keep in mind 'that V and G 'have to be corrected, when removing the mass p from the structure.

Chapter 7

Free Surface Correction

Free surfaces of liquids inside a floating structure can have a large influence on its static stability They reduce the righting stability lever arm

Figure 7.1: Floating Structure with A Tank, Partially Filled with A Liquid

When the structure in figure 7.1 heels as a result of an external moment MH, the surface of the fuid in the tank remains horizontal This means that this free surface heels relative to the structure itself, so that the centre of gravity of the structure (including liquid) shifts. This shift is analogous to the principle of the shift of the centre of buoyancy due to the emerged and immersed wedgs, as, discussed' in chapter 4. Also, the under water geometry of the entire structure and the boundaries of the wedges at the water plane as well as in the tank play a role. They determine, together with the angle of heel,, the amount of shifting of the 'centre of buoyancy.

In case of a vertical boundary of the wedges, so a wall-sided tank as given in figure 7.2, 'this shift' can be calculated easily with the first moment of' volumes with triangular cross

!.1

tg24

y 2

Figure 7.2: Effect of A Liquid. Free ;Surface in A Wall-Sided Tank

In this figure:

p' density of the liquid in the. tank y volume of the liquid in the tank

j _{transverse moment of inertia (second moment of areas)}

of the surface of the fluid in the tank ( = O)

The influence of the .shift of the mass p'v from. b to b on the . position of the centre of buoyancy can be determined by the &s,t moment of vôlumes with triangular cross sections too: direction shift of centre of gravity of fluid shift of centre of grav.ity of total structure horizontal vertical tanq . tan2 q -tanç5

t1tan2q

Stability 27

Figure. 7.3: Metacentric Height Reduction Caused by Wall-Sided Tanks

This means that the righting stability lever arm will be reduced by GG" sin , with:

GG" =

(1-iJtan2)

pV 2

The magnitude GG" is called the frec surface correction or the reduction of the

metacentric height.

For small angles of hee1 for, which tan2 is small relative to 1.0, one can write for the reduction of the metacentric height;

GG" GG'

pV

For more free surfaces (more tanks) the reduction cf the metacentric height becomes: {p'it}

_{(i + tan2)}

pV

GZ

i rad

'Ir.

Figure 7.4: GZ-Curve, Corrected 'for Free Surfaçe

It should be noted that the effect of the 'free surface is independent of the position of the tank in the floating structure. The tank can be at any height in the structure and at; any position in its length or breadth.

The effect is. also independent of the amount of liquid in the tán'k, provided that the

moment of inertia (second moment of areas), of the free surface is substantially unchanged when inclined. A tank which is almost empty or almost full can suffer a change' in moment of' inèrtia of the free ;surfàce and it is for this reason that if, during an inclining experiment tanks cannot be completely full or empty; they are specified to. be' half full.

It is usual .to regard the free surface effect as being a virtual rise df the centre of gravity of the structure, although it will be appreciated that, this is merely a matter of convention, which has no real factual basis.

A similar effect can arise from movements, of granular cargo stowed in a bulk such as grain. Here,, however, there is no simple relation between the 'angle of inclination of the ship and the slope of the surface of the grain At small angles of heel the grain is not likely to shift, so there is no influence on initial stbi'lity. When' the heel angle becomes great enough, a significant shift. can take place very abruptly. Ihe effects can be equally abrupt!

Chapter .8

Exercises

Eight exercises on a variety of stability probemè are given here. Detailed solutions are given in the next chapter.

8.1

No. 1: Scribanti Formula

gm

Figure 8. 1: Rectangular Pontoon

A rectangular pontoon with length L, breadth B and draught T is floating in an upright even keel cnditiòn. Caused by an external heeling moment MH, the pontoon assumes a relatively large angle of heel, .

Question:.

Proie for this case the validity of the 'Scribanti formula; BN, .

(i

+ tan

and show that the term (IT/V) . i represents the effect of the horizontal part of the shift of Ze to zj and that the term (IT/V) tan2 represents the èffect of the vertical part of

the shift of Ze to z..

M N

B.

.L ½..tg

E

-

-

--waterline in submerged condition

cwi

nI

Figure 8.2; Sketch of Float-On Flòat-Off Pontoon

In a harbour with fresh water (p = 1_000 ton/rn3) an empty rectangular Float-On Float-Off Pontoon, as given in figure 8.2 has a midship draught of 1.55 metre in an upright even keel condition.

Questions:

A rough inclining experiment has been carried out, by fully filling a port side tank aft (tank VI), which is bounded by half the length of the pontoon and the longitudinal middlè line plane of the pontoon, with fresh water (pwb = 1.000V ton/rn3). The measured angle of heel q5 was 1.46 degrees.

2-1. Determine the position of the centreof gravity G0 and the initiaI rnetacentric height G0M0 of the empty pontoom

(Solution G0 amidships at middle line plane with KG0 4 02 metre and G0M0 = 45 15 metre)

2-2. Determine the trim angle 01 during the inclination experiment.

(Solution: Oi 0.38°)

2-3. Determine the draughts at the four corners of the pontoon dùring the inclination experiment.

(Solution: T1 = 1.28, 2.00, 2.04 and 2.76 metre, respectively)

Cross section Cross section

A-A B-B 7.0 30.0 owl L = 108.0 m

B =300rn

D

75m

TcwI= 5.3m Tmax= 11.5m DW= 12600 t

.--..

-1 L.... .... I,

I ...

p-... 21.0 27.0 27.0 21.0 -, 1080 4.8v B 18.0 I...

i

:-:

L-Exercise No. 2 ₃₁

2-4. Determine the angle of heel 2 in case of an 80 per cent filled tank, during the

inclination experiment. (Solution: 2 1.160)

During operation, the pontoon will be sunk down in a protected bay by loading, ballast seawater (p Pwb = 1.025 ton/rn3) in all 1'6 tariks. This ballast water is supposed to have

an equal height h in all 16 tanks.

2-5. Determine the initial metacentric height GM at an even keel draught of 7.50 metre, supposing that water will just not cover the deck.

(Solution: GM 9.93 metre)

2-6. Determine the GM-value at an even keel draught of 7.50 metre, supposing that water had just. covered the deck.

(Solution: GM 1.55 metre)

2-i. Determine the GM-value, when the pontoon has been sun:k down until an even keel draught of 11.50 metre.

(Solution: GM = 1.62 metre)

Now, the pontoon picks up a drilling rig, with the following specifications: - upright even kee1 condition

- mass = 4920 ton - = 20.00 metre

- water plane dimensions: 40 x 40 mette - no free surfaces of liquids in any tank

2-8. Determine the initiai metacentric height of the rig. (Solution: GM = 25.94 metre)

2-9. Determine the initial metacentric height of pontoon+rig, suppossing that they just hit each other when de-ballasting the pontoon during loading of the rig (centre of rig above centre of pontoon).

(Solution: GM 5.51 metre)

2-10. Determine the initial metacentric height of pontoon+rig when all water ballast has been removed fôm the pontoon.

(Solution: GM = 10.86 metre)

2-li. letermine the angle of heel when, due to an inaccurate loading of t:he pontoon, the centre of the rig is amidships but 1.0 metre outsidé the middle line pláne of the.pontoon. (Solution: = 2.61°)

E

Q

LOE cs .4

I 5OO rn

base

Figure 8.3: Unlóading A Ship with A Crane

ad

Ùt

A loaded wall-sided ship with its own dérrick on board, as given in figure 8.3, is floating without heel at an even keel conditiön in fresh water (p. = 1.000 ton/rn3) with 'a volume of displacement V of 12000 rn3. The lóad, which is a mass poI 250 ton, is placed in one of the holds on the tanktop of the double 'bottom. The centre of gravity of this mass lies in the' middle line plane, 3.00 metre above the base plane' of the ship.

The suspension point of the cargó in the derrick lles 25MO metre above the base plane 'of the ship. When the derrick is turned outboard fully, thissuspension point lies 15.00 metre from the middle line piane of the ship.

As soon as the mass has been hoisted from the tanktop 'of the double bottom, the ship heels 2.0 degrees.

After hoisting the mass further and turning it outboard fully, the ang1 'of heel becomes

Exercise No. 3 ₃₃

Q uestion:

Determine the initial metacentric height G0M of the ship, before the unloading operations. (Solution: G0M = 0.44 metre)

Horizontal cross section Under deck

Figure 8.4: 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 metre breadth B. = 6. 00 metre

- depth D

2.50 metre

- mass

= 20 ton

The pontoon is coating at an even keel condition in fresh water (p= 1!000 ton/m3),. Tie vertical position of the centre of gravity above the base plane KG0 is 1.20 metre. The pontoon is lòaded with two masses of p = 50 ton each, of which the centres of gravity are positioned at half the length of the pontoon, y, = 1.75 metre from the middle line plane of the pontoon and z, = 1.25 metre above the deck of the pontoon, see figure 8.4.

E o N

/\

N

7/

/

_\

y . 300m 300m

<-T -<T T>

- _;__._ --10.00 m 10.00 m I 1000m 10.00 m

Exercise No. 4 35

Questions:

4-1. Determine the initial metacentric height of pontoon+rnasses. (Solution: GM = 0.975 metre)

4.2. Determine the angle of heel of the pontoon, after unloading one of the masses. It may not be assumed that this angle is smaI1.

(Solution': = 11.3°)

4-3. To obtain an upright position again, two tanks will b.e equally filled with water ballast (pvjb = 1.0.00 ton/rn3). Determine the mass of the water ballast and the reduced initial metacentric height.

8.00

Figure 8.5: Crane Pontoon Lift Operation

A rectangular pontoon has the following principal dimensions: length L = 60.00 metre

- breadth B = 12.00 metre

- depth D = 6.00 metre

One of the double bottom tanks is partly filled with fuel with a density pf = 0.900 ton/rn3. The length of this tank i is 20 00 metre and the breadth b1 is 6 00 metre

The pontoon is floating at an even keel 'condition with a draught T0 = 2.50 metre in fresh water (p = 1.000 ton/rn3).

The vertical' position of the centre ôf gravity of the pontoon, including fuel, above the base plane KG0 is 4.00 metre.

A sketch of the pontoon in this situation is given in figure 8.5.

o,

a)

Exercise No. 5 37

Then, a mass of p 100 ton will: be hoisted from the quay. When the derrick is turned outboard fully, the suspension point of the cargo in the derrick lies 13.00 metre above the

base plane and 8.00 metre from the middle line plane of the pontoon.

Question:

Determine the maximum angle of heel of the pontoon during hoisting this load. (Solution: = 1!6.5)

TOP VIEW

Figure 8.6: Loading A Drilling Platform

A floating drilling platform, as given in figure 8.6 consistsof 2 sets of 3 vertical cylindrical columns with a diameter of 6.00 metres. The 3 columns of each set are connected to each other at the bottom by a horizontal cylinder with a diameter of 6.00 metres.

The platform has a draught of 25MO metres in an upright even keel condition 'in seawater (p 1.025 ton/rn3). The centre of gravity G is situated at 26.00 metres above the base plane.

Fi

_________

D

I,

H

'CL

G01 E

E

CD

= i o:t

E

40m

¿6m

jP=

loot

_SIDEVÍEW

¿6m

_J

4

'E

îA E

'B

I

Exercise No. 6 . 39'

Questions:

A load with a mass p' of 100 ton will be placed at deck above column A. 6-1. 'Determine, the draughts at the columns A, C, D' and F.

(Solution: TA = 38.10 rn, T 29.18 m, TD = 13.05 m and TF = 21.97 rn)

'Io avoid heel and trim, the columns C and E will be partially filled with ballast water. 6-2. Determine the required amount of ballast water.

(Solution: h = 3.45 rn and hE = 6.90 rn)

6-3. Determine.the resulting draught of the platform. (Solution: T = 27.30 m)

6-4. Dermine the initial metacentric height, including the free surface correction. (Solution: G'M = 1.44 m)

Note: The moment of inertia (second moment f areas)' 'of a circular area with a radius R' is:

T0

=17.00m

KG0 = 16.60 m

08

60 80 50. G0 IC

e

w

100 t

IB

o

V y 1.00 t

Figure 8 7: Loading A Sémi-Submérsible with A Crane

A floating structure, as given in fgure. 8.7 consists of 2sets of 2 vertical cylindical columns with a diameter öf 8.00 metres. The 2 columns of each set are connected to each other at the bottom by a horizontal cylinder with a diameter of 10.00 metres.

The structure has a draught of 17.00 metres at an upright even keel condition in seawater (p = 1.025 ton/rn3). The centre of gravity G is situated at 16.60 metres above the base plane of the structure.

e

Exercise No. 7 41

Question:

With a crane on the floating structure, a mass p of loo, ton will be loaded from a supply ship. The top of the crane (suspension point) is in the middle line plane of the structure, 50.00 metres forward of half the length of the structure and 60.00 metres above the base plane of the structure;

Determine the draughts at the four columns A, B, C and D during hoisting the load. (Solution: TA TD = 12.48 m and TB = T = 22.52 rn)

Dimensions are given withoUt ioad P (47.1 t) ø10.0 p. 35.0 CROSS SECTIONA- A 010.0 P 2.5

4.9

Figure 8.8: Loading A Drilling Platform with A Crane.

A floating structure with three cylindrical floaters has a draught of 4.90 metres at an upright even keel condition in fresh water (p = 1.000 ton/m3). Figure 8.8 shows the dimensions and details of the structure, jit before a loading operation,

G0 _5.01

Exercise No. 8 43

Questions:

With a crane on the 'floating structure, a massp of 47.10 ton will be loaded from a supply ship. The positiOn of the top of t!he crane (suspension point) is given in figure 8.8.

8-1. Determine the angleof inclination duiing 'hoisting the load:. (Solution: '= 4.33°)'

8-2. Determine the draughts at the centres 'of the floaters A, B and C. (Solution: TA = 3.56 m, TB = 5?87 m and Ta 5.87 m)

To avoid aheeling angle, ballast water has been pumped i floater A. 8-3. Determine the' height of the ballast water in floater A.

(Solution: hA 0.75 m)

8-4. Determine the reduced initial metacentric height in this condition. ('Solution: G'M = 12.21 m)

Note: The moment of inertia (second moment of areas)' of a circular area with a radius R is:

or expressed in the moment of inertia of the water plane IT, the volume of displacement of the pontoon V and the angle of heel :

Chapter 9

Sólutions

9.1

No. i:, Scribanti Formula

The horizontal part of the shift of Ze of the emerged wedge to z of the inimersed wedge causes a horizontal: displacement of the centre of buoyancy B, equal to BM tan. The first moment of volumes with respect to the middle line plane of the pontoon in the heeledì condition Is given by:

{L!B.T}!{BM.tançl}

=

{LB.T}.{o} +

new = old

+

change

So:

BM

or expressed in' the moment of inertia (second moment of areas) of the water plane 'T' and the volume of displacement of the pontoon V:

B.M=

The vertical part of the shift of Ze of the emerged' wedge to z of the immersed wedge causes a vertical' displacement of the centre of buoyancy B, equal to MN.

The first moment of volumes with respect to a plane parallel to the unheeled water plane through the centre of buoyancy B of the pontoon in the unheeled condition is given by:

{L.B.T}.{MN}

'

{L.B.T}.{o}

+

new =' old'

_±

change

So:

MN =

Jtan2

With this, for the pontoon the Scribanti formula can be written as:

BN = BM+MN#

=

(i+.tan2q)

9.2

No. 2: Float-On Float-Off Pontoon

Volume of displacement V0 of empty pontoon:

V0 = LBT

= 108.0030.00i.55 = 5022m3

Mass of displacement o of empty pontoon:

Lo = p. Vo

= 1.000 5022 = 5022 ton

Solution of Question 2-1

Determine 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 centre of buoyancy of the pontoon and also the centre of gravity of the pontoon are situated amidships at half the 'length of the pontoon.

Mass of water ballast ub in a fully filled tank:

wb = Pwb'lb"h

= l000 27.00. 7.50 . 7.50 =' 151J9 ton

Centre of gravity of water ballast in a fully filled tank with' respect to half the length of the pontoon, t'he middle line plane'and the base piane:

Xwb

13.50 m

Ywb = +3.75m

Zwb = +375 in

The calculation will be carried' out in two steps:

Step 1:. Suppose the water ballast concentrated on 'a vertical line through the centre of the water plane in a h rizontãl plane through the centre of gravity of the water ballast

(parallel sinkage).

Step 2: Shift the centre o'f gravity of the water ballast to the right position by addinga

heeling moment (heel)..

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

pLB.T = 1.000108.0030.00T1 = 3240T1

10 ± wb = 5022 + 1519 = 6541 6541

'So: T1

Centre of buoyancy above keel KB1:

KBI=

Metacentre abôve centre of buoyancy B1M1,

M_JT_

i_;_

B2

LBT Th2T1

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 in a angle of heel çt = 1.46°.

Replace this shift by a heeling moment MH: B 2.02 2 2 G1M1 = (L0 _±twb) .tan qi 1519 . 3.75 = 1'.O'lrn (5022 _± 1519) . tan(1.46°) = 34.18 m This formula for G1M1 can be used, because q5 is very small:

M1 N = B1 M1 . . tan2

B1M 0.000325 0.00

Herewith is the position of the centre of gravity of the pontoon including water ballast

KG1, 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 first moment of masses of the pontoon including water ballast with respect to the base plane:

KG1 = KG0 ₊ wb Zv

(pontoon + wb) (pontoon) (wb)

MH =

wbgywbcos1

The righting stability moment of the pontoon M is:

M =

+

g

G1M1 sin

Because of the equilibrium MH = Ms, it follows for the initial metacentric height GiM:

/-wbYwb

Solution No. '2 49

(5022 ₊ 1519)

3.96 - i59 .375

5022

Centre of buoyancy above keel' KB0:

KB0 =

= 038m

Metacentre above centre of buoyancy BóMò:

'T 1LB3 B2

BM

O 0

_12

LBT0 12.T0 30002 48.39m I2 1.55

Herewith is the initial metacentric height of the' empty pontoon G0M0 known: G0M0 = KB0 + B0Mo - KG0

0.78 + 48.39 4.02 = 45.15 m

Sólution of Question 2-2

Determine the trim angle 01 during the inclination experiment Centre of buoyancy above keel KB1:

KB1 1.01 m (as found before) Longitudinal metacentre above centre of buoyancy BML:

BM

I iL

_{- V1 - LBT1 - 12T1}

'L BL3 L2

108002

= 12 2.02

= 481.19rn Centre of gravity above keel KG1:

KG1, = 3.96 m (as found before)

Herewith is the initi1 longitudinal metacentric 'height of the pontoon including water

ballast G1 MIL known:

G1MÌ,L = KB1 ± 'BIMIL KG1

± LOI + 481.19 3.96 = 478.24 rn

-

=4.02m

A longitudinal shift of the centre of gravity of the water ballast over a distançe Xb will result in a trim angle Oi.

So:

Solution of Question 2-3

Determine the draughts at the four angular points of the pontoon during the inclination experiment.

Half the heel displacement is:

tan 30;00 tan(1.469). 0.38 rn

Half the trim displacement is:

Solution of Question 2-4

Determine the angle of heel 42 in case of an 80 per cent filled tank, during the inclination experiment.

Mass o1 water ballast wb in the 80 per cent filled tank:

wb Ú.80 1519 = 1215 ton

Centre of gravity of water ballast in the 80 per cent filled tank with respect to half t:he length of the pontoon, the middle line plane and the base plane:

Xwb =

13.50 m

Ywb = +3.75 m ane: Xwb =

13.50 m

Ywb = +3.75 m twb = 0.80 . 3.75 = +3.00 m = 0.80 . 3.75 = +3.00 m (5022 + 151.9) . 478.24 = 0.380 1108.00 tan(O.389) = 0.36 m

Draughts at angular points of pontoon:

Starboard aft: 2.02 + 0.38 + 0.36 = 2.76 m Starboard forward:

2.02 + 38 - 0.3.6

= 2.04 m Port side aft: 2.02 - 0.38 + 0.36 = 2.00 m Port side forward: 2.02 - 0.38 - 0.36 = 1.28 m Replace this shift by a trimming moment MHL:

MHL = g xwb

cos1

The longitudinal righting stability moment of the pontoon MSL is: MSL ₌

(o +

wb) g G1 MiL 5x11

Solution No. 2 51

Suppose a parallel sin:kage. The new draught T2 becomes:

O+&vb =

2 =

p.L.B.T2

= i..000108.0030.00T2

5022+1215 So: T2 = 1.92 ni

Centre of buoyancy above keel KB2:

KB2 0.96rn

The position of the centre of gravity KG2 follows from the first moment of masses of the pontoon. and the water ballast with respect to the base plane:

2KG2 = ¿0 KG0 +JXtubZwb

KG0 + wb B2M2 5022 4.02 - 1215 . 3MO

= 3.82m

5022 + 1215

= 0.15m

Metacentre above centre of buoyancy B2M2:

r i 1D3

=

= LBT2 = i2.T2

jj.L1.L) L?

30 002

= 12 1.92

= 39.06m

Herewith is the. initial metacentric height of the pontoon with the 80 per cent filled tank with frozen water ballast G2M2 known:

G2M2 = KB2 + B2 M2 - KG2

= 0.9.6, + 39M6 - 3.82 = 36.20 m

Now we let the water ballast unfreeze and the reduction G2G of the metacentric height becomes

rl. ri,

-

Pwb

pV2

1.000. 2MO 7.50 1.000 (5022 + 1215)

Herewith is the reduced initial metacentric height of the pontoon with the 80 per cent filled tank GM2 known::

GM2 = G2M2G2G

= 36.20 - 0.15 = 36.05. m

So:

Add a heeling moment.

A transverse shift of the centre of gravity of the water ballastover a distance Ywb will result

in a angle öf heel ₂

Replace this shift by a heeling, moment MII:

MH =

wbgy(,bcos2'

The righting stability moment of the pontoon Ms is:

M5 =

2g'GM2sin2

Because of the equilibrium MH M5, it fóllows for the angle of heel 2:

tan ç

= L2GM2

&vb 1215 . 3.15 = or: = 1.160 (5022

₊

1215) 36.05 Note 1:

In this. exercise, the effect of the free surface in 'the tank is very small. In case of frozen cargp the angle of heel

_'2

would be:

tan Ywb

G2M2 12153.75

(,5022 + 1215) . 36.20 or: = 1.15 Note 2;

The small reduction of the metacentric height has been obtained in the design of the pontoon by a subdivision in the' transverse direction in 4 compartments. As a result of this, the transverse moment of inertia (second moment of areas) of the free surface of the water ballast t has been' reduced considerably.

With a tank over the full breadth, the reduction G2G of the metacentric height would be:

Pwbt

-p.v2

1.000 . ₁₂ 27.00. . 30.O0 = 9.78 m (instead of 0.15 m)

Solution of Question 2-5

Determine the initial metacentric 'height GM at an even keel draught of 7.50 metre, sup-posing t:hat water will just not enter the deck.

Volume 'of displacement V of pontoon:

V = LBT

= 108.00 . 30.00 . 7.50 = 24300 rn3 i.000. (5022+ 1215)

Solution No. 2 53 Tötal mass of displacement of pontoon:

= p.v

= L025 24300 24908 ton Mass of water ballast wb in pontoon:

24980.

- 5022 = 19886 ton

Height 4 of water level in baJiast tanks:

Lwb

pLB

19886

1.025 108MO . 30MO = 5.98 m

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

Xwb = 0.00 m

= 0.00m

h 5.98

Zwb =

--, = 2.99 m

For the ballasted pontoon, the position of the centre of gravity k follows from the first moment of masses of the empty pontoon and the water ballast, with respect to the .bae plane:

z0 KG0 +

Zb So:

KG=

¿0.JG0 .+ LwbZwb

50224.02+ 198862.99

-. 24908

Centre of buoyancy above keel K:

GG'

7.50

= 3.75m

Metacentre above centre of buoyancy BM:

'T 1LB3 B2

V

LB.T

12T

30:002

= 12.750

= 10.00 m

The reduction of the metacentric height CG1 due to the free surface of the water ballast becomes: t) pV 16 1.025. 27.00 7.50

= 0.62 n

=

24908

BM=

= 3.20m

Herewith is the reduced ini'tia1 metaceutric height of the pontoon G"M known:.

G'M

KB+BMKGGG'

3.75 + 10.00 -, 3.20

- .62 = 9.93 rn

Solution of Question 2-6

Determine the GM-value at an even keel draught of 7.50 metre, supposing that water had just entered the deck.

With respect to question 5', here BM changes only: 'T

V

The traúsverse moment of inertia (second moment of areas) of the water plane can be found by subtractiòn of moments of inertia of rectangles:

'T

= 2 {9.00. (30.003

-

16.003) + 118.00 . (30.003

-

23.003)}

= 39428 m or by using Steiner's rule:

'T 2 {. (9.00. 7MO3 + 18.00 ..35O)'+9OO. 7.00 11.502+ 18.00 3.5C. 1.252}

= 39428 rn4

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

Ir

39428

BM =

24300 = 1.62 rn

'Herewith is 'the reduced initial metacentric height of 'the pontoon G'M known:

G'M =' KB+BMKCCG'

= 3.75 + 1.62 '3.20 O62 = 1..55m

Solution of Question 2-7

Determine the 'GM-value, when the pontoon has been sinked down until an even keel draught .of Il 150 m'etre

Volumé of displacen ent V of pontoon:

'V 24300+ 2 (9.00. 7.00+ 18.0,0: 3.50):. (11.50 7.50) = 25308 m Total mass of displacement of pontoon

Solution Ño. 2 55

Mass of water ballast. wb in pontoon:

=

-= 25941 - 5022 -= 20919 ton

Height h of water level in ballast tanks: h

pLB

20919

= L025 108.00 30.00 = 6.30 rn

Centre of gravity of total water ballast wi'h respect to half the length of the pontOon, the middle line plane and the base plane:

Xwb

= 0.00m

Ywb = 0.00 iji

h 6.30.

Zwb

==---=3.1'5m

For the ballasted pontoon, the position of the 'centre of gravity k f011ows from the first moment of masses of the empty pontoon and .the water ballast with respect to the base plane:

L

KG

= Lo

KG0 + wb wb

5022 4.02, + 20919. 3.15'

= 25941 3.32m

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

V = 108.00 . 30.00 7.50 3.75 + 2 (9.00. 7.00+ 18.00 .350) (11.50

7.50) 950

= 10070i1 m4 So: 100701 KB

= 25308

= 3.98 m

Metacentre above centre of' buoyancy AÏ:

'T 39428

BM =

= 25308

= 1.56'm

The reduction of the metacentric height due to the free surface of the water bällast

becomes: >'(pwt it) pV 16. 1.025 . 27.00 7.50 25941

= 0.60rn

So:

KG=

Lo KGo + twb Zwb

Herewith is the reduced initial metacentric height of the pontoon G'M known:

G'M = KB+BMKGGG'

= 3.98 + 1.56 3.32 - 0.60 = L62 m

Solution of Question 2-9

Determine the initial metacentric height of pontoon+rig, suppossing that they just hit each other when inflating the pontoon during loading the rig (centre of rig above centre of pontoon).

Draught T at even keel of pontoon:

T = 7.50 + 3.00 = 10.50 m

Volume of displacement V of pontoon:

V = 24300 + 2 (9.00 7.00 + 18.00. 3.50) . (10.50 - 7.50

Total mass of displacement ¿ of pontoon:

L =pV

= 1.025 . 25056 = 25683 ton Mass of water ballast L in pontoon:

&vb O

= 25683 - 5022 = 20661 ton

= 25056 m3

Solution of Question

Determine the initial

Vrig = Trig

=

2-8

metacentric height of the

trig

4920 _à 4800m rig.

= 3.00m

= 1.025 = Vrig 4800 Lrig Brig = 40.00 . 40MO Trig 3.00 1.50m K Brig =

=

Lrig B;jg 40.002 = 44.44 m BMrig

Vrig Lrig Brig

Trig - 12 Trig - 12 3.00

Solution No. 2

_5j

Height h öl water level in ballast tanks:

h

_{- p.L.B}

1'wb

20661

622 1.0251O8.00.30.0O - m

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

Xwb 0.00 m

!íwb = 0.00 rn

h 6.22

zWb,=---_3.11m

For the ballasted pontOon + iig, the position of the centre of gravity

k

follows from the first moment of masses of the empty pontoon, the water ballast and the rig with respect to the base plane. of the pontoon:

(

+ trig)

.X = 0 KG0 + wb

+ trig

With z,.j = 7.50 + 20.00 2750 metre, it is found:

K + /.wb ± 1nig Zn9

+rig

5022 . 4.02 + 20661 3.111 ± 4920 27.50

25683 ±4920

= 7.18rn

The centre of buoyancy above keel

R

follows from the first moment of volumes of the individual :parts of the under water geometry of the pontoon and the rig with respect to the base plane of the pontoon:

(y + Vn9)

k

= 108.00 . 30MO . 7.50 3.75

+ 2

(9.00 . 7.00± 18.00 - 3.50) . 3.00 9.00 + 4800 (7.50 + 1.50)

= 141129 rn4

So:

Metacentre above centre of buoyancy BM

'T 39428 + 40.00 . 40.00

BM

= V + V

25056 ± 4800 = 8.47 m

The reduction of the metacentric height GG' due to the free surface of the Water ballast becomes:

F -

>1(Pwb t)

= 4.73m

pV 16. 1.025 27.00 7.50 = 0.51 rn = 25056 + 4800

KB=

141129 25056 ± 4800

Herewith is the reduced initial metacentric height of the pontoon G'M known::

G'M = KB+BMKGGG'

= 4.73 +8.47 - 7.1'8 - 0.51 = 5.51 m

Solution of Question 2-10

Determine the initial metacentric height of pontóon+rig when all water ballast has been removed' fom the pontoon..

= o ± rjg = 5022 + 4920 = 9942 ton

V =

=

= 9700 m V 9700 T

= L

. B = 108.00 . 3Q.00E = 3.00 m T

₌

3.00

=1.50m

'T B2 30.002

BM =

V = 12 . T = 123.00

25.00 m 5022 . 4.02 + 4920 . 27.50 KG. 5022±4920 15.64m

GG' = 0.00 m

GM = KB + BM - KG

1.50 + 25.00 - 15.64 = 10.86 m

Solution of Question 2-11.

Determine the angle of heel when, due to an inaccurate loading of the pontoon, the centre of the rig is amidships 1.0 metre outside the middle line plane of the pontoon.

KB

MH = 4920g1.00cosqf

Ms = (5022 +4920) g 10.86

sin 4920 . 1.00

tanq =

= 0.0456 (5022 + 4920) 10.86 or: qf = 2.61°

9.3

No.. 3: Unloading A Ship with A Crane

Mass free of tanktop:

. = 2°.

Thiseqiiil.ibrium will be achieved when the righting stability moment M equals theheeling, moment MH, which s zero:

M5

pgVGZ

= pgV GN

sin q

=

pgV.:(M±MN).:sin.

= pgv .

'(GM + BM tan2)

.

=MH =0

Thisequi;librium at 2° gives: the following equation with the unknowns BM and M:

GM-M.tan2

:=

or:

GM +0M0061i0 BM = 0

Mass turned outboard fully:

= i7°.

This equilibrium will be achieved when therightingstabi1ity moment Ms equalsthe heeling

moment. MH:

Ms =

pgV.'(GM+M.tan2q)

= MH = 250g1J5.00.cos.

'This equilibrium at = 17° gives a second equation with the unknowns BM and GM: 1.000 . g. 12Ú00 (GM +

'tan(I7°))

. sin(17°) = 250 g. 15.00 . cos(17°)

or:

GM + 0.046736 BM = 1.0221

The solution of these two equations with' the irnknówn BM and CM gives:

BM = 22.16 m

GM = 0.01 rn

The position of the original centre of gravity G0 follows from the first moment of masses of the ship with hoisted cargo with. respect to the' base plane:

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

G0G = k7 - KG0

250. (25.00 3.00) = 0A6 m

The metacentric height befóre the unloading operations of the cargo becomes: G0M = G0G-f-GM

94

No. 4: Unloading A Pontoon

Solution of Question 4-1

L. = Lo+2p

200±2.50 = 300 ton

300

V =

= 1.000 = 300 m3

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

So: V 300 T

= L B = 40.00

. 6.00 = 1.25 m

KB =

= 0.625m

I

B 6002

BM=--=

=

_=2.40m

V

12.T

12.1.25

The vertical position of the centre of gravity of the loaded pontoon follows from the first moment of masses with respect to the base plane:.

= oKGo + 2.p.(D+z)

So:

¿0.KG0 + 2.p(D±z)

200 1.20 + 2 . 50 (2.50 + 1.25) =

= 2.05m

300 Herewith, the initial metacentric height is known:

GM = KB±BMKG

= 0.625 ± 2.40 - 2.05 = 0.975 m

Solution of Question 4-2

The pontoon is loaded 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 through

the centre of the mass and, as a result of this, let the pontoon sink deeper par1lel to the water plane. Determine in this situation the righting stability moment M in relation to the angle of heel .

The righting stbiIity moment Ms of the pontoon is given by:

pgVGNsinq = pgv.{GM±MN}.sin

= pgV

{GM + tan2 sin q

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

MH = pgccosq

Finally, the equilibrium M5 = Mj1r should be fulfilled.

pgv

{M

± BM

tan2}

. sin q = p . g . c cos or:

{GM +

=

Detailed Further Solution of Question 4-2

Displacements of loaded pontoon:

=

o+p = 200 +50 = 250 ton

V 250 m3

Due to the paraJilel sinkage, the under water hull form remains. a rectangular pontoon at an even keel condition withoùt heel, so:

T=

V 250

P.0

pV

= i.04m

L . B 40.00 6.00 KB T L04

= 0.52m

'T B2 6.002 BM

= V

12T

12I.02

= 2.89rn

The vertical position of the centre of gravity of the loaded pontoon follows from the flrst moment of masses with respect to the base plane:

VKG = V0KG0 ± p(D+z)

V0KG0 + p.(D+z)

V 200 1.20 + 50 (2.50 + 1.25) = 1.71 rn 250 So: KG