UCNS606 Ship Stability IV

PART AUnit 1

2 Marks

Angle of heel when turning, Wind heeling 1

Define ‘Pivot Point’Pivot Point is the point of application of the hydrodynamic forces acting on the ship's hull. It is the point around which the ship rotates.

2

What is the purpose of the turning circle manoeuvre?• The purpose of the turning circle manoeuvre is to determine the turning characteristics of

the ship.• Basic information obtained from the turning test are advance, transfer, tactical diameter,

steady turning diameter, final ship speed and turning rate in the steady state.

3

What is ‘Advance’ with regards to turning circle?• Advance is the distance travelled in the direction of the original course by the centre of

gravity of a ship from the position at which the rudder order is given to the position at which the heading has changed 90° from the original course.

4

What is ‘Transfer’ with regards to turning circle?• Transfer is the distance travelled by the centre of gravity of the ship in a direction

perpendicular to the original heading of the ship to the position at which the heading has changed 90° from the original course.

5

What is ‘Tactical diameter’ with regards to turning circle?• Tactical diameter is the distance travelled by the centre of gravity of a ship from the position

at which the rudder order is given to the position at which the heading has changed 180° from the original course. It is measured in a direction perpendicular to the original heading of the ship.

6

What is Steady turning circle radius with regards to turning circle?Steady turning circle radius is the steady radius of the turning circle when a steady rate of turn is achieved. This state is usually achieved by the time the ship has altered course between 90º and 180º however this will vary from ship to ship

7

What are the information obtained from the turning test?• Basic information obtained from the turning test are advance, transfer, tactical diameter,

steady turning diameter, final ship speed and turning rate in the steady state.

8

What is ‘drift angle’?• Yaw is the angle between the ship’s fore and aft line and the direction of travel of the ship’s

centre of gravity at any instant during the turn.

9

What is the IMO standard requirement for the size of tactical diameter in relation to ship’s length?

10

What is the IMO standard requirement for the size of Advance in relation to ship’s length?

11

State the formula to calculate the centrifugal force, when the ship achieves a steady rate of turn. Centrifugal Force = WV2 tonnes gR

W is ship displacement;V is ship speed in metres per second;g = 9.81 m/sec2;R is the radius of the turning circle.

12

State the formula to calculate the centripetal force, when the ship achieves a steady rate of turn.

13

State the formula to calculate the angle of heel when turning, when the ship achieves a steady rate of turn.Tan = V2 BG

g R GM

14

State the formula to calculate the increase in Draft Due to List/Heel.• Draught when heeled = (½Beam x SineƟ) + (Upright draught x CosƟ)

15

State the formula to calculate the heel due to beam wind.

• P.y = W.GM SinƟ or SinƟ = P.yW.GM

16

When the Turning circle manoeuvre is carried out in a ship’s life?• Turning circles are normally carried out during the sea trials of the vessel prior to handover

from builders to owners. The fact that the manoeuvre may have to be carried out at sea, for collision avoidance purposes, makes this an essential for the ship's Master and Watchkeeping Officers.

17

What is the use of the turning circle diagramme displayed in the bridge?

Unit2Incli 1 What is the purpose of 'Inclining experiment'?

ning experiment

• Chapter 2 of the Code on Intact Stability for all Types of Ships, details the information that must be provided to the master of all ships in order that stability calculations may be accurately conducted to ensure the ship’s safe operation.

• A key element of this information is the Inclining Test Report that details the calculation procedure conducted to determine the ship’s light KG and displacement.

2

Which ships are required to conduct the ‘Inclining experiment’It requires that every passenger ship regardless of size and every cargo ship of 24 m or over be inclined on completion in order to determine the value of the KG in the light condition.

3

Define ‘Lightship condition'Lightship condition:the ship complete in all respects, but without consumables, stores, cargo, crew and effects, and without any liquids on board except that machinery and piping fluids, such as lubricants and hydraulics are at operating levels.

4

What other data are determined during the Inclining experiment apart from light ship KG• During the experiment the longitudinal position of the centre of gravity (LCG) for the light

condition will also be determined.

5

State the formulae to calculate GM during inclining experiment

GM = w x d x Pendulum lengthWxDeflection of the pendulum

6

Why the pendulum weight should be suspended in a trough of hydraulic oil?Trough of oil-plumb weight has vanes to help dampen movement

7

Why the mass of the inclining weight(s) used should be sufficient to provide a minimum list of 1° and a maximum list of 4°?

1. The mass of the inclining weight(s) used should be sufficient to provide a minimum list of 1°

and a maximum list of 4° to each side. This is to ensure that the formula: Tan ƟLIST = GGHGM

remains valid, being applicable to small angles of inclination only.

8

Why the Decks should be free of water, snow or ice when conducting inclining experiment?1. Decks should be free of water. Any water trapped on deck will move during the test and

reduce the accuracy of the result. Snow and ice must also be removed.

9

Why all tanks should be verified as being completely empty or full when conducting inclining experiment?

1. Ideally tanks with rectangular free surfaces should only be slack so that the free surface effect can be accurately determined.

2. Slack tanks must have the contents accurately determined with respect to liquid mass and Kg.

10

When is the first inclining experiment conducted for a ship?

11

How many pendulums are used to conduct inclining experiment?The use of three pendulums (but no less than two) is recommended, one forward, one midships and

one aft to allow bad readings at any one station to be identified.

Unit 3Dry docking and grounding 1

What is the use of ‘shores’ during drydocking?• For ships those have a relatively small percentage of flat bottom area additional measures

must also be taken such as using side shores to support the ship in the upright condition when in the dry dock.

2

What is P force during drydocking?the P force takes over supporting the ship and the role of the buoyancy force in supporting the ship reduces

3State the formula to calculate ‘P’ force at Any Stage During Dry-dockingP force (t) = Reduction in TMD (cms) x TPC

4

State the formula to calculate ‘P’ force, only incritical period During Dry-docking

P= COT (cms ) xMCTCDist  of  LCF  foap

5

what is the 'critical instant' with regard to drydocking• The maximum loss of GM of concern occurs at the instant immediately prior to the ship

settling on the blocks forward and aft - this time being termed the critical instant.

6

what is the 'critical period' with regard to drydockingThe interval of time between the stern post landing on the blocks and the ship taking the blocks overall is referred to as the critical period

7 what is Virtual loss of GM during drydocking

8

State the formula to calculate ‘Loss of GM’ as a Result of a Rise in ‘G’During Dry-docking

• Virtual loss of GM = PxKGW-P

9

State the formula to calculate ‘Loss of GM’ as a Result of a Fall in ‘M’During Dry-docking

GM = P x KMW

10

ShouldFSC be applied while calculating the loss of stability during drydocking?

11

ShouldFSC be applied while calculating the loss of stability of a vessel when grounded?

12

When will be the maximum loss of GM occurs during Dry-docking

13

What will happen if the ship should become unstable during the critical period

Unit 4Bilgi 1 What is a bilged ship?

ng A 'bilged' ship is one that has suffered a breach of the hull through grounding, collision or other means and a compartment holed below waterline to such an extent that the water may flow freely into and out of the compartment.

2

State the formula to Calculate the KM of a box-shaped vessel

KMBOX = Draught

2 + LB 312V

3What is the difference between Added weight method and Lost buoyancy method

4

What are the effects of bilging an empty amidships compartment?• The changes in draught and stability when a compartment becomes flooded due to

damage can be investigated by either of two methods:(1) the lost buoyancy (constant displacement) method, or;(2) the added weight method.

5

Define ‘Permeability’Permeability (μ) is the percentage ratio of the space available for the entry of water into a compartment.For an empty compartment μ =100% andfor a compartment, so full that water cannot enter at all if bilging occurred, μ =0%

6

State the formula to calculate ‘sinkage’ due to Bilging

S = Vol of lost buoyancyIntact water plane area

7 When a DB tank below water level get bilged, how a vessel’s BM is affected?

8

State the formula to calculate permeability if Stowage factor and Broken stowage are given• For any compartment μ may be calculated by:

μ = (Broken Stowage/Stowage Factor) x 100%• Where SF is the stowage factor in cubic metres per tonne and BS is the broken stowage per

tonne of cargo.

9 If a forward compartment is bilged, what will happen to a vessel with even keel?10

If a stbd compartment is bilged, what will happen to an upright vessel?

Unit 5Shear Forces and Bending Moments in

1 define shear force• Shear force is a force which tends to break or shear a beam across (perpendicular to) its

major axis.

ships

2

define bending moment• Bending moment at any point on abeam is the total moment tending to alter the shape of

the beam.

3Show on a load curve, where maximum shear force values will occur?it should be evident that the maximum values of shear force will occur at the bulkhead positions

4Show on a shear force curve, where maximum bending moment values will occur?The maximum bending moment value of occurs at amidships

5

Show on a shear force curve, where the Point of inflexion of the bending moment will occur?It should be noted that a point of inflexion of the bending moment curve will occur in any position where there is a shear force maximum

6

What is type 1 Stability Software on a Loading instrument?Type 1

• Software calculating intact stability only (for vessels not required to meet a damage stability criterion)

7

What is type 2 Stability Software on a Loading instrument?Type 2Software calculating intact stability and checking damage stability on basis of a limit curve (e g. for vessels applicable to SOLAS Part B-1 damage stability calculations, etc.) or previously approved loading conditions

8

What is type 3 Stability Software on a Loading instrument?Type 3

• Software calculating intact stability and damage stability by direct application of pre-programmed damage cases for each loading condition (for some tankers etc.)

9 Which ships are required to carry a Loading instrument?

10

What is a passive Calculation system of a Loading instrument?• A passive system requires manual data entry,

11

What is an active Calculation system of a Loading instrument?An active system replaces the manual entry with sensors reading and entering the contents of tanks, etc.,

12

What is an integrated Calculation system of a Loading instrument?• A third system, an integrated system, controls or initiates actions based on the sensor-

supplied inputs.

PART BUnit 1

6Marks

Angle of heel when

1 Explain the operational conditions that affects Turning ability.Operational conditions that affects Turning ability

• Once operational and a vessel has reason to perform a tight turn, e.g. Man Overboard, it should be realized that a deep laden vessel will experience little effect from wind or sea conditions, while a vessel in a light ballast condition, may experience considerable leeway

turning, Wind heeling

with strong winds prevailing.• Another feature exists with a vessel that is trimmed by the stern. She will generally steer

more easily, but the tactical diameter of a turn could be expected to decrease; while a vessel trimmed by the head will still decrease the size of the circle, but will be more difficult to steer.

• Should the vessel be carrying a list at the time of conducting the circle, the completion time could expect to be delayed. Also, turning towards the list would expect to generate a larger turning circle than turning away from the list side, bearing in mind that a vessel tends to heel in towards the direction of the turn, once helm is applied.

• It should also be realized that a ship turning with an existing list and not in an upright condition, especially in a shallow depth, could experience an increase in draught. Such a situation could also result in reduced buoyancy under the low side causing a degree of sinkage to take place. This increase in draught would not be enhanced if the turning action was also being conducted at high speed.

2 State the Environmental Conditions those are required for turning circle maneuver’?Environmental Conditions required for turning circle maneuver

• IMO prescribes the maximum environmental condition for carrying out the manoeuvring trials as follows:

1. Wind• The trials should not be conducted with a true wind speed greater than Beaufort 5 (19

knots).2. Waves

• The trials should be carried out in sea states less than 4. This means that the significant wave height should be less than about 1.90 meter, with most probable peak periods of about 8.8 seconds.

3. Currents• No specific information on current limitations is given.• Wind, waves and currents can significantly affect the ship manoeuvrability. However, the

IMO suggests a method to account for environmental effects during turning tests only.• The effect of the current velocity on the path of a ship during turning circles is shown in

figure. This type of figures can be used to obtain some information on current velocity and direction.

3

State the ‘Manoeuvring Standards’ required for turning circle manoeuvres.Manoeuvring Standards

1. The standard manoeuvres should be performed without the use of any manoeuvring aids which are not continuously and readily available in normal operation.

2. Conditions at which the standards applyIn order to evaluate the performance of a ship, manoeuvring trials should be conducted to both port and starboard and at conditions specified below:

1. deep, unrestricted water; (IMO standards require that the water depth should exceed four times the mean draft of the ship.)

2. calm environment;3. full load (summer load line draught), even keel condition; and4. steady approach at the test speed.

4 With a neat diagramme explain ‘Transfer’, ‘Tactical diameter’ and ‘Turning Radius’.

• Transfer is the distance travelled by the centre of gravity of the ship in a direction perpendicular to the original heading of the ship to the position at which the heading has changed 90° from the original course.

• Tactical diameter is the distance travelled by the centre of gravity of a ship from the position at which the rudder order is given to the position at which the heading has changed 180° from the original course. It is measured in a direction perpendicular to the original heading of the ship.

Steady turning circle radius is the steady radius of the turning circle when a steady rate of turn is achieved. This state is usually achieved by the time the ship has altered course between 90º and 180º however this will vary from ship to ship

5

A ship heels 15° as it makes a turn. If the Draft when upright is 6 m calculate the Draft when heeled given that the breadth is 10m.

• Draught when heeled = (½Beam x SineƟ) + (Upright draught x CosƟ)• Draught when heeled = (½ x 10 x Sin 15°) + (6 x Cos 15°)

6

A ship with a transverse metacentric height of 0.40 m has a speed of 21 kts. The centre of gravity is 6.2 m above keel whilst the centre of lateral resistance is 4.0 m above keel. The rudder is put hard over to port and the vessel turns in a circle of 550 m radius.

7A ship with BG 2.2 m has a speed of 15 kts, GM 0.32m. The rudder is put hard over to port and the vessel turns in a circle of 300 m radius. Find the angle of heel due to turning.

8A container ship has windage area of 12,000 m2 with its centre 15m above the COB. If displacement is 75,000t & GM 0.9m, calculate the angle of heel if it experiences a beam wind of 50 knots.

9 Describe the forces acting on the ship when turning.Forces acting on the ship when turning

• When the ship is making a turn, then apart of the propeller thrust and water resistance, also rudder force and transverse hydrodynamic force are active.

• Ship is moving along the curvilinear path with the centre at point O. The distance between the centre of curvature and the centre of gravity of the ship is radius of instantaneous turn.

• Ship's centreplane deviates from the tangent to the path of the centre of gravity by the drift angle.

• The line perpendicular to the ship's centreplane through the centre of rotation, marks pivot point (PP).

• At this point, there is no transverse velocity in turning; for the crew on board it appears that the ship rotates around this point.

• Transverse velocity is greatest at stern.

10

What are the assumptions taken with regards to ‘Severe wind and rolling criterion (weather criterion)’ of ‘Intact stability 2008’?

• Assumptions• The ability of a ship to withstand the combined effects of beam wind and rolling is to be

demonstrated for each standard condition of loading, with reference to the figure as follows:• the ship is subjected to a steady wind pressure acting perpendicular to the ship's

centreline which results in a steady wind heeling lever (lw1);

• from the resultant angle of equilibrium (Ɵ0), the ship is assumed to roll owing to wave action to an angle of roll (Ɵ1) to windward;

• the ship is then subjected to a gust wind pressure which results in a gust wind heeling lever (lw2);

• free surface effects are to be accounted for in the standard conditions of loading.• Under the assumptions, the following criteria are to be complied with:• the area "b" is to be equal to or greater than area "a", where:

• Area above the GZ curve and below lw2, between ƟR and the intersection of lw2 with the GZ curve

• Area above the heeling lever lw2 and below the GZ curve, between the intersection of lw2 with the GZ curve and Ɵ2.

• the angle of heel under action of steady wind (Ɵ0) is to be limited to 16° or 80% of the angle of deck edge immersion, whichever is less.

11

Derive the formulae to calculate heel due to beam wind.

CALCULATING HEEL DUE TO BEAM WIND• Beam winds cause a thrust, called windage, on above-water obstructions. The term ‘wind

force’ is not used here as it can be confused with ‘Beaufort Wind Force’ as used in Meteorology. The term ‘Area of windage’ may be used to denote the area that results in windage.

• Windage is experienced by all ships, especially when in ballast when the freeboard is high. Container ships (with full stacks of boxes above deck), car carriers, cruise line ships, etc have a large windage area whether loaded or in ballast.

• P = 2(1O-5) Aw(Ws)2

• Where P = windage in tonnes• Aw = Windage area in m2

• Ws = Wind speed in knots• Wind Heeling Moment = P.y tonne metres

• Wind Heeling Lever = P.yW metres

• Righting Lever to counter this = GZ metres

• Equating above, GZ = P.yW metres

• For small angles, GZ = GM SinƟ

• So P.y = W.GM SinƟ or SinƟ = P.yW.GM

This formula can be used for calculating small values of angle of heel.

12

A ship's speed is 12 knots. The helm is put hard over and the ship turns in a circle of radius 488 m. GM = 0.3 m and BG = 3 m. Assuming that 1 knot is equal to 1852km/hour, find heel due to turning.

13

A ship steaming at 10 knots turns in a circle of radius 366 m. GM = 0.24 m. BM = 3.7 m. Calculate the heel produced.

14

A ship turns in a circle of radius 100 m at a speed of 15 knots. BG = 1 m. Find the heel if the GM = 0.6 m.

15

A ship with a transverse metacentric height of 0.40 m has a speed of 21 kts. The centre of gravity is 6.2 m above keel whilst the centre of lateral resistance is 4.0 m above keel. The rudder is put hard over to port and the vessel turns in a circle of 550 m radius.

Unit

2Inclining experiment 1

Explain why there should be no significant tide during the Inclining experiment

2

On which occasions an Inclining Experiment and Lightweight Survey must be conducted?The Occasions when an Inclining Experiment and Lightweight Survey must be Conducted

1. Every passenger ship regardless of size and every cargo ship having a length of 24 m and upwards should be inclined upon its completion and the elements of its stability determined.

2. Where any alterations are made to a ship so as to materially affect the stability, the ship should be re-inclined.

3. At periodic intervals not exceeding five years, a light-weight survey should be carried out on all passenger ships to verify any changes in lightship displacement and longitudinal centre of gravity.

4. The ship should be re-inclined whenever a deviation from the light-ship displacement exceeding 2% or a deviation of the longitudinal centre of gravity exceeding 1% of L is found, or anticipated.

5. The Administration may allow the inclining test of an individual ship to be dispensed with provided basic stability data are available from the inclining test of a sister ship and it is shown to the satisfaction of the Administration that reliable stability information for the exempted ship can be obtained from such basic data.

6. The Administration may allow the inclining test of an individual ship or class of ships especially designed for the carriage of liquids or ore in bulk to be dispensed with, when existing data for similar ships clearly indicates that, due to the ship's proportions and arrangements, more than sufficient metacentric height will be available in all probable loading conditions.

7. The inclining test prescribed is adaptable for ships with a length below 24 m if special precautions are taken to ensure the accuracy of the test procedure.

• Annex 3 of the Code details a means of approximately determining the initial stability (GM) of small ships up to 70m in length by consideration of the rolling period.

3 List the items that are to be stowed in their normal seagoing positions during the Inclining experimentPreparations for the Inclining Test

1. The ship should be as complete as possible at the time of the test.2. There must be adequate depth of water to ensure that the ship will not contact the bottom

during the inclination.3. The ship should be moored in a quiet, sheltered area free from external forces such as

propeller wash from passing ships.4. The ship should be moored as to allow unrestricted heeling.5. There should be no significant tide. Any tide will prevent moorings from being maintained

slack. If under-keel clearance is small the effect of squat may lead to erroneous draught readings. Ideally the experiment should be conducted in a sheltered dock.

6. All temporary material and equipment such as toolboxes, staging, welding equipment etc. should be removed.

7. All fittings and equipment such as accommodation ladders, lifeboats and derricks/cranes

should be stowed in their normal seagoing positions.8. All tanks should be verified as being completely empty or full.9. The number of slack tanks should be kept to an absolute minimum and their FSC accurately

determined.10. Ideally tanks with rectangular free surfaces should only be slack so that the free surface

effect can be accurately determined.11. Slack tanks must have the contents accurately determined with respect to liquid mass and

Kg.12. Decks should be free of water. Any water trapped on deck will move during the test and

reduce the accuracy of the result. Snow and ice must also be removed.13. Her draft should be such that abrupt changes will not occur in her water plane, when

inclined.14. An accurate list is to be made of any items of weight yet to be placed onboard and those to

be removed from the ship together with their KG and LCG.

4 Derive the inclining experiment formula to calculate KG with suitable diagrammeDerivation of the inclining experiment formulaPrior to starting the experiment the ship must be exactly upright to ensure that the centre of gravity, G, is on the centre line. Figure shows that if a known weight is then shifted transversely across the deck through a certain distance in metres, G will move off the centre line to GH, causing the ship to list.

The distance GGH is calculated by the formula: GGH = wxdW ----------(1)

• If a plumb line is suspended at O such that it crosses a batten at X, then as the ship lists a deflection XY will be observed and can be measured.

• Triangles MGGH and OXY are similar.

• In triangle MGGH :Tan Ɵ = OPPADJ =

GGHGM

• In triangle OXY: Tan Ɵ = OPPADJ =

XYOX

• Therefore: GGHGM =

XYOX

• So TanƟ = GGHGM =

Deflection of the pendulumPendulum length

• Rearranging this gives: GM = GGH xPendulum lengthDeflection of the pendulum -----(2)

• Combining formulae (1) and (2) gives: GM = w x d x Pendulum lengthWxDeflection of the pendulum

• If there is no slack tanks at the time of inclining, the GM in the inclined condition will be a solid GM.then the KG as inclined = KM - GM

• If there were slack tanks at the time of inclining, the GM in the inclined condition will be a fluid GM. then the KGF as inclined = KM - GM

• KM is obtained from the ship’s hydrostatic data for the true mean draught as calculated from the observed draughts.

5 Write short note on the pendulum, used in the Inclining experiment

6

Write short note on the necessity to determine the value of light ship KG?Precaution necessary when conducting the inclining test11. As an alternative to the use of inclining weights, water ballast transfer may be carried out, if acceptable to the Administration. This method will be more appropriate on very large ships.12. The use of three pendulums (but no less than two) is recommended, one forward, one midships and one aft to allow bad readings at any one station to be identified.13. The pendulum wire should be piano-wire and the top connection should allow unrestricted rotation at the pivot point (a washer with the pendulum wire attached suspended from a nail would suffice). The pendulum weight should be suspended in a trough of hydraulic oil to dampen movement.

14. The pendulums should be long enough to give a measured deflection to each side of upright of at least 15 cm. This will require a pendulum length of at least 3 metres. Usually, the longer the pendulum the greater the accuracy of the test; however, if excessively long pendulums are used on a tender ship the pendulums may not settle down and the accuracy of the readings will be questionable.

15. The pendulums should be located as far apart as practicable and should be protected from wind.

Unit 3Dry dock

1 Derive the formula to calculate ‘Loss of GM’ as a Result of a Fall in ‘M’ with suitable diagramme

ing and grounding

2

Explain the following statement: ‘The greater the trim when dry-docking, thegreater is the required GM before entering the dry-dock’

3M.V. VIJAY enters a SW drydock drawing 3 m fwd& 5.2 m aft. KG 9 m. Calculate the force 'P' at the critical instant

4

A ship of 6000 tonnes displacement enters a drydock trimmed 0.3 m by the stern. KM = 7.5 m, KG = 6 m and MCTC = 90 tonnes m. The centre of flotation is 45 m from after perpendicular. Find the effective metacentric height at the critical instant.

5

A ship of 5000 tonnes displacement enters a drydock on an even keel. KM = 6 m, KG = 5.5 m and TPC = 50 tonnes. Find the virtual loss of meta-centric height after the ship has taken the blocks and the water has fallen another 0.24 m.

6

A ship of 5000 tonnes displacement enters a drydock trimmed 0.45 m by the stern. KM = 7.5 m, KG = 6.0 m and MCTC = 120 tonnes m. The centre of flotation is 60 m from aft. Find the effective metacentric height at the criticalinstant, assuming that the KM is 7.575m at the critical instant.

7

A ship being drydocked has a displacement of 1500 tonnes. TPC = 5 tonnes, KM = 3.5 m, GM = 0.5 m, and has taken the blocks fore and aft at 3 m draft. Find the GM when the water level has fallen another 0.6 m.

8

A ship of 8000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres, KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonnes.

9

A ship of 6000 tonnes displacement is 120 m long and is trimmed 1 m by the stern. KG = 5.3 m, GM = 0.7 m and MCTC = 90 t m. Is it safe to dry-dock the ship in this condition? (Assume that the centre of flotation is amidships.)

10

A box-shaped vessel 150 m long, 10 m beam and 5 m deep, has a mean draft in salt water of 3 m and is trimmed 1m by the stern, KG = 3.5 m. State whether it is safe to drydock this vessel in this condition.

Unit 4Bilging 1

Discuss the Added weight method and lost buoyancy (constant displacement) method used in Bilging calculations.

2 During Bilging calculations how you will choose between the Added weight method and lost buoyancy (constant displacement) method?

• The two methods of calculation will give identical answers for the final draughts, trim and righting moment. Despite giving different values for GM (and GML), when the GM values are allied to the appropriate displacement value for the damaged condition they will give equal values of righting moment; remembering that the righting moment is the true measure of a vessels initial stability and not GM alone!

• The lost buoyancy (constant displacement) approach is considered more appropriate when the damage is extensive and the floodwater is very much part of the sea. Because damage stability legislation assumes worst case scenarios. all damaged stability calculations are

conducted using this method.• The added weight method may be considered more appropriate when the breach of the hull

is small and the floodwater is more contained so that the effect of the water introduced is the same as that if it were introduced into an intact ship. Pumps may be able to limit the amount of water admitted and a level below the external waterline may be maintained.

3Explain how water-plane area is calculated for an amidships compartment extending the full breadth of the vessel is bilged. Permeability 25%.

4

Prove the following: If KM changes and KG is assumed to remain constant, any change in KM will be the same as the change in GM

• We must now consider the change in the vessel’s initial stability. Consider the change in effective water plane area in figure

• In this case the part of water plane area of the bilged compartment has been lost. Since:

BM= IV and for a box-shaped vessel: BMBOX =

LB 312V

• BM will reduce directly as a result of the reduced water plane area. V, the volume of displacement of the vessel, has not changed, since if displacement remains constant; so does the volume of displacement.

• In addition, because the draught has increased due to the sinkage, KB will increase. KB = Draught

2• This is still valid for the bilged condition since the KB of each of the end compartments will be

the same. Since: KM = KB + BM; it is most probable that KM will change as a result of the increasing KB and the decreasing BM; the changes in both unlikely to be the same.

If KM changes and KG is assumed to remain constant, any change in KM will be the same as the change in GM, being either an increase or decrease.

5

A box shaped vessel is floating on an even keel in salt water and has the following particulars:Length 220.00 m, Breadth 30.00 m, Draft 12.80 m. KG 9.50 mThere is an empty amidships watertight compartment of 16.00 m length that extends the full width of the vessel.Calculate the Draft if this compartment becomes bilged.

6

A box-shaped vessel, length 100 m, breadth 20 m, depth 8 m is on an even keel Draft of 4.00 m in SW and has a KG of 4.00 m. An empty amidships DB tank, length 25 m, breadth 20 m and depth 1.00 m is bilged. Calculate the new Draft.

7

A box-shapped vessel, length 200 m, breadth 20 m, depth 14 m is on an even keel Draft in SW of 8.00 m with a KG of 6.00 m. An empty amidships deep tank, length 25 m, breadth 20 m extending from the bottom shell to a height of 8.5 m above the keel is bilged. Calculate the final Draft.

8A box shaped vessel 110m long & 16m wide floats in SW at 4m even keel draft. An empty central compartment 16m long & 10m wide is bilged. Calculate the KM before and after bilging

9 Explain with suitable diagrammes, the procedure to calculate the list when an amidships side compartment is bilged

• Consider a box-shaped vessel with side compartments amidships as in the above figure. F is the centre of flotation initially on the centre line. The vessel is floating upright on an even keel when a side compartment becomes bilged. The volume of buoyancy lost is shown.

• The vessel will sink to regain the buoyancy lost. Figure below illustrates the shape of the buoyancy gained.

• Because the water plane area has changed shape the centre of flotation moves off the centre line of the vessel (F to F,). This causes the axis of rotation of the water plane area to move off the centre line as already discussed.

• Figure shows the vessel after it has experienced sinkage but before it lists.• The transfer of the volume of buoyancy (b to bi) causes thea centre of buoyancy, B, to move

off the centre line (and upwards) to B1 The horizontal component of this shift creates the listing lever, which is equal to GX.

• The vessel will now list over to the bilged side

TanƟLIST =OPPADJ =

GXXM =

BBHXM

• Where XM is the GM in the bilged condition.• Because of the symmetry of a box-shaped vessel BBH is equal to the movement of the centre

of flotation off the centre line (FF1) that is found by taking moments of area of the water plane area about one edge.

• Therefore: TanƟLIST = BBHGMBILGED

• Note: When calculating list arising from the loading, discharging or shifting of weights the formula:

• TanƟLIST = GGH GM is used.

• In this instance, GGH is the cause of the list, which represents the distance that the centre of

gravity of the ship is off the centre line at the time for which the list is being calculated.• When calculating the list caused by the bilging of a side compartment the formula:

• TanƟLIST = BBHGMBILGED

• Here the list is being caused by the horizontal component of the movement of the centre of buoyancy and the GM is that which applies to the vessel s damaged condition.

10

Explain with suitable diagrammes, the procedure to calculate the Drafts when an end compartment becomes bilged

11

Explain with suitable diagrammes, the procedure to calculate the draft when an extreme end compartment with a watertight flat (100% permeability) is bilged

12

Explain with suitable diagrammes, the procedure to calculate the righting moment, when an amidships compartment extending the full breadth and depth of the vessel becomes bilged

13

Explain with suitable diagrammes, the procedure to calculate the change in GM when an empty amidships compartment with a watertight flat (double bottom) is bilged.

14

Explain with suitable diagrammes, the procedure to calculatethe moment of inertia of a water plane area of a box-shaped vessel with a bilged side compartment.

15

Explain with suitable diagrammes, the procedure to calculate the change in GM whena Filled DB tank is Bilged.Explain with suitable diagrammes, the procedure to calculate the change in GM whena Centralised DB tank is Bilged.

Unit 5Shear Forces and Bending Moments in ships

1

State the SOLAS regulation regarding ‘Loading instrument’.Regulation 11Loading instrument

• (Unless provided otherwise, this regulation applies to bulk carriers regardless of their date of construction)

1. Bulk carriers of 150 m in length and upwards shall be fitted with a loading instrument capable of providing information on hull girder shear forces and bending moments, taking into account the recommendation adopted by the Organization.

2. Bulk carriers of 150 m in length and upwards constructed before 1 July 1999 shall comply with the requirements of paragraph 1 not later than the date of the first intermediate or periodical survey of the ship to be carried out after 1 July 1999.

3. Bulk carriers of less than 150 m in length constructed on or after 1 July 2006 shall be fitted with a loading instrument capable of providing information on the ship’s stability in the intact condition. The computer software shall be approved for stability calculations by the Administration and shall be provided with standard conditions for testing purposes relating to the approved stability information.

2 Explain the types of Stability Software used in ‘Loading instruments’Types of Stability Software

• Three types of calculations performed by stability software are acceptable depending upon a

vessel’s stability requirements:Type 1

• Software calculating intact stability only (for vessels not required to meet a damage stability criterion)

Type 2• Software calculating intact stability and checking damage stability on basis of a limit curve (e

g. for vessels applicable to SOLAS Part B-1 damage stability calculations, etc.) or previously approved loading conditions and

Type 3• Software calculating intact stability and damage stability by direct application of pre-

programmed damage cases for each loading condition (for some tankers etc.)

3

What are the Pre-programmed input data to be entered in ‘Loading instrument’?Examples of Pre-programmed input data include the following:

1 Hydrostatic data Displacement, LCB. LCF. VCB. KMt and MCT versus draught.

2 Stability data KN or MS(Residual stability arm) values at appropriate heel/ trim angles versus displacement, stability limits

3 Compartment data Volume. LCG, VCG, TCG and FSM/ Grain heeling moments vs level of the compartment's contents

4

What are the output data from ‘Loading instrument’?Examples of output data include the following

1 Hydrostatic data Displacement. LCB. LCF, VCB. KMt and MCT versus draught as well as actual draughts, trim.

2 Stability data FSC (free surface correction). GZ-values. KG. GM, KG/GM limits, allowable grain heeling moments, derived stability criteria, e.g. areas under the GZ curve, weather criteria

3 Compartment data Calculated Volume, LCG, VCG, TCG and FSM/ Grain heeling moments vs level of the compartment's contents

5 How will you ensure that the loading programme working correctly?

6

What are the Periodical testing required for stability calculation program of a ‘Loading instrument’Periodical Testing

• It is the responsibility of the ship’s master to check the accuracy of the onboard computer for stability calculations at each Annual Survey by applying at least one approved test condition. If a Society surveyor is not present for the computer check, a copy of the test condition results obtained by the computer check is to be retained on board as documentation of satisfactory testing for the surveyor's verification.

• At each Special Survey this checking for all approved test loading conditions is to be done in presence of the surveyor.

• The testing procedure shall be carried out.

7 List the System Requirements and Data Representation on modern stress loading programs8 Explainthe procedure to calculate SF values from load curve with suitable diagramme

Producing the Curve of Shear ForcesThe shear force at any position is defined as being the algebraic sum of the loads acting to the left (or right) of the

position in question and is measured in tonnes.

• Integrating the load curve will produce the curve of shear forces.

• The maximum shear force values will arise at the positions where the loads change direction, being at the

bulkhead positions.

• Consider the aforementioned definition of shear force. For our purposes this definition of shear force can be

modified to read as being the area under the load curve to the left of the point in question.

• Therefore; SF at AP - 0 tonnes (since there is no area to the left of the AP under the curve!)

• Placing a sheet of paper over the curve and moving it to the right at 5 metre intervals, calculate the net area

to the left of the edge of the sheet for each point in question.

• SF at 5 m foap (forward of After Perpendicular) = 2t/m x 5m=10 tonnes.

• Now calculate the SF value at 10 m foap (by moving the paper further to the right and revealing more of the

curve to the left.

• SF at 10 m foap = 2 t/m x 10 m = 20 tonnes.

• SF at 15 m foap (Bulkhead 3/2) = 2 t/m x 15 m = 30 tonnes.

• At 20 m foap there is area revealed above and below the baseline and this is treated as positive and negative as per the load scale.

• SF at 20 m foap =(2 t/m x 15 m) + (-4 t/m x 5 m) = 10 tonnes.Continuing with this method gives:• SF at amidships (22.5 m foap) = (2t/m x 15 m) + (-4 t/m x 7.5 m) = 0 t• SF at 25 m foap = (2 t/m x 15 m) + (-4 t/m x 10 m) =-10 tonnes.• SF at 30 m foap (bulkhead 2/1) = (2 t/m x 15 m) +(-4 t/m x 15 m) = -30 t• SF at 35 m foap = (2t/m x 15 m) + (-4 t/m x 15 m) + (2 t/m x 5 m) = -20t• SF at 40 m foap - (2 t/m x15 m)+ (-4 t/m x15 m) + (2 t/m x 10 m) = -10t• SF at 45 m foap = (2t/m x 15 m) + (-4 t/m x 15 m) + (2 t/m x 15 m) = 0

From AP(m) 0 5 10 15 20 22.5 25

SF(t) 0 10 20 30 10 0 -109 Explainthe procedure to calculate BM values from SF curve with suitable diagramme

Producing the Curve of Bending Moments• The bending moment values are calculated in exactly the same way as the shear force

values, by considering the areas under the shear force curve to the left of the position in question.

• The area of a triangle is given by:• Area = ½ Base x Perpendicular height

• Area of a trapezium is given by: Area = (a+b)2

x base

The bending moment values are calculated as follows:• BM at AP= 0 tonnes (since there is no area to the left of the AP under the SF curve!)• Placing a sheet of paper over the curve and moving to the right as far as the first bulkhead

position (Bulkhead 3/2), calculate the areas as before.

BM 5 m foap = ½ x 5 m x10t= 25 t-m

BM 10 m foap=½ x10mx20t = 100 t-m

BM 15 m foap (bulkhead3/2) = ½ x 15 m x 30t = 225 t-m

• Once past bulkhead 3/2 it is necessary to consider the area of a trapezium formed by the area under the shear force curve to the right of the bulkhead as seen in the figure.

• BM 20 m foap = 225 + (30t + 10t)2 x 5m)=325 t-m

• BM at amidships (22.5 m foap) = ½ x 22.5 x 30 = 337.51 m• BM at 25 m foap = 337.5 t-m + (½ x 2.5 m x -10t) = 325 t-m

(Since we know the area from 0 to 22.5 m foap, being 337.5t-m)• BM at 30 m foap (bulkhead 2/1) = 337.5 + (½ x 7.5 m x -30 t) = 225 t-m

• BM at 35 m foap = 225 + (-30t + -20t)2 x 5 m) = 100 t-m

(Since we know the area from 0 to 30 m foap, being 225 t-m)• BM at 40 m foap

= 225 + (-30t + -10t)2 x10 m)

• = 25t-m• BM at FP = 0 t-m

From AP(m) 0 5 10 15 20 22.5 25

BM(t-m) 0 25 100 225 325 337.5 325

10

Sketch a load curve, a shear force curve and bending moment curve to same scale showing the places of maximum values of shear force and bending moment

11

A box shaped vessel has L = 100 m, B = 20 m and floats at a Draft of 4.0 m when light in FW. It has five equal holds. Nos. 2 and 4 are each loaded with 1000 t, trimmed level.

12

A box shaped barge is 45 m long, 8m wide and floats at a Draft of 3.0 m in FW when empty. It is subdivided into three equal compartments. 90 t of cargo is loaded into No.2 hold and trimmed level. Construct the load curve

13

A vessel of length 30 m and beam 7 m is of rectangular cross-section. It has a light Draft of 2.0 m in FW and has three holds of equal length. 60 t of cargo is uniformly distributed between holds Nos. 1 and 3. Construct the load curve

14

A box shaped barge with three equal holds is 45 m long and 9 m wide. The Draft in FW is 3.0 m when empty. 180t is loaded, being uniformly distributed between the two end compartments. Construct the luoad curve

15

A box vessel has L = 48 m, B = 8 m and draws 3.0 m in FW when empty. It has four equal holds. Nos. 1 and 4 are each loaded with 480 t. Construct the load curve.

PART CUnit 1

10Marks

Angle of heel when turning, Wind heeling

1

(a) Describe any one type of ‘passive system’ methods Adopted to Minimise a Ship’s Rolling Motion at Sea with suitable diagrammes. Or(b) Describe any one type of ‘active system’ methods Adopted to Minimise a Ship’s Rolling Motion at Sea with suitable diagrammes.BILGE KEEL

• Bilge keels form the simplest method of controlling roll. They consist of narrow steel strips extending along a portion of the length of the hull.

• Bilge keels are employed in pairs (one for each side of the ship).• Bilge keels increase the hydrodynamic resistance when a vessel rolls, thus limiting the

amount of roll a vessel has to endure.• Bilge keels should be fitted throughout the length of the parallel mid-body of the ship at the

turn of the bilge. Roll amplitudes may be reduced by up to 35% and are therefore a very cost-effective means of limiting roll amplitude.

• Bilge keels should always be fitted to ships having a large well-rounded bilge radius, whereas ships having a more square bilge shape will be more resistant to rolling.

• They are mounted at the turn of the keel and project no further than the breadth and depth of the ship thereby preventing contact damage.

• They are attached to the hull by a relatively weak joint, such as riveting to a second fixing which is welded to the hull, or by stitch welding allowing the keel to be torn off without further hull damage. Size of bilge keels depend upon the ship.

• The web should not be so deep so as to sustain damage due to force of water during rolling.• In addition the ends of the keel should be tapered or well-rounded so that they blend

smoothly with the lines of the hull, this reduces eddies which could lead to vibration and/or erosion damage.

Active Systems(These include active tank systems and stabilising fins.)

• Active tanks• These tanks have a positive means of directing water to one side or the other.• There are two tanks, one each side of the ship and the level of water is controlled by a pump

or pumps acting in response to a gyro controlled roll sensing system.• In the systems illustrated in the figure, water is distributed so that the greatest quantity will

be in the tank on the high side at the extent of the roll.

• (a) Water is pumped from one tank to the other so as to keep the greater quantity in the higher tank.

• (b) This system is similar to (a) but each tank has its own pump to add and remove water from the tank as the ship rolls.

• (c) Water level is controlled indirectly by means of air pressure above the water in each tank, the tanks are open to the sea at the bottom.

•

•

•

2

(a)With a neat diagramme explain advance, transfer, tactical diameter, steady turning diameter, Tactical diameter angle and Yaw. Or(b)Draw and label the turning circle diagramme displayed in the bridge.

• Advance is the distance travelled in the direction of the original course by the centre of

gravity of a ship from the position at which the rudder order is given to the position at which the heading has changed 90° from the original course.

• Transfer is the distance travelled by the centre of gravity of the ship in a direction perpendicular to the original heading of the ship to the position at which the heading has changed 90° from the original course.

• Tactical diameter is the distance travelled by the centre of gravity of a ship from the position at which the rudder order is given to the position at which the heading has changed 180° from the original course. It is measured in a direction perpendicular to the original heading of the ship.

• Steady turning circle radius is the steady radius of the turning circle when a steady rate of turn is achieved. This state is usually achieved by the time the ship has altered course between 90º and 180º however this will vary from ship to ship.

• Yaw is the angle between the ship’s fore and aft line and the direction of travel of the ship’s centre of gravity at any instant during the turn.

3

(a)Derive the formula to calculate the angle of heel when turning with suitable diagrams. Or(b)Draw and label the reference figure given in ‘intact stability 2008’ and explain the criteria those are to be complied, to demonstrate the ability of a ship to withstand the combined effects of beam wind and rolling for each standard condition of loading.

4 (a)Compare the effect of ship parameters on turning circle Manoeuvre of a full bodied ship (Cb=0.8) such as a Tanker and a slender ship (Cb=0.6) such as a Container ship of the same length with suitable diagrammeOr(b)Compare the effect of ship’s turning circle Manoeuvrefor 20° and 35° rudder angles for a turn at a constant rpm for slow ahead and for same ship conducting a turn at a constant rpm for full ahead with suitable diagrammes.Effect of ship parameters on turning performance:Comparison of a full bodied ship (Cb=0.8) such as a Tanker and a slender ship (Cb=0.6) such as a Container ship of the same length.

1. Two ships have nearly the same transfer2. Tactical diameter for both ships is almost the same3. Radius of the steady turning circle is much smaller for tanker

4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.4.

4. Drift angle is much larger for tanker5. Pivot point is closer to the bow in tanker

Following is the result of a ship’s Turning circle Manoeuvre for 20° and 35° rudder angles. The advance and transfer can be measured from the scale for both 20° and 35° turns. it may be surprising to note that the turning circles are virtually identical for the slow ahead turn and full ahead turn.

• a turn at a constant rpm for slow ahead

• same ship conducting a turn at a constant rpm for full ahead

Unit2Inclining experiment

1 (a)Explain why there is fluctuation in a ship's lightweight over a period of time?Fluctuations in a ship's lightweight over a period of timeOver the years in service, there will be increases in the lightweight due to:

• Accretion of paintwork. Formation of oxidation or rust.• Build up of cargo residue. Sediment in bottom of oil tanks. Mud in bottom of ballast tanks.

Dunnage.• Gradual accumulation up of rubbish. Lashing material.• Retrofits on accommodation fittings and in navigational aids.• Barnacle attachment or animal growth on the shell plating.• Vegetable growth on shell plating.• Additional engine room spares.• Each item in the above list will change the weight of an empty ship. It can also be

accumulative.• One example of increase in lightweight over a period of years is the 'Herald of Free

Enterprise,' which capsized in 1987. At the time of capsize it was shown that the

lightweight had increased by 270t, compared to when newly built.• Regular dry-docking of the ship will decrease the animal and vegetable growth on the shell

plating. It has been known to form as much as 5 cm of extra weight around the hull.• Regular tank-cleaning programmes will decrease the amount of oil sediment and mud in

the bottom of tanks. Regular routine inspections should also decrease the accumulation of rubbish.Over years in service, there will also be decreases in the lightweight due to:

• Oxidation or corrosion of the steel shell plating, and steel decks exposed to the sea and to the weather. Wear and tear on moving parts.

• Galvanic corrosion at localities having dissimilar metals joined together.• Corrosion and loss of weight is prevalent and vulnerable in the boot topping area of the

side-shell of a vessel, especially in way of the machinery spaces. Feedback has shown that the sideshell thickness can decrease over the years from being 18 mm thickness to being only 10 mm in thickness. This would result in an appreciable loss of weight.

• Wear and tear occurs on structures such as masts and derricks, windlass, winches, hawse pipes and capstans.

• These additions and reductions will all have their own individual centres of gravity and moments of weight. The result will be an overall change in the lightweight itself, plus a new value for the KG corresponding to this new lightweight.

• It has been documented that the lightweight of a vessel can amount to an average addition of 0.5% of the lightweight for each year of the ship's life.

• The above indicate that sometimes the lightweight will increase, for example due to plate renewal or animal and vegetable growth. Other times it will decrease, for example due to wear and tear or build up of corrosion. There will be fluctuations. It would seem judicial to plan for an inclining experiment perhaps every 5 years. This will re-establish for the age of the ship exactly the current lightweight.

•

Or(b)Explain the precautions necessary when conducting the inclining experiment?Precaution necessary when conducting the inclining test

1. The ship must be upright at the commencement of the test2. The ship’s trim should be such that the deviation from her designed trim does not exceed

1% of L.3. Moorings should be slack and any shore side gangways landed to allow unrestricted

heeling.4. Excess crew and personnel not directly involved in the test should be sent ashore.5. There must be no significant wind, especially on the beam, which might cause the ship to

heel.6. Access ramps should be removed and power lines hoses etc. connected to shore should be

minimized.7. It is recommended that at least five freeboard readings approximately equally spaced on

each side of the ship be taken or that all draught marks (forward, aft and amidships) be read on each side of the ship to determine her displacement accurately.

8. The mean value of specific gravity of the water should be obtained accurately by taking water samples forward, midships and aft, at a sufficient depth.

9. The test weight used should be compact such that their VCGs can be accurately determined. Their weight to be recoded. Ideally, six equal weights are placed on deck,

three on each side, at measured equal distances off the centre line.10. The mass of the inclining weight(s) used should be sufficient to provide a minimum list of

1° and a maximum list of 4° to each side. This is to ensure that the formula: Tan ƟLIST = GGHGM remains valid, being applicable to small angles of inclination only.

11. As an alternative to the use of inclining weights, water ballast transfer may be carried out, if acceptable to the Administration. This method will be more appropriate on very large ships.12. The use of three pendulums (but no less than two) is recommended, one forward, one midships and one aft to allow bad readings at any one station to be identified.13. The pendulum wire should be piano-wire and the top connection should allow unrestricted rotation at the pivot point (a washer with the pendulum wire attached suspended from a nail would suffice). The pendulum weight should be suspended in a trough of hydraulic oil to dampen movement.

14. The pendulums should be long enough to give a measured deflection to each side of upright of at least 15 cm. This will require a pendulum length of at least 3 metres. Usually, the longer the pendulum the greater the accuracy of the test; however, if excessively long pendulums are used on a tender ship the pendulums may not settle down and the accuracy of the readings will be questionable.

15. The pendulums should be located as far apart as practicable and should be protected from wind.

2 (a)Explain the Preparations necessary for the Inclining experimentPreparations for the Inclining Test

• The ship should be as complete as possible at the time of the test.• There must be adequate depth of water to ensure that the ship will not contact the bottom

during the inclination.• The ship should be moored in a quiet, sheltered area free from external forces such as

propeller wash from passing ships.• The ship should be moored as to allow unrestricted heeling.• There should be no significant tide. Any tide will prevent moorings from being maintained

slack. If under-keel clearance is small the effect of squat may lead to erroneous draught readings. Ideally the experiment should be conducted in a sheltered dock.

• All temporary material and equipment such as toolboxes, staging, welding equipment etc. should be removed.

• All fittings and equipment such as accommodation ladders, lifeboats and derricks/cranes should be stowed in their normal seagoing positions.

• All tanks should be verified as being completely empty or full.• The number of slack tanks should be kept to an absolute minimum and their FSC accurately

determined.• Ideally tanks with rectangular free surfaces should only be slack so that the free surface

effect can be accurately determined.• Slack tanks must have the contents accurately determined with respect to liquid mass and

Kg.• Decks should be free of water. Any water trapped on deck will move during the test and

reduce the accuracy of the result. Snow and ice must also be removed.• Her draft should be such that abrupt changes will not occur in her water plane, when

inclined.

• An accurate list is to be made of any items of weight yet to be placed onboard and those to be removed from the ship together with their KG and LCG.

Or(b)Explain the Inclining experiment procedure in detail with suitable diagrammeInclining experiment procedure

• The initial position of the pendulum is noted against each batten. Using a shore crane, one weight is shifted port to starboard and the deflection noted on each pendulum. A second weight is shifted port to starboard and the deflection on the pendulums noted again. The third weight is also then shifted port to starboard and the deflection of each pendulum noted. All the three shifted weights are then returned to their original position on the port side and the deflection, if any, on the pendulums noted. This entire procedure is then repeated with the three weights from the starboard side.

• The standard test employs eight distinct weight movements. A plot is then made with the heeling moment (w x d) on the X axis and tan Ɵ (deflection/length of plumb line) on the Y axis.

• The plot of all readings on each pendulum should lie on a straight line. If a straight-line plot is not achieved, those weight movements that did not give an acceptable plot must be repeated.

• By taking moments about the keel, allowance is then made for any weights on board, including the inclining weights, which do not form part of the light ship, equipment those are yet to fitted and FSM if any, to obtain her ‘Light ship KG’.

• As well as calculating the lightship displacement and KG, draught and trim readings at the time of the experiment will be used to determine the ship’s longitudinal centre of gravity for the inclining condition. This will then be corrected by calculation to obtain the true lightship LCG.

• On completion of the test a report will be written and included as part of the ship’s stability data book.

3 (a)M.V. Hindship' was floating with all compartments empty except as follows: No. 2 (P&S) DB tanks full with water ballast No. 1 DB tank contained 100 tonnes of H.F.O. An inclining experiment was conducted in this condition. A weight of 10 tonnes KG 10.2 m, shifted transversely through a distance of 17.6 m, caused a deflection of 8.3 cms in plumb line 8.5 m in length. Calculate the GM (Solid) and the KG of the light ship.

• Light Displacement of M.V. 'Hindship'= 5499.8 t• Water ballast in No. 2 (P & S) = 404.8 x 1.025 = 414.92 t• H.F.O. in No. 1 DB tank = 100.00 t• Inclining weight = 10.00 t• Displacement at the time of Inclining Experiment = 6024.72 t• KM for displacement of 6024.72 t = 11.113 m

• GM as inclined = w x d x Pendulum lengthWxDeflection of the pendulum =

10x 17.6x 8.56024.72x 0.083 = 2.992m

• KG as inclined = 11.113-2.992= 8.121m• Volume occupied by 100t of HFO = 100/0.95=105.263m3

• Volume of the No. 1 DB tank= 157.6m3

• Hence No. 1 DB tank is slack• FSM of No. 1 DB tank= 419x0.95= 398.05t-m

Take moments about the keel to calculate the lightship KG and displacement

Weights (t) KG (m) Moments (t-m)Displacement 6024.72 8.121 48926.751No. 2 DB tanks (-) 414.92 0.65 (-) 269.7No. 1 DB tank (-) 100.0 1.14 (-) 114.0FSM for No. 1 DB tank (-) 398.05Inclining Wt (-) 10.0 10.2 (-) 102.0Light ship 5499.8 8.735 48043.001Lightship displacement = 5499.8 t (Ans)Lightship KG = 8.735 m (Ans)

Or(b)An inclining experiment is to be carried out on a ferry- near completion. The ship is upright and has a displacement of 14260 tonnes with a KM of 10.92m.During the experiment 14 persons will be on the vehicle deck (allow 75kg/person; Kg 2.90 m).The vessel has the following tank contents:No. 4 DB ballast tank 436t. Kg 2.50m (tank full) No. 7 Fuel oil tank 128t. Kg 1.68m (tank full) Fresh Water Tank 66t, Kg 4.30m (fsm 1020 t)The vessel is inclined using weights totalling 50 t (Kg 2.90 m) and the plumb lines are secured to the vehicle deck having an effective length of 7.62 m. The inclining weights are shifted a number of times through a transverse distance of 4.60 m and the mean horizontal deflection of the plumb line was found to be 28.2 cms. If a Marine Escape System of weight 62 t is still to be fitted at Kg 10.40 m, calculate EACH of the following:(i) the vessel’s KG as inclined;(ii) the vessel’s lightship KG and displacement.

4 (a)An inclining experiment is performed on a ship in the following condition:Displacement 12200 t, including 40 t of inclining weights at Kg 16.2 m; KM 13.24 m.Successive movements of 20t of weights through a distance of 15m to port and starboard cause the following deflections of two pendulums each 14m in length:

Pendulum 1 Pendulum 2Movement to port 30.6 cm 30.2 cmMovement to stbd 29.8 cm 30.0 cmThe following must be accounted for to put the ship in the completed light displacement condition:(i) Inclining weights to be discharged;(ii) 26 t of equipment, Kg 10.4 m, to be discharged;(iii) 48 t of contractor’s machinery, Kg 24.0 m, to be discharged;(iv) 16 t of ER machinery to be fitted, Kg 6.0 m;(v) Labour force on board (40 men), Kg 18.0 m (allow 75 Kg per person).Calculate the ship’s light KG and displacement.Or

(b)A ship is to be inclined at a displacement of 11163 tonnes.Free surface moments exist in a rectangular double bottom tank containing diesel oil of relative density 0.88. The tank is 16 m long. 18 m wide and the sounding is 2.2 m.The movement of 11 t through a transverse distance of 19.6 m causes a 7 cm deflection of a 9 m long pendulum. If KM is 10.16 m. calculate the KG in the inclined condition.The following changes are required to bring the ship to the light condition:Discharge: 22 t inclining weights. Kg 16.1 m, 44 t builder’s equipment. Kg 11.8 m, The diesel oil in the double bottom

Load: 18 t machinery. Kg 5.4 mCalculate the lightship displacement and KG.

5

(a)During the course of an inclining experiment in a ship of 4000 tonnes displacement, it was found that, when a mass of 12 tonnes was moved transversely across the deck, it caused a deflection of 75 mm in a plumb line which was suspended from a point 7.5 m above the batten. KM = 10.2 m KG = 7m. Find the distance through which the mass was moved.Or(b)A ship of 8000 tonnes displacement is inclined by moving 4 tonnes transversely through a distance of 19 m. The average deflections of two pendulums, each 6 m long, was 12 cm 'Weights to load' to complete this ship were 75t centred at KG of 7.65 m 'Weights to discharge' amounted to 25t centred at KG of 8.16 m.(i)Calculate the GM and angle of heel relating to this information, for the ship as inclined.(ii)From Hydrostatic Curves for this ship as inclined, the KM was 9 m. Calculate the ship's final Lightweight and KG.

6

(a)A ship of 8000 tonnes displacement is inclined by moving 4 tonnes transversely through a distance of 19 m. The average deflections of two pendulums, each 6 m long, was 12 cm 'Weights on' to complete this ship were 751 centred at KG of 7.65 m 'Weights off' amounted to 251 centred at KG of 8.16 m.(i)Calculate the GM and angle of heel relating to this information, for the ship as inclined.(ii)From Hydrostatic Curves for this ship as inclined, the KM was 9 m. Calculate the ship's final Lightweight and KG at this weight.OrAs a result of performing the inclining experiment it was found that a ship had an initial metacentric height of 1 m. A mass of 10 tonnes, when shifted 12 m transversely, had listed the ship 3y degrees and produced a deflection of 0.25 m in the plumb line. Find the ship's displacement and the length of the plumb line.

Unit 3Dry docking and grounding 1

(a)Explain the Sequence of Events During Dry-Docking with a suitable diagrammes. Or(b)Explain with a suitable diagramme, why it is essential that the righting moment afforded by the upward acting buoyancy force remains greater than the capsizing moment afforded by the upthrust of the P force acting at the keel at all times prior to the ship touching the blocks forward and aftduring Dry-docking.

2 (a)The following details are of a ship preparing to dry-dock:displacement: 6600 tonnes KM: 7.2 m KG: 6.4 m MCTC: 110;present trim: 2.25 m by stern LCF: 65 m forward of AP.Calculate the change of trim required before entering dry-dock in order that an effective GM of 0.5 m is achieved at the instant of taking the blocks fore and aft.Or(b)Calculate, from the following details of a ship about to dry-dock:

(a) the GM at the instant the vessel takes the blocks fore and aft;(b) theDraft fore and aft at the same instant.Displacement: 14,400 t Length BP: 168 m

KM = 9.32 mKG = 8.70 m

Drafts: Forward 6.84 m Aft 8.40 mMCTC: 262

LCF: 88 m forward of APTPC: 39 (average for the range of Drafts concerned)

3

(a)MV. Hindship' displacing 9540 tonnes and trimmed 0.78 m by the stern is to be drydocked for bottom inspection. KG 7.826 m, FSC 0.164 m. Calculate:-The GM (Fluid) of the vessel before entering dry dock.The virtual GM of the vessel when her keel takes the blocks all along the length of the vessel.The ford and after draft, at which the virtual GM of the vessel becomes zero.Or(b)MV Hindship at a draft of F 3.82m A 5.46m, in water of density 1.015 is being docked. KG 8.38m, FSC 0.12m. Assuming the KM, MCTC, LCF and FSC remain unchanged over the range of drafts involved, calculate:The virtual GM of the vessel on taking blocks all overHer righting movement at a heel of 2°, at the critical instantThe fore and aft drafts at which her virtual GM becomes nil.

i. The virtual GM of the vessel on taking blocks all overii. Her righting movement at a heel of 2°, at the critical instant

iii. The fore and aft drafts at which her virtual GM becomes nil.F 3.82 mA 5.46mM 4.64mTrim 1.64 m by stern

• LCF for TMD 4.64 m = 73.008m

• TMD = dA – [( dA  – dF )LBP x LCF foap]

• TMD = 5.46 – [(5.46 – 3.82)143.16 x 73.008] = 0.836m

• Hydrostatic particulars in water of density 1.015

• Displacement = 8962.56 x 1.0151.025 = = 8875.1 t

• MCTC = 163.992  x 1.0151.025 = 162.392m

• LCF = 73.012 m• KM = 9.176m• KG = 8.380m• GM (Solid) = 0.796m• FSC = 0.120m• Initial GM (Fluid) = 0.676m• Calculating Virtual loss of GM on taking the blocks all over

• P= COT (cms)xMCTCDist LCF foap =

162.392 x 16473.012 = 364.766 t

•

4 (a)A ship about to be dry-docked displacing 9l00 t, KG 7.32 m, KM 8.56 m, is floating upright. A double bottom tank of rectangular cross section contains oil of relative density 0.9 and is slack. The tank is 18 m long, with a total breadth of 15 m, and has a centre-line division.

The Drafts are: Forward 3.14 m, Aft 5.20 m Length BP 120 mLCF: 63 m forward of APMCTC: 195Calculate:

(i) the effective GM when the ship is afloat;(ii) the GM at the instant the vessel takes the blocks fore and aft.

OrA vessel which is about to enter dry-dock is required to have a minimum GM of 0.60 m on taking the blocks overall.Displacement: 20300 t LCF: 98 m forward of AP.

MCTC 310 KG (Solid) 9.20 m KM 10.38 mPresent trim: 3.05 m by the stern.Allowing for a free surface effect on GM of 0.14 m for slack tanks, calculate the trim necessary before entering the dry dock.

5

(a)M.V. VIJAY has W = 10333 t in SW, KG = 8 m, FSM = 970 tm. Find the maximum trim with which she may enter a SW drydock, if the GM at the critical instant is to be not less than 0.3 metre. 26 Or(b) M.V.Vijay enters a SW drydock drawing F3.4m, A5.8m. KG 8.0m. Find the virtual GM when the level of water has fallen one metre after the stern has taken to the blocks

6

(a)M.V.Vijay in SW, while drawing F4.8m, A8.0m runs aground lightly on a sandy shoal, during a fall of tide. External soundings indicate that the depth of water near the after perpendicular is 1 m more than near the forward perpendicular. If KG is 7.5 m & FSM is 1300 tm, find;(i) the drop in water level at which the ship would sit overall on the sea-bed;(ii) the virtual GM when the ship sits overall on the sea-bed; 38Or(b)M.V.Hindship sailed from port in condition No.8. Soon after departure she grounded on an isolated rock, without damage to the hull. The drafts were then observed to be F 5.90m A 9.30m. Calculate the following:The upthrust provided by the rockThe position with respect to AP, where the grounding occurredThe virtual GM of the ship then 42

Unit 4Bilging

1

(a) A box-shaped vessel floating on an even keel in salt water has the following particulars; length 110 m, breadth 22 m and Draft 5.00 m. There is an empty amidships bottom compartment 20 m in length extending the full breadth of the vessel with a watertight flat 4.80 m above the keel. Calculate the change in GM if this compartment becomes bilged.17Or(b)A box-shaped vessel is floating on an even keel in salt water and has the following particulars: length 100 m, breadth 20 m and Draft 5.50 m. There is an empty amidships bottom compartment of 18 m length that extends the full breadth of the vessel with a watertight flat 6.20 m above the keel. Calculate the change in initial GM if this compartment becomes bilged.

2 (a)A box shaped vessel 100m long and 25m wide floats at 5m SW draft. KG 10m. A DB tank amidships 20m long, 25m wide and 1.8m high, full of SW gets bilged. Find the GM before and after bilging. 41Or(b)A box-shaped vessel 220 x 36 m is afloat in SW at an even keel draft of 10 m. KG = 12 m. An

empty DB tank 1.8 m high, 20 m long & 18 m wide, situated centrally, is bilged. Find the GM before and after bilging.

3

(a)A box-shaped vessel has length 75 m, breadth 12 m and is floating on an even keel Draft in salt water of 2.5m. In this condition the KG is 3.00 m. An empty forward end compartment of length 6 m extending the full breadth and depth of the vessel is bilged. Calculate the Drafts in the flooded condition. 49Or(b)A box-shaped vessel has length 100 m, breadth 18 m and is floating on an even keel Draft in salt water of 4.0 m. In this condition the KG is 6.8 m. An empty forward end compartment of length 10 m below a watertight flat 3 m above the keel and extending the full breadth of the vessel is bilged. Calculate the Drafts in the flooded condition.

4

(a)A box shaped vessel 180m long & 18m wide floats at an even keel draft of 8m in SW. The forward-most compartment has a DB tank, 15m long, 18m wide and 1.5m high full of SW, gets bilged. Find the new drafts F&A 50Or(b) (a)A box shaped vessel 160m long & 20m wide floats at an even keel draft of 7.6m in SW. The forward-most DB tank, 16m long, 20m wide and 1.2m high full of SW, gets bilged. Find the new drafts F&A

5

(a)A box shaped vessel, length 100 m, breadth 18 m, is on an even keel Draft of 7.50 m in salt water. KG 4.0 m. An empty amidships compartment, length 15 m on one side of a centreline longitudinal bulkhead is bilged. Calculate the list given that the second moment of area (the moment of inertia) of the waterplane in the damaged condition is 44730 m4.OrA box shaped vessel, length 100 m, breadth 10 m, on an even keel Draft of 5 m, has a centreline longitudinal bulkhead. KG 3 m. Calculate the angle of heel developed when an empty amidships compartment, length 20 m is bilged on one side of the bulkhead

6

(a)A box-shaped vessel floating upright on an even keel in saltwater has the following particulars: Length 120 m Breadth 25 m Draft 6.00 m KG 5.80 m. The vessel has a centre line watertight bulkhead with an empty amidships side compartment of 20m length. Calculate the angle of list when this compartment becomes bilged.76OrA vessel 200 m long & 20 m wide is box-shaped and afloat in SW at an even keel draft of 8 m. A DB tank amidships on the starboard side is rectangular, 12 m long, 10 m wide, and 1.2m deep & empty. Calculate the list if this tank is now bilged, given that KG = 7.5 m and FSM of the other tanks of the ship = 820 tm.

Unit 5Shear Forces and Bending Moments in

1 (a)A box-shaped barge is 36 m long and has light displacement = 198 t. It has four identical holds into which cargo is loaded and trimmed level as follows: 135 t in No: 1,162 t in No: 2, 162 t in No: 3 and 135 t in No: 4. Draw the SF & BM curves to scale.State the maximum values of SF and BM and the points in the length where they occur.Or

(b)A rectangular barge of length 40 m and light displacement 200 t, has five identical holds into which bulk cargo is loaded and trimmed level as follows: 360 t in No: 1, 720 t in No: 2, 720 t in No: 4 and 360 t in No: 5. No: 3 hold is left empty. Draw the SF & BM diagrams to scale.State the

ships maximum values of SF and BM and the points in the length where they occur.

2

(a)A box shaped vessel has length 80 m and breadth 10 m and is floating in the light condition at a Draft of 3.0 m in water RD 1.010. It is divided into four holds of equal length. Cargo is loaded as follows:No. 1 120 tonnes,No. 2 120 tonnes,No. 3 empty,No. 4 160 tonnes.Construct the curves of shear force and bending moment, calculating the maximum values and stating the positions where they occur.17Or(b)A box vessel L = 80 m, B = 12 m, Draft 3.0 m in FW when light, has five holds of equal length. Cargo is loaded as follows:No.1 EmptyNo.2 150 tNo.3 EmptyNo.4 50 tNo.5 50 tConstruct the curves of loads SF and BM. State the maximum values of SF and BM and the points in the length where they occur.

3

(a)A box-shaped vessel 100 m long, 15 m wide, light displacement 1200 t, has five identical holds. 3000 t of bulk cargo is loaded and trimmed level: 1500 t in No: 2 and 1500 t in No4. Construct the curves of Load, shear force and bending moment.Identify the positions where the maximum shearing forces and bending moments occur.OrIn the light condition a box-shaped vessel is 45 m in length, 8 m in breadth and floats at a Draft of 3.0 m in fresh water. The vessel has three holds each 15 m in length. 90 tonnes of bulk cargo is loaded into number 2 hold and is trimmed level. For the loaded condition construct the following:a. load curve;b. curve of shear forces,c. curve of bending moments.d. Identify the positions where the maximum shearing forces and bending moments occur.7

4 (a)Explain in detail, the IACS guidelines for loading instrument.Or(b)What are the IACS guidelines ‘Functional requirements’ of loading programme of loading instrument?Functional requirements:

• 1. The calculation program shall present relevant parameters of each loading condition in order to assist the Master in his judgement on whether the ship is loaded within the approval limits. The following parameters shall be presented for a given loading condition:

– deadweight data;– lightship data;– trim;– draft at the draft marks and perpendiculars;– summary of loading condition displacement, VCG, LCG and, if applicable, TCG;– downflooding angle and corresponding downflooding opening;– compliance with stability criteria: Listing of all calculated stability criteria, the limit

values, the obtained values and the conclusions (criteria fulfilled or not fulfilled).• 2. If direct damage stability calculations are performed, the relevant damage cases

according to the applicable rules shall be pre-defined for automatic check of a given loading condition.

• 3. A clear warning shall be given on screen and in hard copy printout if any of the loading limitations are not complied with.

• 4. The data are to be presented on screen and in hard copy printout in a clear unambiguous manner.

• 5. The date and time of a saved calculation shall be part of the screen display and hard copy printout.

• 6. Each hard copy printout shall contain identification of the calculation program including version number.

• 7. Units of measurement are to be clearly identified and used consistently within a loading calculation.

5

(a)Explain the procedure for constructing the curves of Load, shear force and bending moment with suitable sketches.Or(b) Explain in details what are Sheer force & Bending moment and how they affect the ship with suitable sketches

6

(a)A rectangular barge of length 40m and light displacement 200t, has five identical holds, into which bulk cargo is loaded and trimmed level as follows: 360t in No.1, 720t in No.2, 720t in No.4 and 360t in No.5. No.3 hold is left empty. Construct the curves of Load, shear force and bending moment.State the maximum values of SF and BM and the points in the length where they occur.OrA box shaped vessel 100m long, 15m wide, light displacement has five identical holds, into which 3000t bulk cargo is loaded and trimmed level as follows: 1500t in No.2, 1500t in No.4. Construct the curves of Load, shear force and bending moment.State the maximum values of SF and BM and the points in the length where they occur.