MARCH 2017

NAVAL ARCHITECTURE TIME ALLOWED - 3 HOURS

Instructions 1. Answer SIX questions only. THREE from each section

2. All questions carry equal marks. 3. Neatness in handwriting and clarity in expression

carries weightage 4. Illustration of an answer with clear sketches /

diagrams carries weightage 5. All unused pages of answer script must be

cancelled out by two lines (X) across the page.

Q1.A. Discuss the need for adequate support of engine room gantry cranes, detailing the following i) Sketch section through the engine room casing showing how the crane is supported by the ship structure; ii) State what restricts the forward and aft limits of the crane and what is fitted to prevent the crane damaging the forward and aft bulkheads or casing. B. State the Second Engineer’s responsibilities for the engine room gantry crane and state the instructions he should give to junior engineers to ensure safe working practices are adopted for its use.

SAnswer:- (A4) Efficient lifting gear is essential in the engine room to allow the removal of machinery parts for inspection, maintenance and repair. The main equipment is a travelling crane of 5 or 6 tonne lifting capacity on two longitudinal rails which run the full length of the casing. The rails are formed by rolled steel joists, efficiently connected to the ends of the casing or the engine room bulkheads by means of large brackets. Intermediate brackets may be fitted to

reduce movement of the rails, as long as they do not obstruct the crane.

Fig. shows a typical arrangement.

The height of the rails depends upon the height and type of the machinery, sufficient clearance being allowed to remove long components such as pistons and cylinder liners. The width between the rails is arranged to allow the machinery to be removed from the ship.

The rails are provided with steel stop brackets at the forward and aft ends, which will prevent the movement of the crane carriage beyond the brackets. In cranes provided with electrical traverse arrangements , limit switches are also provided.

(B) The second engineer is responsible for the efficient maintenance of the crane and these include:

* on a regular fortnightly routine , the crane grease points should be augmented by grease.

* The crane falls must be unwound and examined for any broken strands and then carbon-grease coated. If the falls are damaged it should be changed.

* The auto safety brake should be tested whenever lifting any heavy component of the engine.

* The motor insulation resistances should be checked every six months and recorded.

* The crane should not be used for any other purpose other than lifting engine components during maintenance work.

Q2 Explain what is meant by permissible length of compartments in passenger ships; B. Describe how the position of bulkheads is determined. C. Describe briefly the significance of the factor of subdivision.

Answer:- Bulkhead deck :- In every passenger ship the lowest continuous deck aft of the collision bulkhead is termed the bulkhead deck. The volume under this deck is sub divided by water-tight steel transverse bulkheads making them into water tight compartments for machinery spaces comprising of separate main motor room, boiler room and generator room, The other spaces being store rooms, baggage room and cargo holds in passenger cum cargo ships. This deck is called the bulkhead deck since the water-tight bulkheads extend up to this deck.

Floodable length:- The floodable length of a water-tight compartment as described above is defined as the maximum length of a compartment with reference to the midpoint of the considered length, when flooded by bilging will cause the ship to sink to a level coinciding with an

imaginary line called the margin line which is 76 mm below the bulk head deck line and parallel to it.

(B)This length is theoretically determined by applying the principles of stability and floatation considering symmetrical bilging, and an assumed permeability as given in the regulations.

(A) Permissible length:- The maximum permissible length of a compartment having its centre at any point on the ships length as obtained above , when multiplied further by a constant less than one , the permissible length of the compartment is obtained. This safety provision is made so the ship will remain afloat with the margin line above the waterline when the compartment is flooded

(C) The factor of subdivision shall depend on the length of the ship , and for a given length shall vary according to the nature of the service , for which the ship is intended. It shall decrease in a regular and continuous manner as follows:

1. As the Length of the ship increases, and 2. From a factor A, applicable to ships primarily engaged in the carriage of

cargo , to a factor B, applicable to ships, primarily engaged to carriage of passengers. The variations of the factors A and B shall be expressed by the following formula (1) and (2) where L is the length of the ship as defined in regulation to ( length between perpendiculars)

A = 58.2 + 0.18 ( L = 131 m and upwards ) (1) L - 60

B = 30.3 + 0.18 (L = 79m and upwards) (2) L – 42

The above two factors A and B when combined mathematically with the criterion of Service ( Cs) gives the appropriate factor of subdivision.

Criterion of Service:-

For a ship of given length the appropriate factor of subdivision shall be determined by the criterion of service numeral ( Herein after called the criterion Numeral ) as given by the following formula (3) and (4) where :

Cs = Criterion NumeralL = Length of the ship in meters as defined aboveM = The volume of the machinery space ( cubic meters) as defined in regulation 2 with the addition thereto of the volume of any permanent oil fuel bunkers which may be situated above the inner bottom and forward of or abaft the machinery space.P = The whole Volume of the passenger spaces below the margin line ( cubic meters) as defined in regulation 2 V = The whole Volume of the Ship below the margin Line ( cubic meters)

P1 = KN (3) Where:

N = the number of passengers for which the ship is to be certified, and K = 0.056 L

Q3. Explain how the period of roll varies with - A. The amplitude of roll; B. The radius of gyration; C. The initial metacentric height; D. The location of masses in the ship.

Answer:- The frequency of the rolling period of a loaded ship is given by:

F = 1/2∏*√(GM*g)/k2 -------------(1)

Where F is the natural frequency;

g is the gravitational constant;

k is the radius of gyration caused by mass distribution in the transverse plane.

GM is the initial metacentric height.

The period of roll is the reciprocal of the natural frequency of the ship.

(A) In the above formula the amplitude of roll does not figure , since the formula is obtained by applying the constraint that the roll amplitude is small . The derivation of the natural frequency of roll is based on the assumption that sinƟ =Ɵ wjich limits the natural angular amplitude to about 50. The ship rolls because of the continuous wave action which makes the oscillation of the ship a forced oscillation in which the period of roll of the ship will be the same as the wave period . The amplitude of the roll becomes the vector sum of the wave amplitude and the natural roll amplitude of the ship which cannot be more than 5 0 as based on the assumption. The vector sum in the minimum condition is when the phase difference between the wave amplitude and the ship amplitude is 1800 and the resultant amplitude at this position is wave amplitude minus ship amplitude. In the maximum condition the phase difference is 00 and the resultant is wave amplitude plus ship amplitude. Comparing both these conditions the difference is negligible since the ship amplitude is small. Hence the amplitude of roll will be same as the wave amplitude for all practical purposes

(B)The radius of gyration is the result of mass distribution in the ship and the variation is negligible when compared to the variation of GM. Hence the radius of gyration , for all practical purposes is considered constant in the formula stated above defining the period of roll.

(C) The initial metacentric height is GM and it varies inversely as the period of roll. Smaller the GM greater the period of roll and the ship is comfortable and safe. Whereas greater the GM, the ship is called a stiff ship and though very stable it will be uncomfortable , can cause movement of cargo and lead to structural damage

(D) The location of masses in the ship determines the radius of gyration . If we consider two examples of cargo loading in a bulk carrier say grain of o.6 relative density and iron ore of say relative density 6, the radii of gyration in both cases differs negligibly as compared to the variation in initial GM. Hence the radius of gyration is considered a constant .

Q4. A. Describe the relationship between frictional resistance and: (i) Ship speed; (ii) the wetted area; (iii) surface roughness; (iv) The length of the vessel.

Answer:- As the ship moves through the water, friction between the hull and the water causes a belt of eddying water adjacent to the hull to be drawn along with the ship, although at a reduced speed. The belt moves aft and new particles of water are continually set in motion, the force required to produce this motion being provided by the ship.

The frictional resistance of a ship depends upon:

(i) The speed of the ship(ii) The wetted surface area(iii) The length of the ship(iv) The roughness of the hull(v) The density of the water.

William Froude developed the formula:

Rf = f S Vn N

Where f is a coefficient which depends upon the length of the ship L, the roughness of the hull and the density of the water

S is the wetted surface area in m2

V is the ship speed in knots

N is an index of about 1.825

The value of f for a mild steel hull in sea water is given by:

f = 0.417 + 0.773/(L+2.862)

Thus f is reduced as the length of the ship is increased.

In a slow or medium-speed ship the frictional resistance forms the major part of the total resistance, and may be as much as 75% of Rt(total resistance.). The importance of surface roughness may be seen when a ship is badly fouled with marine growth or heavily corroded, when the speed of the ship may be considerably reduced.

Q5. If a ship is seriously damaged under water in way of a large fuel side bunker tank what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety?

Answer:- Answer:- I am depicting two different tank system to show the effect and their comparison for safety and convenience.

Fig 1 shows the older system without the provision of the overflow tank. Fig 2 shows the modern system with overflow tank.

In both these cases , since seawater is heavier than fuel oil the oil will not gush out of the damage area but the seawater will enter the tank and the equilibrium level attained will be in the air pipe of the bunker tank which is higher than the sea level outside. These conditions are based on the reality that the sea is relatively calm. Hence in both the systems overflow of oil will not occur

Considering the older system if an attempt to empty the damaged tank is done by transferring its contents using the fuel oil transfer pump,, only sea water will be removed through the damaged tank , without removing the oi and it will be an endless process since the sea is connected to the pump.

Whereas in the second system the extraction of oil from the overflow tank will give a continuous flow of oil to the pump till all the oil is extracted, and when water appears at the pump

suction, the pumping can be stopped . The contaminated oil collected in an alternate empty tank can be purified used.

The damage which may be caused by corrosion or mechanical damage may be attended at the due dry docking which may have to be advanced.

SECTION – II Q6. A box barge 45 m long and 15 m wide floats at a level keel

draught of 2 m in sea water, the load being uniformly distributed over the full length. Two masses, each of 30 tonne, are loaded at 10 m from each end and 50 tonne is evenly distributed between them. Sketch the shear force diagram and give the maximum shear force.

Answer:- Refer to solution of EX 60 CLASS1 examination questions. REEDS NA.

Q7. A ship of 2890 tonne displacement and speed of 14 knots has a machinery mass of 410 tonne. The mass of ship’s machinery is given by the formula: m=𝑘∆ 2/3 V3 tonne;(i) Calculate the mass of the machinery of a similar ship of 3000 tonne displacement at the corresponding speed; (ii) if the 2890 tonne ship required 2920 Kw shaft power, calculate the shaft power required by the 3000 tonne ship.

Answer:- Using the formula for the mass of the machinery , the constant k is found and applied to the second ship of 3000 tons displacement.

Applying the formula for the first ship we have

410 = k*(2890)2/3*143

Hence k =410/(28902/3*143)

The corresponding speed for the second ship is given by

V1 = 14* (3OOO/2890)1/3=14.175 knots.

Hence m2= 410/(28902/3*143)*30002/3 *14.1753

Hence m2 = 436 tonnes.

The shaft power of the second ship is obtained by applying the admiralty constant principle.

P2*(28902/3*143)/2920= 30002/3*14.1753

Hence P2=2920* (14.175/14)3*(3000/2890)2/3

P2 = 3107KW

Q8. A ship of 3000tonnes displacement is 100m long has KM=6m, KG=5.5m. The centre of floatation is 2m.aft of amidships. MCTC=40 tonnes-m. Find the maximum trim for the ship to enter a dry dock if the metacentric height at the critical instant before the ship takes the blocks forwarded and aft is to be not less than 0.3m.

Answer:- The critical situation during the docking is depicted in the sketch given under

There are three conditions of stability in this situation and they are

(1) ∆*KG> (∆-p)* KM ---------(1)upsetting moment the ship will slip offthe blocks.

(2) ∆*KG< (∆-p)* KM ----------(2) restoring moment the ship will not fall off the blocks

(3) ∆*KG= (∆-p)* KM ----------(3) critical moment it is this least metacentric height which is still positive is called the critical metacentric height and it is less than the metacentric height when floating.

Using the third equation for criticality we have

∆*KG= (∆-p)* KM which is

∆*(KM—GM) = (∆-p)* KM ,or

∆*GM = P*KM or

GM/KM = P/∆--------------(4)

In the final equation GM is the critical metacentric height and KM is the original KM

Hence P /3000= 0.3/6 or

OR P =150 tonnes

The trim moment is 150*(50-2)= 7200 TM.

The required maximum trim therefore allowed = 7200/40 =18 cms by stern

In this solution we have considered that the location of Mis constant and there is a virtual rise in location of G

There is an alternate solution which considers the location of G constant but the location of M is lowered virtually. This leads to the solution

P/3000= 0.3/(5.5+0.3)

Which gives P = 155.18 TONNES which is higher than the previous result leading to a larger trim and hence must be rejected to be on the safer side.

Q9. A ship 150 metres long arrives at the mouth of a river with drafts 5.5m. F and 6.3m A MCT 1 cm. 200 tonnesm. TPC 15 tonnes. Centre of flotation is 1.5m. aft of amidships. The ship has then to proceed up the river where the maximum draft permissible is 6.2m. It is decided that SW ballast will be run into the forepeak tank to reduce the draft aft to 6.2m. If the canter of gravity of the forepeak tank is 60 metres forward of the centre of flotation, find the minimum amount of water which must be run in and also find the final draft forward.

Answer:- Let M tons be the mass of sea water filled in the fore peak tank.

The trim moment caused by the addition of this water in FP tank is 60*M TM---------------(1)

The total trim is M60/ 200 CMS

NEGATIVE TRIM AT AFT END = 60M/200 *73.5/150 = 0.147M ---(2)

Bodily sinkage is M/15 cms

The final draft aft is 6.2 m

We can form the draft equation for aft side using m as the unknown quantity.

Hence 6.3 – 60/200*73.5/150 *M +M/15 = 6.2 ----------------(1)

Hence 10 cms =0.147 M-0.066M

Therefore M= 123 .456 tons

The draft foreward will be

5.5 + 60/200* 76.5 /150 *123.456 + 123.456/15 =5.5+.188 +0.0823 =5.77m

Q10. A ship 75m long has semi ordinates at the load water plane commencing from forward as follows: 0, 1, 2, 4, 5, 5, 5, 5, 4, 3, 2 and 0 metres respectively. The spacing between the first three semi-ordinates and the last three semi-ordinates is half the spacing at the remaining ordinates. Find the position of the Centre of Floatation relative to amidships.( Note the problem as copied is lacking the spacing of the first three and last three ordinates . I have modified it as half the spacing at other ordinates)

Answer:- Using the 1-4-1 rule of simpson and modified to suit the semiordinate distribution , the table is set

Semi ordinate

Simpsons multiplier

Product for area

Lever for moment from forward station

Moment product

0 1*1/2=1/2

0 0 0

1 4*1/2=2 2 1/2 1 2 1*1/2 +1 3 2 6 4 3 12 3 36 5 3 15 4 60 5 2 10 5 60 5 4 20 6 120 5 2 10 7 70

4 4 16 8 128 3 1+1/2 4.5 9 40.5 2 2 4 9.5 38 0 1/2 0 10 0 TOTAL ----- 96.5 ------- 559.5

The centre of floatatiom is (559.5/96.5)*7.5 m from aft station and is therefore 43.48 -37.5 m from mid ship , which is

5.98 m aft of mid ship.

( this problem has been solved by converting the sm for the first 3 and last 3 semi ordinates to suit the common distance interval. In addition the semi ordinates from 3 to 10 stations have been combined by using rule i and 2. Since the areas are not required the location of centre of area is given by

Sum of moment product/ sum for area * common distance.)