warship design experiments

8/3/2019 Warship Design Experiments

1/20

AML 794 WARSHIP DESIGN

LAB BOOK

NAVAL CONSTRUCTION WINGDEPARTMENT OF APPLIED MECHANICS


2/20

EXPERIMENT 1DIMENSION MEASUREMENT

OBJECTIVE

To measure the dimensions of a rectangular barge model

EQUIPMENT USED

(i) Metre scale(ii) Rectangular barge model

INTRODUCTION

Molded Base Line; Molded Dimensions

The molded base line, drawn in the sheer plan and body plan as astraight horizontal line, represents an important reference datum, both fordesign and construction purposes. The line, in fact, represents a plane inspace to which many vertical heights are referred. It also represents thebottom of the vessel's molded surface, and so is coincident with the topsurface of the flat plate keel on most straight-keel ships with a singlethickness of shell plating. The molded depth of a vessel is the verticaldistance between the molded base line and the molded depth line of theuppermost deck at side as shown in Fig 1.1

Fig 1.1 Midship section, molded form


3/20

A ships size and capacity can be defined in two ways: lineardimensions or tonnages. A ship is a three dimensional structure havinglength, width and depth. Dimension measurement is the initial step in thestability analysis of ships.

(i) Length: Most commonly used length measurements arelength overall, length between perpendiculars and length on loadwaterline.

(ii) Width: A ships width or more properly a ships breadth isexpressed in a number of ways like length. A ships extreme breadthcommonly called beam is measured from the most outboard point onone side to the most outboard point on the other side. The dimensionmust include any projections on either side of the vessel.

(iii) Depth: It involves several vertical dimensions like

freeboard, draught, draught marks and load lines. The vesselsdepth is measured vertically from the lowest point of the hull.

Fig 1.2: Ships Dimensions


4/20

Fig 1.3: How to read draught marks

PROCEDURE

(i) Keep the model on a flat surface and measure all the maindimensions using meter scale.

(ii) For measuring dimensions on the waterline, place the modelgently in water and mark the draught. Then carefully measure the length onload waterline. Threads may also be used for this purpose.

DATA SHEET

LOA : _____________________LBP : _____________________LWL : _____________________Beam : _____________________Depth : _____________________

PRECAUTIONS

(i) All measurements should be carefully and accurately taken.(ii) Ensure minimum error in the scale being used.

(iii) While taking measurements on the waterline, ensure that vesselis upright.


5/20

EXPERIMENT 2WEIGHT ESTIMATION - ARCHIMEDES' PRINCIPLE

OBJECTIVE

To estimate the weight of a rectangular barge model using twomethods:

(i) Physical measurement(ii) Displacement methods

EQUIPMENT USED

(i) Meter scale(ii) Pan balance(iii) Model

(iv) Tank / water front

INTRODUCTION

Mass is a physical property which all objects possess, but objects ofthe same size can have different masses and weights. This difference ischaracterized by another property, density. Density is defined as the ratio ofan object's mass to its volume: Weight of a body is given as the product of itsmass and acceleration due to gravity.

W = m x g

(Note that in this experiment the balance gives mass in grams rather thanweight)

The Displacement method makes use of Archimedes Principle whichstates that when a body is fully or partially submerged in a liquid, it displacesits own volume of liquid and the weight of this volume displaced (orunderwater volume) is the weight of the body.

W = Volume of water displaced x density of water

PROCEDURE

DETERMINATION OF WEIGHT BY DIRECT MEASUREMENT

(i) Use the pan balance, determine and record the mass of thebarge model

(ii) Calculate the weight by multiplying the mass recorded byacceleration due to gravity.


6/20

WEIGHT ESTIMATION BY DISPLACEMENT METHOD

(i) Lower the barge in water and record the water level.

(ii) Find out the underwater volume; = L x B x T

(v) Calculate the weight of the barge: = x

DATA SHEET

Mass of barge : _____________________

Length : _____________________

Breadth : _____________________

Weight from Pan : _____________________

Water level 1 : _____________________

Water level 2 : _____________________

UW Volume : _____________________

Density of liquid : _____________________

PRECAUTIONS

(i) Readings should be taken accurately.

(ii) Ensure no trim or heel of the barge while recording the data.


7/20

EXPERIMENT 3INCLINING EXPERIMENT

OBJECTIVE

To perform Inclining experiment on a rectangular barge model forfinding the vertical centre of gravity (KG) for different conditions of loading.

EQUIPMENT USED

(i) Barge model(ii) Plumb bob(iii) Metre stick on batten

INTRODUCTION

Terminologies:

(i) Lightship Displacement () is the quantity of water displacedby the vessel in tonnes (1 tonne = 1000 kg)

(ii) Transverse Metacentre If a vessel is inclined transverselythrough a small angle, the centre of buoyancy B will move slightly fromthe middle towards the side. A line perpendicular to the new WLthrough the new centre of buoyancy will intersect the centreline of theship at a point Mt referred as the transverse metacentre.

(iii) Metacentric Height GMT in any condition of loading is thedistance between the transverse metacentre and the centre of gravityof the vessel, in the condition of loading under consideration.

The experiment is carried out when the ship is as near to completion aspossible; that is as near to light condition as possible. When all is readyand the ship is upright, a weight is shifted across the deck transverselycausing the ship to list. A little time is allowed for the ship to settle and thedeflection of the plumb line along the batten is noted. If the weight is nowreturned to its original position the ship will return to the upright condition.

She may now be listed in the opposite direction. From the deflections, theGM is obtained as follows

Fig 3.1: Inclining experiment


8/20

Let a mass of w tonnes be shifted across the deck through a distanceof d meters. This will cause the centre of gravity of the ship to move from Gto G1 parallel to the shift of centre of gravity of the weight. The ship will then

list to bring the G1 vertically under M, i.e. to degrees.

From triangle MGG1, thus, we can write;

The key points of reference are:

K Point at the keelB Centre of buoyancyG Centre of gravity

M MetacentreThe height KM can be obtained from displacement tables.

PROCEDURE

(i) With the vessel steady, record the measurement or mark thezero point in line with the pendulum. Call this Experiment 0

(ii) Move weight A from port side to starboard side over weight C.When the vessel settles, mark the position of the pendulum line 1 ormeasure the deflection. Call this Experiment 1.

(iii) Move weight B to the starboard side over weight D. When thevessel is steady, mark the position of the pendulum. Call thisExperiment 2. The position of key reference points now differs fromFigure 1. B and C are no longer in line and the righting lever GZ can beclearly seen.

(iv) Move the weights A and B back to their original positions onthe port side and when all is steady, mark the position of the pendulum4 which is zero but it may not be coincident with the original zeroposition. Call this Experiment 3.

(v) Move the weight C to the port side and place it over weight A.When settled, mark the position no 5. Call this Experiment 4

(vi) Place weight d over weight B and when all is settled mark theboard no D or enter the distance moved from the new Zero position.Call this Experiment 5.

(vii) Move weights C and D back to their original positions on thestarboard side. Mark the new zero position if not the same as before orenter the distance moved by the pendulum from position no 6. (RefFig3.2)


9/20

(viii) Repeat the movements of the weights at least three times andalways be careful to record the correct zero position.

Fig 3.2: Inclining experiment procedure


10/20

CALCULATION

Weights moved (w) : ____________________ Distance moved by the weight (d) : ____________________Length of pendulum (l) : ____________________

Displacement () : ____________________

The deflection of the pendulum being recorded as follows:

EXPERIMENTNO

MEASUREMENTFROM METER

STICK ONBATTEN

DEFLECTION OF PENDULUM

01 & 08 0

02 _______weight A to B

03 _______weight B to D

04 _______weights A+B returned toport side (2 moves)

05 _______weight C to A

06 _______weight D to B

07 _______weight C+D returned tostarboard side (2 moves)

Total deflection_________ Mean deflection () _________

To obtain GM: GM = (w*d) / (W*tan )

= ((w*d*l) / (W*)

KM for a particular draught can be obtained from hydrostatics

Hence KG = KM - GM

As there are total 8 deflections henceMean deflection = total deflection / 8

PRECAUTIONS

Necessary conditions for the experiment:(i) There should be little or no wind as it may influence the inclinationof the barge.

(ii) The barge should be floating freely. This means that nothingoutside the barge should prevent her from listing freely.

(iii)Any loose weights within the barge should be removed or securedproperly.

(iv)There should be no free surfaces within the barge.

(v) The barge must be upright at the commencement of experiment.


11/20

EXPERIMENT 4FREE SURFACE EFFECT

OBJECTIVE

To calculate the effect of free surface of liquids on stability.

EQUIPMENT USED

(i) Test set up(ii) Ballast water

INTRODUCTION

When a tank is completely filled with a liquid, the liquid cannot movewithin the tank when the ship heels. For this reason as far as stability is

concerned, the liquid may be considered as a static weight having its centre ofgravity at the centre of gravity of the liquid within the tank. Figure 4.1 shows aship with a double bottom tank filled with a liquid having its centre of gravity at

g. The effect when the ship is heeled to a small angle is shown. No weightshave been moved within the ship, therefore the position of G is not affected.The centre of buoyancy will move out to the low side indicated by BB1.

Moment of statical stability = x GZ

= x GM x sin

Fig 4.1: Inclination of a ship with fully filled tank

Now consider the same ship floating at the same draught and havingthe same KG, but increase the depth of the tank so that the liquid now onlypartially fills it as shown in (c) and (d). When the ship heels, the liquid goes to

the low side of the tank such that its centre of gravity shifts from G to G1,parallel to gg1.


12/20

Fig 4.2: Inclination of a ship with partially filled tank

Moment of statical stability = x G1Z1= x GvZv= x GvM sin

This indicates that the effect of free surface is to reduce the effectivemetacentric height from GM to GvM. GGv is therefore the virtual loss of GMdue to the free surface. Any loss in GM is a loss in stability.

Fig 4.3: Effect of free surface on GM

It can be derived that


13/20

Where

I

second moment of the free surface about the centreline in m

4

W ships displacement in tonnes.

density of the liquid in the tank in tonnes / m3

N number of longitudinal compartments in the tank

PROCEDURE

Consider the barge model with the dimensions and displacement ascalculated in the previous experiments. Take the values of GM from theprevious experiment for the given loading condition. It has a tank withdimensions __________

(i) Fill water in the central tank and calculate the GM for thiscondition using inclining experiment.

(ii) Calculate the change in GM after free liquid has been added tothe central tank.

(iii) This experiment can be repeated for checking the effect oflongitudinal subdivisions in the tank.

CALCULATION

(i) Same as inclining experiment

(ii) Calculate the change in GM(iii) Loss in GM = GMEmpty GMLiquid

PRECAUTIONS

(i) Ensure that no external factors are affecting the stability of themodel.

(ii) Take all measurements in static condition.


14/20


15/20

EXPERIMENT 5ANGLE OF LOLL

OBJECTIVE

To study the effect of negative GM on ships stability

EQUIPMENT USED

(i) Barge model(ii) Pendulum.

INTRODUCTION

A special case arises when GM is negative but GZ becomes positive at

some reasonable angle of heel. This is illustrated in Fig. 5.1 as 1. If the ship

is momentarily at some angle of heel less than 1, the moment acting due to

GZ tends to increase the heel. If the angle is greater than 1, the moment

tends to reduce the heel. Thus the angle 1 is a position of stable equilibrium.

Unfortunately, since the GZ curve is symmetrical about the origin, as 1 isdecreased, the ship eventually passes through the upright condition and will

then suddenly lurch over towards the angle 1 on the opposite side andovershoot this value before reaching a steady state. This causes anunpleasant rolling motion which is often the only direct indication that the heelto one side is due to a negative GM rather than to a positive heeling moment

acting on the ship.

Fig 5.1: Angle of loll

GZ is zero when sin = 0. This merely demonstrates that the upright

condition is one of equilibrium. GZ is also zero when GM+ BM tan2

= 0.


16/20

PROCEDURE

(i) Shift the movable weight up till GM becomes negative.

(ii) Slightly disturb the model and measure the angle of loll usingthe pendulum.

(iii) Calculate the angle of loll theoretically and verify theexperimental results.

CALCULATION

GMT ________________Angle of loll ________________


17/20

EXPERIMENT 6DAMAGE STABILTY

OBJECTIVE

To study the effect of bilging of compartments on stability.

EQUIPMENT USED

(i) Barge model(ii) Meter scale

INTRODUCTION

(i) BILGING AMIDSHIPS COMPARTMENT

When a vessel floats in water, it displaces its own weight of water.Figure bellow shows a box shaped vessel floating in waterline WL. The weightof the vessel acting vertically downwards through G, force of buoyancy B =W. Now let an empty compartment amidships be holed below the waterline tosuch an extent that the water may flow freely into and out of the compartment.The vessel is thus said to be bilged. Fig 6.1 below shows the vessel in thebilged condition. Buoyancy provided by the bilged compartment is lost.Draught has increased and now vessel floats at W1L1, where it is againdisplacing its own weight of water. X represents the increase in draught dueto bilging.


18/20

Fig 6.1: Bilging of midship compartments

The volume of lost buoyancy v is made good by the volumes y and z.Hence, v = y + z

Let A be the area of waterplane before bilging, and let a be the area ofbilged compartment. Then

y + z = Ax - ax.

Or v= x (A - a)

Increase in draught = x = v / (A - a)= Volume of lost buoyancy / Area of intact waterplane

Note: KG after bilging will be same as KG before bilging.

(i) BILGING THE END COMPARTMENTS

When the bilged compartment is located in a position away frommidships, the vessels mean draught will increase to make good the lostbuoyancy but the trim will also change. Consider the box shaped vesselshown in the figure below. The vessel is floating upright on even keel, WLrepresenting the waterline. The centre of buoyancy is the centre of displacedwater and the vessels centre of gravity is vertically above B. There is notrimming moment.


19/20

Fig 6.2: Bilging of end compartments

Now let the forward compartment which is X metres long be bilged. Tomake good the loss in buoyancy, the vessels mean draught will increase asshown in figure (b), where W1L1 represents the new waterline. The centre ofgravity will remain at G. It has already been shown that the effect of meandraught will be similar to that of loading a mass in the compartment equal tothe mass of water entering the bilged space to the original waterline. Thevertical component of the shift of centre of buoyancy (B to B 1) is due to theincrease in the mean draught. KB1 is equal to half the mean draught.Horizontal component of the shift of centre of buoyancy (B1B2) is equal to X/2.

It can be shown thatw x d = Trimming moment

Thus the effect of trim is similar to that which would be produced if a

mass equal to the lost buoyancy are loaded in the bilged compartment.PROCEDURE

(iv) Calculate the permeable volume of the compartment upto theoriginal waterline.

(v) Calculate the TPI, longitudinal and lateral positions of CF for thewaterplane with the damaged area removed.

(vi) Calculate the original undamaged and revised second momentof areas of the waterplane about the CF in the two directions and

hence new BMs.

(vii) Calculate parallel sinkage and rise of CB due to the verticaltransfer of buoyancy from the flooded compartment to the layer.

(viii) Calculate new GMs.

(ix) Calculate the list and trim due to the eccentricity of the loss ofbuoyancy from the new CFs.

CALCULATION


20/20

Before flooding:Draught ________________ Displacement ________________ KG & KB ________________

KMT & KML ________________WP area, m2 ________________ LCF & LBP ________________

After flooding:Permeable volume= Permeability * initial compt. Volume

= ________________

Damaged waterplane area ________________Movement of CF aft ________________ Movt. Of CF to port ________________

Original IT ________________Damaged IT ________________Damaged BMT ________________ Original IL ________________ Damaged IL ________________Damaged BML ________________Parallel sinkage ________________ Rise of B ________________ Damaged GMT ________________Damaged GML ________________Angle of heel ________________ Angle of trim ________________ Change of trim ________________

warship design experiments

Documents