## The Theorem of Parallel Axes

The second moment of an area about an axis through the centroid is equal to the second moment about any other axis parallel to the first reduced by the product of the area and the square of the perpendicular distance between the two axes. Thus, in Figure 29.4, if G represents the centroid of the area (A) and the axis OZ is parallel to AB, then Iqz Iab _ Ay2 parallel axis theorem equation To find the second moment of a ship's waterplane area about the centre line. It has been shown in chapter 10...

## Example

A box-shaped vessel 100 metres long X 20 metres wide X 12 metres deep is floating in salt water on an even keel at 6 metres draft. A forward compartment is 10 metres long, 12 metres wide and extends from the outer bottom to a watertight flat, 4 metres above the keel. The compartment contains cargo of permeability 25 per cent. Find the new drafts if this compartment is bilged. Increase in draft w TPC 123 20.5 6 cm Increase in draft 0.06 m

## Moments of Forces

The moment of a force is a measure of the turning effect of the force about a point. The turning effect will depend upon the following (a) The magnitude of the force, and (b) The length of the lever upon which the force acts, the lever being the perpendicular distance between the line of action of the force and the point about which the moment is being taken. The magnitude of the moment is the product of the force and the length of the lever. Thus, if the force is measured in Newtons and the...

## Exercise

1 (a) Define permeability, 'p'. (b) A box-shaped vessel 100 m long, 15 m beam floating in salt water, at a mean draft of 5 m, has an amidships compartment 10 m long which is loaded with a general cargo. Find the new mean draft if this compartment is bilged, assuming the permeability to be 25 per cent. 2 A box-shaped vessel 30 m long, 6 m beam, 5 m deep, has a mean draft of 2.5 m. An amidships compartment 8 m long is filled with coal stowing at 1.2 cu. m per tonne. 1 cu. m of solid coal weighs...

## The effect on boxshaped vessels

New mass of water displaced Old mass of water displaced New volume X new density Old volume X Old density New volume Old density Old volume New density But volume L X B X draft L X B X New draft Old density L X B X Old draft New density New draft Old density Old draft New density A box-shaped vessel floats at a mean draft of 2.1 metres, in dock water of density 1020 kg per cu. m. Find the mean draft for the same mass displacement in salt water of density 1025 kg per cubic metre. New draft Old...

## Strength curves for ships

Strength curves consist of five curves that are closely inter-related. The curves are 1 Weight curve tonnes m run or kg m run. 2 Buoyancy curve either for hogging or sagging condition tonnes m 3 Load curve tonnes m run or kg m run. 4 Shear force curve tonnes or kg. 5 Bending moment curve tonnes m or kg m. Some forms use units of MN m run, MN and MN.m. Buoyancy curves A buoyancy curve shows the longitudinal distribution of buoyancy and can be constructed for any wave formation using the Bonjean...

## To find the change of draft forward and aft due to change of trim

When a ship changes trim it will obviously cause a change in the drafts forward and aft. One of these will be increased and the other decreased. A formula must now be found which will give the change in drafts due to change of trim. Consider a ship floating upright as shown in Figure 15.4(a). F1 represents the position of the centre of flotation which is I metres from aft. The ship's length is L metres and a weight 'w' is on deck forward. Let this weight now be shifted aft a distance of'd'...

## To find Transverse BM

The Transverse BM is the height of the transverse metacentre above the centre of buoyancy and is found by using the formula 1 The second moment of the water-plane area about the centre line, and V The ship's volume of displacement The derivation of this formula is as follows Consider a ship inclined to a small angle (0) as shown in Figure 12.3(a) Let 'y' be the half-breadth. Since 0 is a small angle then arc WWj arc LL1 Area of wedge WOWj Area of wedge LOL1 Consider an elementary wedge of...

## The Deadweight Scale

The Deadweight Scale provides a method for estimating the additional draft or for determining the extra load that could be taken onboard when a vessel is being loaded in water of density less than that of salt water. For example, the vessel may be loading in a port where the water density is that of fresh water at 1.000 t cu.m. This Deadweight Scale (see Figure 35.1) displays columns of scale readings for Dwt in salt water and in fresh water. Draft of ship (mean). Displacement in tonnes in salt...

## To find the Resultant Thrust

In Figure 30.1, G represents the centroid of an area which is immersed, though not necessarily vertical, in a liquid, and Z represents the depth of the centroid of the area below the surface. Let w be the mass density of the liquid. If an element (dA) of the area, whose centroid is Z1 below the surface, is considered, then Thrust on the element dA Pressure intensity x Area Resultant thrust Density X g X Depth of centroid X Area It should be noted that this formula gives only the magnitude of...

## Revision oneliners

The following are sixty-five one-line questions acting as an aid to examination preparation. They are similar in effect to using mental arithmetic when preparing for a mathematics exam. Elements of questions may well appear in the written papers or in the oral exams. Good luck. 1. What is another name for the KG 2. What is a hydrometer used for 3. If the angle of heel is less than 10 degrees, what is the equation for GZ 4. What are the formulae for TPC and MCTC for a ship in salt water 5. Give...

## J V 1 W P

A ship of 6000 tonnes displacement enters a drydock trimmed 0.3 m by the stern. KM 7.5 m, KG 6 m. MCTC 90 tonnes m. The centre of flotation is 45 m from aft. Find the effective metacentric height at the critical instant before the ship takes the blocks overall. Note. Assume that the trim at the critical instant is zero. From these results it would appear that there are two possible answers to the same problem, but this is not the case. The ship's ability to return to the upright is indicated by...

## Method a

In Figure 28.3 consider the two parallel forces P and (W P). Their resultant W will act upwards through Mt such that (W P)x MMj x sin 0 P x KMt x sin 0 (W P)x MMj P x KMt W x MMj P x MMj P x KMt W x MMt P x KMt + P x MMt P (KMt + MMt) P x KM There are now two forces to consider W acting upwards through Mt and W acting downwards through G. These produce a righting moment of W x GMt x sin 0. Note also that the original metacentric height was GM but has now been reduced to GMt. Therefore MMt is...

## What exactly is ship squat

When a ship proceeds through water, she pushes water ahead of her. In order not to have a 'hole' in the water, this volume of water must return down the sides and under the bottom of the ship. The streamlines of return flow are speeded up under the ship. This causes a drop in pressure, resulting in the ship dropping vertically in the water. As well as dropping vertically, the ship generally trims forward or aft. The overall decrease in the static underkeel clearance, forward or aft, is called...

## The Five Eight Rule Simpsons Third Rule

This rule may be used to find the area between two consecutive ordinates when three consecutive ordinates are known. The rule states that the area between two consecutive ordinates is equal to five times the first ordinate plus eight times the middle ordinate minus the external ordinate, all multiplied by -2 of the common interval. Thus Area 1 (5a + 8b - c) or X CI X E3 h1 Also Area 2 (5c + 8b - a) or XC X Z3 12 12 E3 is used because it is a total using Simpson's Third Rule. Consider the next...

## Draft Surveys

When a ship loads up at a port, departs from this port and travels to another port, a Draft Survey is carried out. This is to check that the cargo deadweight or 'constant' is satisfactory for the shipowner at the port of arrival. It is virtually a check on the amount of cargo that left the first port against that arriving at the second port. This Draft Survey may be carried out by a Master, a Chief Engineer or a Naval Architect. Prior to starting on a Draft Survey the vessel should be in...

## B d Sagging Condition

Planes of these two sections will remain perpendicular to the plane AB but will now be inclined at an angle d0 to each other. The parts of the beam above the layer AB are in compression and those below the layer AB are in tension. Thus the layer AB is neither in compression or tension and this layer is called the Neutral Axis. Let the radius of curvature of the neutral axis layer be R. Consider a layer of thickness dy which is situated at distance y from the plane of the Neutral Axis. Original...

## Notice to Shipowners Masters and Shipbuilders

1 It has become evident that the masters' task of ensuring that his ship complies with the minimum statutory standards of stability is in many instances not being adequately carried out. A feature of this is that undue traditional reliance is being placed on the value of GM alone, while other important criteria which govern the righting lever GZ curve are not being assessed as they should be. For this reason the Department, appreciating that the process of deriving and evaluating GZ curves is...

## Effect of removing or discharging mass

Consider a rectangular plank of homogeneous wood. Its centre of gravity will be at its geometrical centre that is, half-way along its length, half-way across its breadth, and at half depth. Let the mass of the plank be Wkg and let it be supported by means of a wedge placed under the centre of gravity as shown in Figure 2.2. The plank will balance. Now let a short length of the plank, of mass w kg, be cut from one end such that its centre of gravity is d metres from the centre of gravity of the...

## Unresisted rolling in still water

A ship will not normally roll in still water but if a study be made of such rolling some important conclusions may be reached. For this study it is assumed that the amplitude of the roll is small and that the ship has positive initial metacentric height. Under the conditions rolling is considered to be simple harmonic motion so it will be necessary to consider briefly the principle of such motion. Let XOY in Figure 33.1 be a diameter of the circle whose radius is 'r' and let OA be a radius...

## The moment of statical stability at a small angle of heel

At small angles of heel the force of buoyancy may be considered to act vertically upwards through a fixed point called the initial metacentre (M). This is shown in Figure 14.2, in which the ship is inclined to a small angle (0 degrees). Moment of statical stability W X GZ But in triangle GZM GZ GMsin 0 .'. Moment of statical stability W X GM X sin0 From this formula it can be seen that for any particular displacement at small angles of heel, the righting moments will vary directly as the...

## What are the factors governing ship squat

Squat varies approximately with the speed squared. In other words, we can take as an example that if we halve the speed we quarter the squat. In this context, speed Vk is the ship's speed relative to the water in other words, effect of current tide speed with or against the ship must be taken into account. Another important factor is the block coefficient Cb. Squat varies directly with Cb. Oil tankers will therefore have comparatively more squat than passenger...

## Correcting an angle of loll

If a ship takes up an angle of loll due to a very small negative GM it should be corrected as soon as possible. GM may be, for example 0.05 to 0.10 m, well below the D.Tp. minimum stipulation of 0.i5m. First make sure that the heel is due to a negative GM and not due to uneven distribution of the weights on board. For example, when bunkers are burned from one side of a divided double bottom tank it will obviously cause G to move to Gi, away from the centre of gravity of the burned bunkers, and...

## Drydocking and grounding

When a ship enters a drydock she must have a positive initial GM, be upright, and trimmed slightly, usually by the stern. On entering the drydock the ship is lined up with her centre line vertically over the centre line of the keel blocks and the shores are placed loosely in position. The dock gates are then closed and pumping out commences. The rate of pumping is reduced as the ship's stern post nears the blocks. When the stern lands on the blocks the shores are hardened up commencing from aft...

## Ship Construction and Stability

Attempt all questions, Marks for each question are shown in brackets, 1. A ship has a light displacement 435 tonnes with KG 3.25 m. On board this ship there are 40 tonnes of fuel, water, stores and crew effects at Kg 3.80 m with a free surface moment of 35 tm. The single hold is rectangular of length 40 m, width 9.4 m and depth 5.1 m and is to be filled with a bulk cargo (stowage factor 1.4 m3 tonne). When filled the Kg of the hold is 3.34 m. Calculate the GM of the ship in its loaded...

## The effect on shipshaped vessels

It has already been shown that when the density of the water in which a vessel floats is changed the draft will change, but the mass of water in kg or tonnes displaced will be unchanged, i.e. New volume X new density Old volume X old density New volume Old density Old volume New density With ship-shapes this formula should not be simplified further as it was in the case of a box-shape because the underwater volume is not rectangular. To find the change in draft of a ship-shape due to change of...

## The coefficient of fineness of the waterplane area Cw

The coefficient of fineness of the water-plane area is the ratio of the area of the water-plane to the area of a rectangle having the same length and maximum breadth. In Figure 9.1 the area of the ship's water-plane is shown shaded and ABCD is a rectangle having the same length and maximum breadth. Coefficient of fineness Cw ---- - A Area of the water-plane L X B X Cw

## Bilging amidships compartments

When a vessel floats in still water it displaces its own weight of water. Figure 21.1 a shows a box-shaped vessel floating at the waterline WL. The weight of the vessel W is considered to act downwards through G, the centre of gravity. The force of buoyancy is also equal to W and acts upwards through B, the centre of buoyancy. b W. Now let an empty compartment amidships be holed below the waterline to such an extent that the water may flow freely into and out of the compartment. A vessel holed...

## Effect of hog and sag on draft amidships

When a ship is neither hogged nor sagged the draft amidships is equal to the mean of the drafts forward and aft. In Figure 25.1 d the vessel is shown in hard outline floating without being hogged or sagged. The draft forward is F, the draft aft is A, and the draft amidships KX is equal to the average of the drafts forward and aft. Now let the vessel be sagged as shown in Figure 25.1 d by the broken outline. The draft amidships is now K1X, which is equal to the mean of the drafts forward and aft...

## Heel due to turning

When a body moves in a circular path there is an acceleration towards the centre equal to v2 r where v represents the velocity of the body and r represents the radius of the circular path. The force required to produce this acceleration, called a 'Centripetal' force, is equal to -, where M is the In the case of a ship turning in a circle, the centripetal force is produced by the water acting on the side of the ship away from the centre of the turn. The force is considered to act at the centre...

## Angle of loll

When a ship with negative initial metacentric height is inclined to a small angle, the righting lever is negative, resulting in a capsizing moment. This effect is shown in Figure 24.1 a and it can be seen that the ship will tend to heel still further. At a large angle of heel the centre of buoyancy will have moved further out the low side and the force of buoyancy can no longer be considered to act vertically upwards though M, the initial metacentre. If, by heeling still further, the centre of...

## True mean draft

In previous chapters it has been shown that a ship trims about the centre of flotation. It will now be shown that, for this reason, a ship's true mean draft is measured at the centre of flotation and may not be equal to the average of the drafts forward and aft. It only does when LCF is at average Consider the ship shown in Figure 25.1 a which is floating on an even keel and whose centre of flotation is FY aft of amidships. The true mean draft is KY, which is also equal to ZF, the draft at the...

## List with zero metacentric height

When a weight is shifted transversely in a ship with zero initial metacentric height, the resulting list can be found using the 'Wall sided' formula. The ship shown in Figure 38.1 has zero initial metacentric height. When a weight of mass 'w' is shifted transversely through a distance 'd', the ship's centre of gravity shifts from G to G1 where the direction GG1 is parallel to the shift of the centre of gravity of the weight shifted. The ship will then incline to bring the centres of gravity and...

## Bonjean Curves

Bonjean Curves are drawn to give the immersed area of transverse sections to any draft and may be used to determine the longitudinal distribution of buoyancy. For example, Figure 41.11 a shows a transverse section of a ship and Figure 41.11 b shows the Bonjean Curve for the same section. The immersed area to the waterline WL is represented on the Bonjean Curve by ordinate AB, and the immersed area to waterline W1L1 is represented by ordinate CD. In Figure 41.12 the Bonjean Curves are shown for...

## Using the hydrostatic curves

After the end drafts have been taken it is necessary to interpolate to find the 'mean draft'. This is the draft immediately below the LCF which may be aft, forward or even at amidships. This draft can be labelled dH. If dH is taken as being simply the average of the two end drafts then in large full-form vessels supertankers and fine-form vessels container ships an appreciable error in the displacement can occur. See Fig. 16.8. Displacement in metric tonnes 15000 20000 25000 30000 35000 40000...

## Effect of trim on tank soundings

A tank sounding pipe is usually situated at the after end of the tank and will therefore only indicate the depth of the liquid at that end of the tank. If a ship is trimmed by the stern, the sounding obtained will indicate a greater depth of liquid than is actually contained in the tank. For this reason it is desirable to find the head of liquid required in the sounding pipe which will indicate that the tank is full. In Figure 27.1, 't' represents the trim of the ship, 'L' the length of the...

## The virtual loss of GM after taking the blocks overall

When a ship takes the blocks overall, the water level will then fall uniformly about the ship, and for each centimetre fallen by the water level P will be increased by a number of tonnes equal to the TPC. Also, the force P at any time during the operation will be equal to the difference between the weight of the ship and the weight of water she is displacing at that time. 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...

## Maximum Permissible Deadweight Moment Diagram

This is one form of simplified stability data diagram in which a curve of Maximum Permissible Deadweight Moments is plotted against Displacement in tonnes on the vertical axis and Deadweight Moment in Tonnes metres on the horizontal axis, the Deadweight Moment being the moment of the Deadweight about the keel. The total Deadweight Moment at any Displacement must not, under any circumstances, exceed the Maximum Permissible Deadweight Moment at that Displacement. Diagram 3 Figure 44.1 illustrates...

## Examples Of Hogging In Civil Engineering

Mw Mb SWBM b-B-L2-5 x 10 3 WBM Fig. 42.5. Line diagram for solution using Murray's method. To find the Still Water Bending Moment SWBM To find the Still Water Bending Moment SWBM Mean Buoyancy Moment MB 'LCB 2 '25 Still Water Bending Moment SWBM MW - MB 229 225 - 211875 SWBM 17 330 t m Hogging because MW gt MB Wave Bending Moment WBM b-B-L25 X 10-3tm WBM Hogging 9.-95 X 30 X 2002'5 X 10-3 tm 166 228 tm WBM Sagging 11.02 X 30 X 2002'5 X 10-3 tm 18-01-tm TBM Hogging WBM hogging SWBM hogging 166...

## 2 Statical Stability curves

The curve of statical stability for a ship in any particular condition of loading is obtained by plotting the righting levers against angle of heel as shown in Figures 16.4 and 16.5. Curve for a ship with positive initial metacentric height. Curve for a ship with positive initial metacentric height. From this type of graph a considerable amount of stability information may be found by inspection The range of stability. This is the range over which the ship has positive righting levers. In...

## The Moment to Change Trim one centimetre MCT 1 cm or MCTC

The MCT 1 cm, or MCTC, is the moment required to change trim by 1 cm, and may be calculated by using the formula W the vessel's displacement in tonnes GML the longitudinal metacentric height in metres, and L the vessel's length in metres. The derivation of this formula is as follows Consider a ship floating on an even keel as shown in Figure 15.3 a . The ship is in equilibrium. Now shift the weight 'w' forward through a distance of 'd' metres. The ship's centre of gravity will shift from G to...

## The moment of statical stability at a large angle of heel

At a large angle of heel the force of buoyancy can no longer be considered to act vertically upwards through the initial metacentre M . This is shown in Figure 14.3, where the ship is heeled to an angle of more than 15 degrees. The centre of buoyancy has moved further out to the low side, and the vertical through B1 no longer passes through M , the initial metacentre. The righting lever GZ is once again the perpendicular distance between the vertical through G and the vertical through B1, and...

## General Ship Knowledge Btecsqahnd Part

a General ideas on ship construction and on plans available onboard ship. General definitions of main dimensions. The names of the principal parts of a ship. The candidate will be expected to show his practical acquaintance with Longitudinal and transverse framing Beams and beam knees Watertight bulkheads Hatchways and closing appliances Rudders Steering gear Shell and deck plating Double bottoms and peak tanks Bilges Propellers and propeller shafts Stern tubes Sounding pipes Air pipes The...

## The midships coefficient Cm

The midships coefficient to any draft is the ratio of the transverse area of the midships Section Am to a rectangle having the same breadth and depths. In Figure 9.3 the shaded portion represents the area of the midships section to the waterline WL, enclosed in a rectangle having the same breadth and depth. The prismatic coefficient of a ship at any draft is the ratio of the volume of displacement at that draft to the volume of a prism having the same length as the ship and the same...

## Murrays Method

Murray's Method is used to find the total longitudinal bending moment amidships on a ship in waves and is based on the division of the total bending moment into two parts a the Still Water Bending Moment, and The Still Water Bending Moment is the longitudinal bending moment amidships when the ship is floating in still water. When using Murray's Method the wave bending moment amidships is that produced by the waves when the ship is supported on what is called a 'Standard Wave'. A Standard Wave...

## Ship to ship Interaction

Consider Figure 36.5 where a tug is overtaking a large ship in a narrow river. Three cases have been considered G centre of gravity B centre of bouyancy m metacentre Fig. 36.4. Transverse squat caused by ships crossing in a confined channel. Suction effect takes place here Ships are drawn to each other, both heel slightly Vessel squats and grounds. Bilge keel may be bent or broken off Fig. 36.4. Transverse squat caused by ships crossing in a confined channel. Both ships go to port Tug heads for...

## The variable immersion hydrometer

The variable immersion hydrometer is an instrument, based on the Law of Archimedes, which is used to determine the density of liquids. The type of hydrometer used to find the density of the water in which a ship floats is usually made of a non-corrosive material and consists of a weighted bulb with a narrow rectangular stem which carries a scale for measuring densities between 1000 and 1025 kilograms per cubic metre, i.e. 1.000 and 1.025 t m3. The position of the marks on the stem are found as...