The moment of the thrust  wEgEbEx 2 Esin yEdx on the strip about AB The moment of the   a O wEgEbEx 2 Esin y total thrust about AB  wEgEbE a 3 3 Esin  Let " X be the distance of the centre of pressure (P) from AB, then: " X  Total thrust  Total moment about AB " X  wEgEbE a 2 2 Esin y  wEgEb a 3 3 Esin y or " X  2 3 a (unless sin y=0) (ii) For any Plane Area immersed in a liquid. In Figure 30.3, let OY be the line in which the plane cuts the surface of the liquid. Let the plane be inclined at an angle y to the horizontal. Let h  the x co-ordinate of the centroid (G), and let w  the mass density of the liquid. Depth of the element dA  xEsin y Thrust on dA  wEgExEsin yEdA Moment of thrust about OY  wEgEx 2 Esin yEdA Moment of total thrust about OY   wEgEx 2 Esin yEdA  wEgEsin yE  x 2 EdA Liquid pressure and thrust. Centres of pressure 269 Fig. 30.3 Total thrust on A  wEgEAEDepth of centroid  wEgEAEhEsin y Moment of total thrust about OY  wEgEAEhEsin yE " X ; wEgEAEhEsin yE " X  wEgEsin yE  x 2 EdA or " X   x 2 EdA hA (Unless sin y  0) Let I OY be the second moment of the area about OY, then I OY   x 2 EdA and " X  I OY hA or " X  Second moment of area about the waterline First moment of area about the waterline Centres of pressure by Simpson's Rules Using Horizontal Ordinates Referring to Figure 30.4: Thrust on the element  wEgExEyEdx Moment of the thrust about OY  wEgEx 2 EyEdx Moment of total thrust about OY   wEgEx 2 EyEdx  wEg  x 2 EyEdx Total Thrust  wEgEAEDepth of centroid  wEgE  yEdxE  xEyEdx  yEdx  wEgE  xEyEdx Moment of total thrust about OY  Total Thrust  " X 270 Ship Stability for Masters and Mates where " X  Depth of centre of pressure below the surface ; Moment of total thrust about OY  wEgE  xEyEdx  " X or wEgE  xEyEdx  " X  wEgE  x 2 EyEdx and " X   x 2 EyEdx  xEyEdx The value of the expression  x 2 EyEdx can be found by Simpson's Rules using values of the product x 2 y as ordinates, and the value of the expression  xEyEdx can be found in a similar manner using values of the product xy as ordinates. Example 1 A lower hold bulkhead is 12 metres deep. The transverse widths of the bulkhead, commencing at the upper edge and spaced at 3 m intervals, are as follows: 15.4, 15.4, 15.4, 15.3 and 15 m respectively. Liquid pressure and thrust. Centres of pressure 271 Fig. 30.4 Find the depth of the centre of pressure below the waterplane when the hold is ¯ooded to a depth of 2 metres above the top of the bulkhead. Area  1 3  h  S 1  3 3  184X0  184 sq m Referring to Figure 30.5: CG  S 2 S 1  h  366X8 184  3  5X98 m CD  2X00 m z  7X98 m I OZ  1 3  h 3  S 3  1 3  3 3  975X6  8780 m 4 I CG  I OZ À A (CG) 2 , i.e. parallel axis theorem I WL  I CG  Az 2  I OZ À A (CG 2 Àz 2 ) I WL  8780 À184 5X98 2 À 7X98 2   13 928 m 4 y  I WL Az  13 928 184 Â7X98 y  9X5m. Ans. The Centre of Pressure is 9.5 m below the waterline. 272 Ship Stability for Masters and Mates Ord. SM Area func. Lever Moment func. Lever Inertia func. 15.4 1 15.4 0 0 0 0 15.4 4 61.6 1 61.6 1 61.6 15.4 2 30.8 2 61.6 2 123.2 15.3 4 61.2 3 183.6 3 550.8 15.0 1 15.0 4 60.0 4 240.0 184X0  S 1 366X8  S 2 975X6  S 3 Using vertical ordinates Referring to Figure 30.6. Thrust on the element  wEgE y 2 EyEdx  wEgEy 2 2 Edx Moment of the thrust about OX  wEgEy 2 2 EdxE 2 3 y  wEgEy 3 3 Edx Moment of total thrust about OX  w 3 EgE  y 3 Edx Total Thrust  wEgEAEDepth at centre of gravity  wEgE  yEdxE 1 2  y 2 Edx  yEdx  w 2 EgE  y 2 Edx Liquid pressure and thrust. Centres of pressure 273 Fig. 30.5 Let " Y be the depth of the centre of pressure below the surface, then: Moment of total thrust about OX  Total Thrust  " Y wg 3  y 3 Edx  wg 2  y 2 EdxE " Y or " Y  1 3  y 3 Edx 1 2  y 2 Edx The values of the two integrals can again be found using Simpson's Rules. Example 2 The breadth of the upper edge of a deep tank bulkhead is 12 m. The vertical heights of the bulkhead at equidistant intervals across it are 0, 3, 5, 6, 5, 3 and 0 m respectively. Find the depth of the centre of pressure below the waterline when the tank is ®lled to a head of 2 m above the top of the tank. Area  1 3  CI  S 1 Area  1 3  2  68  45 1 3 sq m 274 Ship Stability for Masters and Mates Fig. 30.6 Referring to Figure 30.7: CG  S 2 S 1  1 2  316 68  1 2  2X324 m CD  2X000 m DG  4X324 m  "z Liquid pressure and thrust. Centres of pressure 275 Ord, SM Area func. Ord. Moment func. Ord. Inertia func. 01 0 0 0 0 0 3 4 12 3 36 3 108 5 2 10 5 50 5 250 6 4 24 6 144 6 864 5 2 10 5 50 5 250 3 4 12 3 36 3 108 01 0 0 0 0 0 68  S 1 316  S 2 1580  S 3 Fig. 30.7 I OZ  1 9  CI  S 3  1 9  2  1580  351 m 4 I CG  I OZ À A (CG) 2 I WL  I CG  Az 2  I OZ À A (CG) 2  Az 2  I OZ À A (CG 2 Àz 2 )  351 À 45X33 2X324 2 À 4X324 2  I WL  953X75 m 4 y  I WL Az  953X75 45X33 Â4X324 y  4X87 m Ans. The Centre of Pressure is 4.87 m below the waterline. Summary When using Simpson's Rules to estimate the area of a bulkhead under liquid pressure together with the VCG and centre of pressure the procedure should be as follows: 1 Make a sketch from the given information. 2 Make a table and insert the relevant ordinates and multipliers. 3 Calculate the area of bulkhead's plating. 4 Estimate the ship's VCG below the stipulated datum level. 5 Using the parallel axis theorem, calculate the requested centre of pressure. 6 Remember: sketch, table, calculation. 276 Ship Stability for Masters and Mates Liquid pressure and thrust. Centres of pressure 277 Exercise 30 1 A fore-peak tank bulkhead is 7.8 m deep. The widths at equidistant intervals from its upper edge to the bottom are as follows: 16, 16.6, 17, 17.3, 16.3, 15.3 and 12 m respectively. Find the load on the bulkhead and the depth of the centre of pressure below the top of the bulkhead when the fore peak is ®lled with salt water to a head of 1.3 m above the crown of the tank. 2 A deep tank transverse bulkhead is 30 m deep. Its width at equidistant intervals from the top to the bottom is: 20, 20.3, 20.5, 20.7, 18, 14 and 6 m respectively. Find the depth of the centre of pressure below the top of the bulkhead when the tank is ®lled to a head of 4 m above the top of the tank. 3 The transverse end bulkhead of a deep tank is 18 m wide at its upper edge. The vertical depths of the bulkhead at equidistant intervals across it are as follows: 0, 3.3, 5, 6, 5, 3.3 and 0 m respectively. Find the depth of the centre of pressure below the top of the bulkhead when the tank is ®lled with salt water to a head of 2 m above the top of the bulkhead. Find also the load on the bulkhead. 4 A fore-peak bulkhead is 18 m wide at its upper edge. Its vertical depth at the centre line is 3.8 m. The vertical depths on each side of the centre line at 3 m intervals are 3.5, 2.5 and 0.2 m respectively. Calculate the load on the bulkhead and the depth of the centre of pressure below the top of the bulkhead when the fore-peak tank is ®lled with salt water to a head of 4.5 m above the top of the bulkhead. 5 The vertical ordinates across the end of a deep tank transverse bulkhead measured downwards from the top at equidistant intervals, are: 4, 6, 8, 9.5, 8, 6 and 4 m respectively. Find the distance of the centre of pressure below the top of the bulkhead when the tank is ®lled with salt water. 6 A square plate of side `a' is vertical and is immersed in water with an edge of its length in the free surface. Prove that the distance between the centres of pressure of the two triangles into which the plate is divided by a diagonal, is a  13 p 8 Chapter 31 Ship squat 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 ¯ow 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 ship squat. It is not the difference between the draughts when stationary and the draughts when the ship is moving ahead. If the ship moves forward at too great a speed when she is in shallow water, say where this static even-keel underkeel clearance is 1.0 to 1.5 m, then grounding due to excessive squat could occur at the bow or at the stern. For full-form ships such as supertankers or OBO vessels, grounding will occur generally at the bow. For ®ne-form vessels such as passenger liners or container ships the grounding will generally occur at the stern. This is assuming that they are on even keel when stationary. It must be generally, because in the last two decades, several ship types have tended to be shorter in LBP and wider in Breadth Moulded. This has led to reported groundings due to ship squat at the bilge strakes at or near to amidships when slight rolling motions have been present. Why has ship squat become so important in the las t thirty years? Ship squat has always existed on smaller and slower vessels when underway. These squats have only been a matter of centimetres and thus have been inconsequential. However, from the mid-1960s to the late 1990s, ship size has steadily grown until we have supertankers of the order of 350 000 tonnes dwt and above. These supertankers have almost outgrown the ports they visit, resulting in small static even-keel underkeel clearances of 1.0 to 1.5 m. Alongside this development in ship size has been an increase in service [...]... draft ˆ 4X00 ‡ 0X 19 ˆ 4X 19 m say draft d2 (b) Find the Shift of the Centre of Buoyancy B 4 BB1 ˆ LB À ma 30 10  15  5  56 X 25 4 ˆ ˆ 100 30  15  5 477X5 50  10 À 100 ˆ 0X12 m ma  (c) To Find IOZ LB 3 ml b 3 À 12 3 50  10 3 30 15  5 3 À  ˆ 12 3 100 ˆ 4166X7 À 187X5 ICL ˆ ˆ 397 9 m 4 300 Ship Stability for Masters and Mates IOZ ˆ ICL À AEBB 2 1 ˆ 397 9 À 477X5  0X12 2 ˆ 397 9 À 7 ˆ 397 2 m 4 (d) To... 9  7X5 ˆ 0X61 ˆ 100  18 À 15  9 Bodily increase in draft ˆ New draft ˆ 7X50 ‡ 0X61 ; New draft ˆ 8X11 m ˆ draft d2 298 Ship Stability for Masters and Mates (b) Find the Shift of the Centre of Buoyancy a  Ba4 LB À a 15  9  18a4 607X5 ˆ ˆ 100  18 À 15  9 16 65 BB1 ˆ ˆ 0X37 m (c) To Find IOZ ICL  3  3 B L B …L À l† ˆ E ‡ E 2 3 2 3 ICL ˆ 9 3  100 9 3  85 ‡ ˆ 24 300 ‡ 20 655 3 3 ˆ 44 95 5 m 4... m GM ˆ 0X50 m New GM ˆ 0X57 m 294 Ship stability or K2 ˆ 400  9X81  0X5 4  p2 ˆ 49X 69 ; K ˆ 7X 05 I (Originally) ˆ MEK 2 Io ˆ 10 000  49X 69 Io ˆ 496 90 0 tonnesÁm 2 I of discharged mass about G ˆ 50  14 2 ˆ 98 00 tonnesÁm 2 New I of ship about the original C of G ˆ Original I À I of discharged mass ˆ 496 90 0 À 98 00 ˆ 487 100 tonnesÁm 2 By the Theorem of parallel axes: Let I2 ˆ New I of ship about.. .Ship squat 2 79 speed on several ships, for example container ships, where speeds have gradually increased from 16 knots up to about 25 knots Ship design has seen tremendous changes in the 198 0s and 199 0s In oil tanker design we have the Jahre Viking with a dwt of 56 4 7 39 tonnes and an LBP of 440 m This is equivalent to the length of 9 football pitches In 199 7, the biggest container ship to... New I of ship about À W  GG 2 ˆ 1 the original C of G the new C of G I2 ˆ 487 100 À 9 95 0  0X07 2 I2 ˆ 487 100 À 49 I2 ˆ 487 051 tonnesÁm 2 I2 ˆ M2 EK 2 2 I2 M2 487 051 ; K2 ˆ 2 9 95 0 r 487 051 K2 ˆ New K ˆ 9 95 0 ; New K 2 ˆ Let K2 ˆ 7 m Let 2pK2 T2 ˆ New T ˆ p gEGM2 2p7 T2 ˆ p 9X81  0X57 Ans T2 ˆ 18X6 s Unresisted rolling in still water 2 95 Procedure... been estimated to be in the region of $28 million In May 199 7, the repairs to the Sea 282 Ship Stability for Masters and Mates Empress were completed at Harland & Wolff Ltd of Belfast, for a reported cost of £20 million Rate of exchange in May 199 7 was the order of £1 ˆ $1 .55 She was then renamed the Sea Spirit 2 The ship of®cers could load the ship up an extra few centimetres (except of course where... 2 A ship of 10 000 tonnes displacement has GM ˆ 0 .5 m The period of roll in still water is 20 seconds Find the new period of roll if a mass of 50 tonnes is discharged from a position 14 m above the centre of gravity Assume g ˆ 9X81 m/sec 2 W2 ˆ W0 À w ˆ 10 000 À 50 ˆ 9 95 0 tonnes 2pK T ˆ p gEGM 2pK 20 ˆ p 9X81  0X5 400 ˆ 4Ep 2 EK 2 9X81  0X5 wÂd W2 50  14 ˆ 9 95 0 GG1... 299 from end to end and for the full depth of the vessel A compartment amidships on the starboard side is 15 m long and contains cargo with permeability `m' of 30 per cent Calculate the list if this compartment is bilged KG ˆ 3 m Fig 34.3 (a) Find the New Mean Draft Volume of lost buoyancy Area of intact W.P 30  15  5  4 90 100 ˆ ˆ 30 477X5  15  5 50  10 À 100 Bodily increase in draft ˆ ˆ 0X 19. .. to 0.2 65 Equivalent `B' Width of influence ˆ FB ˆ in open water b Vk ˆ speed of ship relative to the water, in knots Aw ˆ Ac À AS `B' ˆ 7X7 ‡ 20…1 À CB † 2 ship breadths Fig 31.3 284 Ship Stability for Masters and Mates Ship type Typical CB , fully loaded Ship type Typical CB , fully loaded ULCC Supertanker Oil tanker Bulk carrier 0. 850 0.8 25 0.800 0. 750 General cargo Passenger liner Container ship Coastal... the ship, the weight being a unit of force, i.e W ˆ Mg MEv 2 ;MEg  GZ ˆ  B1 Z r Heeling couple ˆ 288 Ship Stability for Masters and Mates Fig 32.1 or GZ ˆ but at a small angle and v2  B1 Z gEr GZ ˆ GMEsin y B1 Z ˆ BGEcos y ; GM sin y ˆ and tan y ˆ v2 BGEcos y gEr v 2  BG gErEGM Example A ship turns to port in a circle of radius 100 m at a speed of 15 knots The GM is 0.67 m and BG is 1 m If g ˆ 98 1 . 8780 À184 5X98 2 À 7X98 2   13 92 8 m 4 y  I WL Az  13 92 8 184 Â7X98 y  9X5m. Ans. The Centre of Pressure is 9 .5 m below the waterline. 272 Ship Stability for Masters and Mates Ord. SM. f7X7  201 À 0X830 2 g 55 ; `B'  455 mY i.e. arti®cial boundaries in open water or wide rivers 284 Ship Stability for Masters and Mates Ship type Typical C B , Ship type Typical C B , fully. 0X98 m y o  2X50 m Hence, remaining underkeel clearance at bow  y o À d max  y 2 . ; y 2  2X500 À 0X98  1X52 m @ V k of 11 kts. Ship squat 2 85 286 Ship Stability for Masters and Mates Exercise