Ship Stability III by Capt. Subramaniam

5/22/2018 Ship Stability III by Capt. Subramaniam

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ABOUT THE AUTHORBorn in Sept. 1942, Capt. H. Subramaniam was educated in the Lawrence School,Lovedale, one of the best scho ols in India. He passed out of the T.S. Dufferin in 1960

winning several prizes including 'Runner-up forthe President s Gold Medal for all roundproficiency. He then served at s ea until Aug 1968when he took to teaching at the L.B .S. Nautical

Eng. College, Bombay. In Dec 1990, he tookover charge as the Principal of the same college .In Nov 1991, he was transferred to commandthe T.S. Rajendra which conducted three-yearB.Sc. (Nautical Sciences) degree courses,approved by Bombay University, for pre-seacadets. In July 1992, he did a stint of four monthsin command o f a bulk carrier from India to Japan,Australia and back to prove to himself that he'Practices what he teaches'. In Aug 1993, he wasin charge of the transfer of training from T.S .Rajendra to a new shore based nauticalacademy called T.S. Chanakya. Havingestablished T.S. Chanakya, he again took overcharge as the Principal of the LBS College, in

May 1995, by then renamed 'L.B.S. College of Advanced Maritime Studies Research.He has thus been associated with the four great maritime training institutionsin Indian hist ry - he was a cadet on T.S. Dufferin, the last Captain Superi ntendentof T.S. Rajen la, the first Captain Superintendent of T.S. Chanakya and Principal ofL.B.S. College. He retired from Govt service on 30th Septemb er 2002 after 34 gloriousyears including 2 years as head of maritime training institutions.His achievem 1ts/distinctions i n l u d ~

Extra Mastel certificate (UK); external examiner of Masters and Mates since1977; member of the 'Extra Masters Examination Board' of India since its inceptionn 1985; nautical s s s s o r in a formal investigation into a major ship collision; leaderof the Indian delegation to the IMO on two occasions; qualified 'Lead Assessor'through Lloyds Register's QSM; examiner in two subjects and overall moderator forNautical Science degree exams of Bombay University; Chief Examiner of ExtraMasters from Dec. 99 to Dec. 01 ; Chairman of the committee on HRD for In, andWater Transport, Govt of India; Member of the Executive Committee governing theIndian Institute of Port Management at Kolkata; Chairman of the Northern AcademicCouncil of the Govt of India for the inspection of maritime training institute s; title of'Principal Emeri tus' of LBS College on retirement; 'Ma n of the year Award' in 2001 bySailor Today magazine for his 'Conceptio n and impl ementation o f IN Dos'; 'LifetimeAchievement Award' in 2002 by Marine World magazine; Chairman of the NauticalInstitute, India (West) Branch ; Master of the Company of Master Mariners of India.

is qualifications, experience and devotion to teaching enable him to put eachsu 1ect in a 'Nutshell . All his eight books in the Nutshell Series - PRACTICALNA IGATION, MARINE METEOROLOGY, SHIPBORNE RADAR ARPA, SHIPSTAB ILITY I, II Ill , NAUTICAL WATCHKEEPING SPHERICAL TRIGONOMETRYhave been great successes.

SHIP STABILITY

NUTSHELL SERIESBOOKS

BY CAPT H SUBRAMANIAMEXTRA MASTER F.R.Met.S. R.I.N. F.N.I. F.C.M.M.I. M.I.Mar.Tech. M.I.Met.S.

VIJAYA PUBLIC TIONS


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SHIP ST BILITY Ill

NUTSHELL SERIESOOKS

BYCAPT. H. SUBRAMANIAMExtra Master F.R.Met.S., M.R.l.N., F.N.I. F.C.M.M.I. M.l.Mar.Tech. M.l.Met.S.

Principal EmeritusL.B.S. College o Advanced Maritime Studies ResearchMumbai.

VIJAVA PUBLICATIONS9122 25217044; e-mail: [email protected] CHAITRA, 550 ELEVENTH ROAD,CHEMBUR, MUMBAI 400 071.


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First edition: Mar 1986Reprinted:Sep 1988 Mar 1991 Mar 1995Aug 1997 Aug 1999 Oct 2000Aug 2004

CopyrightAll rights reserved

Price in India: Rs 240/

Printed and published by rs Vijaya arry for Vijaya Publications of 2 Chaitra550 l l 1h Road Chembur Mumbai 400 071 at the Book Centre Ltd. th Road

Dedicated to my motherwithout whose patient ndconstant encoura.gementthis book would not havebeen possible.


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Capt/Dr.P.S.VANCHISWAR 15 th Feb 1986Ph.D., Extra MasterResident Pro fesso r ,World Marit ime U n i v e r s i ~ yMalmo, Sweden.

F 0 R E W 0 R D

Enthusiasm, perseverance ded ica t ionare the q u a l i t i e s t ha t come to my mindwhen I th ink o f Capta in H. Subramaniam.Enthusiasm because of the way in whichhe goes about , not only his work, buta l so h i s 'hobby' of wri t ing t echnica lbooks Perseverance because of hiss t a r t i n g to wr i t e the next book as soonthe e a r l i e r one i s complete , therebywri t ing s i x books in l e s s than tenyears Dedicat ion to his work because hedoes not l e t h i s book wri t ing i n t e r f e r ewith his main occupat ion - t eaching . Inf ac t , I have heard from those who havehad the good for tune to have been hiss tudents , t ha t h i s ded ica t ion toteaching i s t o t a l .

Capt .Subramaniam's Nutshel l Se r i e s ofbooks a re very popula r today, espec i a l l yamong o f f i c ~ r s of deve loping c oun t r i e s ,because of two reasons . F i r s t l y , hisbooks a re wri t t en in s imple , s t r a i g h t -forward Engl i sh . He has managed to avoidusing d i f f i c u l t expres s ions and complexsen tences . Second ly , he has mainta ined aprac t i c a l approach th roughout , bear ingin mind the ac tua l use to which thes tuden t can apply the knowledge ga inedf r om these books.

S t a b i l i t y I I I dea l s with the s ub j e c ta t the Maste r ' s l eve l . The au tho r ' sprac t i c a l - method of ca l cu l a t i ng thes t a b i l i t y pa r t i cu l a r s of a sh ip in adamaged cond i t i on i s unique . His chap te ron s t a b i l i t y of sh ips ca r ry ing bulkgra in i s ce r t a in ly what a sh ipmas te rneeds to know. His dec i s ion to includechap te rs on shea r fo rce and b ~ n ~ i n gmoment, in t h i s book on sh ip s t a b i l i t y ,w i l l be welcomed by a l l s tuden t s for theCer t i f i c a t e of Competency as Master F.G.The a u t h o r ' s e xpe r t i s e in teaching i sc l ea r ly ev iden t in t h i ~ book.

I c o n ~ t a t u l a t e the au tho r for h i se f f o r t s and hope t ha t he w i l l con t inueto w r i t e on o the r mari t ime sub j ec t s .

Q J ~ ~(P .S .Vanchiswar)*

* Res iden t Pro fesso r , World Mari t imeUnive rs i ty , Malmo, Sweden. PreviouslyNau t i ca l Advise r to the Govt. of Indiaa nd Chief Examiner of Ma s te r s and Mates.Was Chairman of the Committee - "Masterand Deck Depar tment" - a t the ST WConference (1978) held by theI n t e r n a t i o n a l Mari t ime Organisa t ion .


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P R -E F C E

The sub jec t of Ship Stab i l i t y has beencovered in three par t s :Ship S t a b i l i t y I (Nutshel l Ser ies Book4 cons is t s of 18 chapters and adequatelycovers the syl labuses for Second MateF . G Navigat ional Watch Keeping Off ice r .Ship S t a b i l i t y I I (Nutshel l Ser iesbook 5 s t a r t s with chapter 19 and goeson to chapter 32 such t ha t both , Ship

~ a b i l i t y I and I I , cover the sy l l abuses, for F i r s t Mate F.G Mate H. T.Ship S t a b i l i t y I I I (Nutshel l Ser ies

Book 6 s t a r t s with chapter -33 and goeson to chapter 49 such tha t a l l three ,Ship Stab i l i ty I , I I and I I I , cover morethan the sy l l abuses fo r Master F,G. apdMaster Home Trade.As f a r as poss ib le , a pr ac t i ca lapproach has been maintained to enables h i p s of f i ce r s to put the knowledge topr ac t i ca l use on board ship .t i s hoped tha t Marine Engineers a l somay f ind these books use fu l .

Bon:ibay1 s t March 1986 ( H. Subrama niam

SHIP STABILITY I I I

CONTENTS

Chapter Page33 Bilging of an end compartment. 1Exercise 31.34 Bilging of an in termediate 20compartment. Exerc ise 32.35 'Bilging of a s ide compartment. 33Exercise 33.36 Bilging - Prac t i ca l sh ipboard 44ca lcula t ions . Exerc ise 34.37 Calcula t ion of list by GZ curve. 64Exercise 35.38 Angle of l o l l by GZ curve. 75Exercise 36.39 Li s t when i n i t i a l GM i s zero . 8040 - Dynamical s t a b i l i t y . S t a b i l i t y 84requirements under Loadl ineRules . Exerc ise 37.41 Effec t of beam and f reeboardon s t a b i l i t y .42 Change of t r im due to changeof densi ty .4 3 Centre of pressu re .Exercises 38 and 39.

94

97

100


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44 The inc l in ing exper iment . 12245 S t a b i l i t y informat ion supp l ied 127to sh i p s .46 S t a b i l i t y of sh ips car ry ing 132gra in in bulk.47 Shear force bending moment 150in beams. Exerc ise 40.48 SF and BM in box-shaped v e s s e l s . 171Exerc ise 41.49 SF and BM in sh i p s . 188Answers to exerc i ses 194AppendicesI Hydros ta t i c t ab le - m.v.VIJAY 204I I KN Table - m.v.VIJAY 205I I I Hydros ta t i c t ab le - m.v.VICTQRY 206IV . Cross curves - m.v.VICTORY 207V Table of hee l ing momentscargo o f bulk gra in - m.v.VIJAY 208VI Grain loading diagram -

No: 3 LH and TD - m.v.VIJAY 209VII KN t ab le fo r loading bulk gra in- m.v.VIJAY 210

-oOo-

1CHAPTER 33

BILGING OF ANEND COMPARTMENT

The inc rease in dr a f t , and the e f f ec ton GM r e s u l t i n g from b i lg ing an endcompartment i s the same as t ha t causedby b i lg ing an amidships compartment ofs im i l a r dimensions. Since the s tuden t i sexpected to have s tud ied the e f f e c t s ofb i lg ing amidships compartments inChapte r 32 in Ship S t a b i l i t y I I ,needless r e p e t i t i o n has been avoidedhere .The b i lg ing of an end compartmentcauses the COB of the sh ip to bel o n g i t u d i n a l l y d isp laced away from theb i lged compartment . Since the pos i t ionof the COG of the sh ip remainsunaf fec ted by b i lg ing , the consequentl ong i tud ina l s ~ r t i o n of the COB andthe COG i . e . , BG r e s u l t s in a t r immingmoment W.BG). Bilging an end compar tmen t , t he re fo re , causes a change inthe t r im of the sh i p .

The foregoing s t a t ement can bei l l u s t r a t e d s imply by the workedexamples t h a t fol low. However beforeproceeding with the worked examples ti s necessary to know the theorem ofp a r a l l e l axes a s app l i cab le tor ec tang les . The theorem o f par a l l e l axeswith r espec t to o ther r egu la r shapes i sexp la ined in qhapter 41 before theworked examples on cen t re of pres sure .


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2T h ~ o r e m of pa ra l l e l axes : I f I*GG i s themoment of i n e r t i a of an area about anax i s pass ing through i t s geomet r iccen t r e , and ZZ i s an ax is pa r a l l e l toGG, then

I*ZZ = I*GG + Aywhere y i s the d i s t ance between theax is GG and the ax is ZZ.

Case lI f Ly =I*ZZ

1 0 m , . ' 8 = 6 m12 m, I*ZZ = ?= I*GG + Ay 2

= 180 + 86408820 m 4

Case 2I f Ly =I*ZZ

= 8 m, B = 4 m1 m, I *ZZ = ?= I*GG + Ay 2= BL3+ LB y 2) 'T

= 170.667 + 32= 202.667 m 4

G

rI 0 m t ~ t 1 2 m.__ - m ~

Gz

G

- - 4 m ~

r8 m.

z

z

Gz

3Example lA box-shaped vesse l 100 m long 12 mwide f l oa t s a t an even kee l d r a f t of 6 min SW . The compartment a t t ~ forwardend, 10 m long and 12 m broad, i s empty.Find the new dra f t s fwd and a f t i f t h i scompartment ge t s b i lged .Plan view a t wate r l ine

A

~ - - - - 9 0 : i - - - ~ 1 }I 0 m

Side e l eva t i on

15 mlI I

d

Before b i lg ing , AB = AG = AF = 50 m.V = 100 x 12 x 6 = 7200 m3 ) unchanged byW = 7200 x 1.025 = 7380 t ) b i lg ing .

s =S = volume of l o s t bouyancyi n t a c t water -p lane area

lbdLB-lb = 10 x 12 x 61200 - 120 = 7201080 = 0.667 m


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4Fina l hyd ra f t = 6.000 0.667 = 6.667 m.Note: Since the i n t a c t underwater volumei s r e c t a ngu l a r and 90 m long, i t s AB andAF a re both = 45 m.BG = AG - new AB = SO - 45 = 5 metres .S ince the COG i s fwd of the new COB, thet r im caused w i l l be by the head.-TM = W.BG = 7380 x 5 = 36900 tm by head.Trim caused o r Tc = TMthe new MCTC has to MCTC. Howeverbe c a l c u l a t e d .MCTC = (W.GML) 7 lOOL or (W.BML) lOOL

GML = KML - KG. Since KG i s not givenhere , BML may be used ins tead of GML.Note: The L to be used in t h i s formulai s the t o t a l l en g th of the water-p lanei.e. the LBP of the s h i p .BML = I*COF o f i n t a c t WP v = 12 ( 90 3 )12 ( 7 200)

BML = 1 0 1 . ~ S m.

MCTC = W.BML100L = 7380 101 .25) = 74.723 tm.100 100)Tc =TM 7 MCTC= 36900 7 74.723 493.8 cm

Tc = 4.938 m by the head.Ta = AF Tc)

L= 45(4 .938)100 = 2 222 m

5Tf = Tc - Ta 4.938 2.222 = 2.716 m.

New hydra f tTf J r TaNew dra f t s

fwd6.667 m+2.7169.383 m

a f t6.667 m-2 .2224.445 m

No t e : I f the t r a ns ve r.se GM in the bilg.edc o n d i t i o n i s asked, the method ofc a l cu l a t i on i s the same a s t ha t inChapte r 32, in Ship. S t a b i l i t y IIExample 2I f in example 1 the permeab i l i t y ofcompartment i s 40 , f ind ehe d r a f t sand a f t .

thefwd

Before b i l g i ng , AB = AG = AF = 50 metresvw 100 x 12 x 6 - 7200 u n c ~ a n g e d by7200 x 1.025 = 7380 t )b i lg ing .

s = volume of l o s t buoyancyi n t a c t wate r -p lane a rea= 10 x .12 x 6 40/100)100 12) - 10(12)(40/100)

s = 0 . 250 me


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6u n i f o r ~ a t a l l dr a f t s a f t e r b i l g i ng ,AB = new AF. newPlan view a t water l ine

l F b cI II I I~ ~ ~ ~ ; ; ~ ~ 1 L ~ ~I I

A I I CargoSide e l eva t i on

sd

As exp la ined in chap te r 3 2, in ShipS t a b i l i t y I I ,

le = 1/100 - I = 10,/0J> 7 .746 metres .100

be = b/100 - I = 12/Q.6 9.295 metres .100whei;-e le be a re the length and breadthof the cargo .Area of cargo = 7.746 x 9.295 = 72 m2

To f ind new AB AFAl t e r na t i veTo f ind new AB, moments of vol about A.Rec t ang l e i t s AB) =90 12)6 .25 45) = 6750 45) =Submerged ca rgo i t s AB)=72 x 6.25)95 = 450 95)Tota l volume i t s AB) =7200 new AB) =

303 7 5042750

346500

m 4 .

4m New AB = 346500 7 7200 = 48.125 metres.In th i s case , new AF a l s o = 48.125 m

l t e rnat iveTo f ind new AF, moments of area about A.Rec tang le i t s AF) =90 x 12)45 = 1080 45) = 48600 m3.Submerged ca r go i t s AF) =72 x 95 = 6840 m3.I n t a c t WP a r e a i t s AF) =1152 x new AF . . . . = 55440 m 3.New AF 55440 + 1152 = 48.125 metres.In t h i s case , new AB a l so = 48.125 m

l t e rnat ive 3To f ind new AF, moments of area about A.Origina l WP a rea i t s AF) =1200 x 50 . .. .. . . . . . . . . . . . = 60000 m 3.Area l o s t i t s AF) =48 x 95 . .. . . .. . .. . . . . . . = 4560 m 3.I n t a c t WP a rea i t s AF) =1152 x new AF .. . .. .. . = 55440 m 3


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8New AF = 55440 . 1152 = 48.125 metres .In t h i s case new AB a l so = 48.125 m

BG = AG - new AB = 50 - 48.125 = 1.875 mSince the COG i s fwd of the new COB, thet r im caused wi l l be by the head.TM W.SG 7380(1.875) = 13837.5 tm.Tc = TM MCTC However the MCTC in thebi lged condi t ion has ~ be ca lcu la ted .MCTC = W GML 7 lOOL or W BML 7 lOOL.Since KG i s not given, BML has to beused i n s t ead of GML.Note: In t h i s formula, L represen ts LBP.

BML = I*COF of i n t a c t WP area 7 V90 mI

4 5 m I F< I48 .125 m I I 46 .875 mI I

I*COF i n t a c t W area = I*COF rec t angle +I *COF cargo areaI*COF rec t angle =

12 ( 90 3 ) + ( 12) 90 ( 3 . 1 25 2 )12= 729000 + 10546.875 = 739546.875 m4

9I*COF cargo =

9.295(7.746 3 ) + 72(46.875 2)12= 360 + 158203.125 = 158563.125 m4

I*COF i n t a c t WP area =739546.875 + 158563.125 898110 m4

BML = 898110 7200 = 124.738 metres .MCTC = ( 7 380 x .124. 7 38) (100 x 100)

MCTC = 92.057 tm.Tc TM MCTC = 13837.5 92. 0 57Tc = 150.3 cm = 1.503 m by the head.Ta = AF(Tc)

L= 48.125(1 .503)100 = 0.723 m.

Tf = Tc - Ta = 1.503 - 0. 7 23 = 0.780 mFwd AftFina l hydra f t 6.250 m 6. 250 mTf or Ta +o. 780 -0 . 7 23Fina l d r a f t s 7. 030 m 5. 5 27 m

xampleA box-shaped vesse l 150 m long and 20 mwide f l oa t s a t an even keel d ra f t of 8 min SW. The af te rmos t compartment, 15 mlong 20 m wide, has a wa te r - t ig h t f l a t1 m above the kee l separa t ing the LHthe DB t ank . Calcu la t e tht d ra f t s fwd


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10Plan view a t water l ine

I 5 m 1 3 5 m tB F : F I B 1 20 m11

1I I11A

Side e leva t ion.

sd

v = 150 x 20 x 8 =w = 24000(1.025) 24000 m3 ) unchanged24600 t ) by bi lg ing .Before bi lg ing AG = AB = AF = 75 metres.After bi lg ing AF = 75 m but new AB >75 mS = volume of l os t buoyancy =15 x 20 x li n t ac t water-plane area 150 x 20

S = 0.100 metre.New hydraf t = 8.000 + 0.100 = 8.100 m.To f ind new AB, moments of vol about A:Orig ina l vol i t s AB) =24000 x 75 = 1,800,000 m4 Volume l o s t i t s AB) =300 x 7.5 = - 2 , 2 50 m

11Balance a f t e r l o s s = 1,797,750 m 4 Volume rega ined i t s AB) =300 x 75 . . + 22,500 m4 New volume (new AB) =24000 x new AB . . = 1,820,250 m4New AB = 1,820,250 24000 = 75.844 m.BG = new AB - AG = 75.844 - 75 = 0.844 mSince AG < new AB, Tc wi l l be by s t e rn .TM = W.BG = 24600 x 0.844 = 20762.4 tm.MCTC = W.GML 7 lOOL or W.BML 7 lOOLSince KG i s not given, BML may be used.BML = I*COF i n t a c t WP area = 20(150 3)Volume 12(24000)

BM = 2 3 4 3 7 5 mMCTC = 24600(234.375) =

100 x 150384.375 tm.

It i s obvious from the foregoing c a l cu l a t i on t ha t , in t h i s case , BML MCTCa re unchanged by b i lg ing .Tc = __'. '.kL = 20762.4 = 54.0 cm = 0.540 m.

MCTC 384.375Ta = AF(Tc)

L= 75 0 .54) =

1500 . 270 metre.

Tf = Tc - Ta = 0.540 - 0.270 = 2 7 m.


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Final hydraf tTf or TaFinal dra f t s

Example 4

12Fwd8.100 m-0 . 2707 . 830 m

Aft.8.100 m+0. 2708.370 m

I f in example 3, the b i lg ing occurred inthe LH fu l l of general cargo: p = 60 )and not in the DB tank, f ind the newdra f t s fwd and a f t .Plan view a t wate r l ine

F &

1 1s mSide e levat ion

1 m

13g 1m - - - - - - - - - - - - - -8 F 1I I II II I II I II I I

t II I II I II

V = 150 x 20 x 8 = 24000 m 3) n c h a n g ~ dW = 24000(1.025) = 24600 t ) by bi lg ingBefore bi lg ing , AG = AB = AF = 75 metresAfter bi lg ing , AB>75 m, AF>75 m, but ABand Af a re NOT equal .

S = volume of l o s t buoyancyin tac t water -p lane area

13s = 15 x 20 x 7(60/100) = 0.447 m(150 x 20) - 15 x 20(60/100)New hydra f t = 8.000 + 0.447 = 8.447 m.le => 1 J100 - f 15j0: 4 = 9 .4 87 m.100be = b J100 - E = 2q /0 :4 = 12.649 m.100Area of cargo = 9.487 x 12.649 = 120 m2 To f ind new AF, moments of area about A:Cargo area i t s AF) =

120 x 7 5 . . . . . . . .= 900 m3.Rectangle i t s AF) =135 x 20 x 82 . 5 . = 224,750 m3I n tac t WP area i t s AF) =2 ( F) 223,650 m328 0 new A . . =New AF = 223,650 7 2820 = 79.309 m.To f ind new AB, moments of vol about A:Rectangle i t s AB) =135(20)8 .447(82.5) = 1,881,569.25 m4 Submerged cargo i t s AB =

120 x 7 447 x 7 5 = 6702.3 m4 DBT its AB =

15 x 20 x x 7 5 =Tota l vol i t s AB) =24000(new AB =

2250.00 m4 41,890,521.55 m


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14New B = 1,890,521.55 24000 = 78.772 mBG= Qew AB - AG= 78 . 772 - 75 = 3.772 mSince G < new AB Tc wil l be by s te rn .TM= W.BG = 24600 x 3.772 = 92791.2 tm.Tc = 7 MCTC and MCTC = W.BML lOOL.BML = I*COF i n t a c t water-plane 7 VolumeI*COF WP = I*COF rec tangle + I*COF cargo

i20 m

JI*COF r ec t = 20 135 3 + 20)135 3.191 212 = 4128117.698 m4.I*COF cargo area =12.649 9.4873) + 120 71.8092)12 = 619683 937 m4.

I*COF l n t ac t WP area = 4747801.635 m4BML = 4747801.635 + 2 4 ~ = 19 7 . 825 m.MCTC 24600 x 197.825 = 324.433 tm.100 x .150Tc = ,22791.2 = 286.0 cm = 2.860 metres.

3 ~ 4 4 3 3-Ta = _M(Tc )L = 79 .309 2 .86 ) 1 .512 m150

15Tf = Tc - Ta = 2.860 - 1.512 = 1.348 m.AftFina l hydra f tTf or TaFinal dr a f t s

Fwd8.447 m- 1 . 487.0998.447 m+l .512m . 9.959 m

I f in example 4 , the DB tank a lso wasbi lged , f ind the new dra f t s fwd and a f t .Plan view a t wate r l ine

iG B F . B l 20 m1 I1LL L L LLJ = r

Side e l eva t i on II III 1I II

15 m - - - - - - - - - - 135 m - - - - - - - - - - - -- -

sd

v = 150 x 20 x 8 = 24000 m 3w = 24000 1.025) = 24600 t Unchangedby b i lg ingBefore b i l g i ng , AG = AB = AF = 75 metresAf te r b i l g i ng , AB>75 m AF>75 m bu t ABand AF a re NOT equal .

s = volume of l o s t buoyancyi n t a c t water -p lane area


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16

S = 15(20)1 + 15( 20)7(60/100) = 0 .553 m.150(20) - (15)20(60/100)New hydraf t = 8.000 + 0.553 = 8.553 m.The fol lowing pa r t i cu l a r s of the vesse l ,a l r eady ca lcu la ted in example 4, areappl icab le here without any change:AF = 79.309 m, BML = 197.825 m, MCTC =324.433 tm, le = 9 .4 8 7 m, be = 12.649 m,cargo area = 120 m2 To f ind new AB, moments of vo l ab ou t A:Rectangle i t s AB =135(20)8.553(82.5) = 1 ,90 5,180.75 m4 Submerged cargo i t s AB =1.20(7.553)7.5 = 6,797.70 m Tota l volume (new AB =

24000(new AB) = 1,911,978.45 m4 New AB = 1,911,978.45 7 24000 = 79.666 mBG = new AB - AG = 79.666 - 75 = 4.666 mSince AG


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18wide i s f l oa t i ng in SW a t an evenke e l d r a f t of 6 .6 m. The fo rward-mos tcompartment i s 18 m long m wideand has a DB tank 1 m high. The LH ,f u l l o f ca rgo , ge t s b i lged . Ca lc u l a t et he f i n a l d r a f t s fwd a f t , assumingp = 36 .

6 In example 5, if the empty DB tanka l s o ge t s b i lged , f ind the new dr a f t sfwd and a f t .7 A box-shaped ves se l 200 m long 20 mbroad i s a f l o a t in SW a t an even kee ld r a f t o f 8 m. There i s a DB tank a t

the fo rward end , 18 m long, 20 m wideand 1 .4 m high, which has HFO of RD0.95 to a sounding of 0 . 5 m. Find thenew dra f t s fwd and a f t if th i s tankge t s b i lged .1Hin t : Bi lg ing means t h a t sea wate rcan f r e e l y go in and out of the compar tment and so can HFO. The HFO i sNOT, t h e r e fo r e , i n the sh ip anymore .So f i ~ the new w the new G and thenew hydra f t assuming t ha t the HFO i spumped ou t . Then b i lge the t ank, f indt he new even kee l AB, as i n p a s t que s t i ons , comple t e the c a l c u l a t i o n . )

8 A box - s haped tank v e s s e l , 200 m l o ngand 20 m wide , i s a f l o a t in SW a t aneven kee l d r a ft o f 6 . 2 m. The a f t e r mos t t a nk s (No: 8 P C S ) a re ea ch 1 6m lo ng . Th e wi n s t anks a r e 6 m b r oadeac h and the . c e n t r e t a nk i s 8 m. b road . All t h r e e ' t anks a r e emp ty .Find the new d r a f t s fwd a f t if bo ththe wing tanks - No:8 P and No:8 S -a re now b i lged .

199 A box-shaped t ank v e s s e l , 160 m longand 18 m wide , i s a f l o a t in SW a t aneven kee l d r a f t of 9 m. The fo rwa rdmost t anks Nos: l P C S) a re each 20m long. The wing tanks a re 5 m broad

each whi le the cen t re tank i s 8 mbroad. No:l c, which had SW b a l l a s tto a sounding of 15 m, ge t s b i lged .Find the new d r a f t s fwd and a f t .H in t : Find the new w the new AG ,t he new hydra f t assuming t h a t No : 1 Ci s d ischarged . Then b i lge it , f indthe new even kee l AB, as done in pa s tq u e s t i o n s , and comple te thec a l cu l a ti on) .

10 A box-shaped v e s s e l , lOOm long 12 mwide , f l o a t s a t an even ke e l d r a f t of6 m i n SW . The forward-most compa r t m e n t , 10 m long 12 m wide; fu l lo f genera l c a rgo , ge t s b i lged . I f thed r a f t s a re t hen obse rved to be 7 . 03 mfwd 5 . 53 m ft , c a l c u l a t e theapprox ima te permeab i l i t y of thecompar tmen t .

-oOo-


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20CHAPTER 34

BILGING OF AN \IINTERMEDIATE C O ~ P A R T M E N T

Having s t ud i ed the e f f ec t s of bi lg ingan amidsh ips compartment chap te r 32)and an end compartment chap te r 33) , theca l cu l a t i ons i nvo lv ing the bi lg{ng of anin te rmedia te compartment should notpre s en t any d i f f i cu l t y to the tudent .Hence not many worked examples are givenhere .

The ca l cu l a t i ons in e a r l i e r chapterss t a r t e d with the vesse l on an even keel .In t h i s chapte r , bi lg ing of a compar tmen t has been considered a l s o whenthe vesse l has an i n i t i a l t r im. Thisprovides an i n t e r e s t i ng aspec t of s tudy .Example 1A box-shaped vesse l 100 m long and 12 mwide f l oa t s a t an even keel d ra f t of 6 min SW. No:2 hold , 10 m long 12 mbroad , i s empty. The forward bulkhead oft i s hold i s 10 m from the forward endof ~ h e sh ip . Find the d r a f t s fwd and a f tif t h i s hold i s bi lged .

v 100 ' x 12 x 6w 7200 x l . 025 7 200 mJ ) hUnc anged by= 7380 t ) b i l g i ng .Before b i l g i ng , AB = AG = AF = 50 metresAfte r b i l g i ng , AB = AF = < 50 metres .

21Plan view a t wate r l i ne

ASide e l eva t i on

s = volume of l o s t ~ u o y a n c yi n t a c t wate r -p l ane areas lbdLB-lb = 10 x 12 x 61200 - 120 720 o ~ 6 6 7 m1080Fina l hydra f t 6.000 + . 0.667 6.667To f ind new AF, moments o f a rea about ALarge r e c t i t s AF) =80 x 12 x 40 = 38,400 m3.Smal l r e c t i t s AF)10 x 12 x 95 = 11,400 m 3.I n t a c t WP i t s AF) =1080 (new AF) 49,800 m3.


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22In t h i s case , new AB a l so = 46.111 m.BG = AG - new AB = SC - 46.111 = 3.889 mSince AG > new AB, Tc wi l l be by head.TM = W BG = 7380 x 3.889 = 28,700 tm.Tc = T = MCTC and MCTC = W BML lOOL.BML = I*COF i n t a c t WP 7 Volume of d i sp l .I*COF i n t ac t WP a rea =

=*COF la rge r e c t + I*COF smal l r ec t .Plan view

1 0 m

12 m

~ . : . . ~ ~ l L . i l . . _ _ J lI*COF l a rge r ec t =

12(803 )+(12)80(6.1112 )=547,850 .548 m4.12I*COF smal l r ec t =l?.(10 3)+120(48.8892) =12 287,816 .119 m4.

I*COF i n t ~ t WP 835 6 4 = 66.667 m BML =835,666.667 7 7200= 116.065 metres .MCTC = W BML = 7380(116.065) = 85.656 tmlOOL 100 x 100c = -1'. : _ =

MCTC 28,700 =85.656 335. l cm - 3.351 m

Ta = AF(Tc)L

Tf = Tc Ta

23= 46 .111 3 .351)100= 3.351 1 .545

Fwd

==

AftNew hydr a f t n.667 m 6 .667Tf o r Ta +l .806 -1 .545New d r a f t s 8 .473 m 5.122

Example 2

1 .545 m.

1.806 m.

mm

I f in example 1 , the b i l ge d compartmenthad pe r m e a b i l i t y = 65 , f i nd the d r a f t sfwd and a f t .V = 100 x 12 x 6 = 7200 m3) Unchanged byW = 7200 x 1.025 = 7380 t ) b i l g i ng .Before b i lg ing , AB = AG = AF = 50 metres

S = volume of l o s t buoyancyi n t a c t wate r -p lane a reas = 10 12 )6 \65 /100)1200 - 120(65/100) = 468 = 0.417 m.1122New hydr a f t = 6 .000 + 0.417 = 6.417 m.l e = ljl - e = l oJ(5":"'35 = 5.916 met res .100lb = bf ioo - I = 12JO:Ts = 7 .099 met res .100Area of cargo = 5.916 x 7.099 = 42 m 2


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f12 m

~ - - - _ . . ; . . _ ~ . . . L . . L . . L . + - - - - f J .- - - - - - - - - -8 0 m - - - - - - - - - 10 m 10 mSide e leva t ion

Large re c t i t s AF) =80 x 12 x 40 =Small r e c t i t s AF) =10 x 12 x 95 =38,400 m3 11,400 m3 Cargo area i t s AF) =

42 x 85 = 3 , 570 m3 I n t a c t WP i t s AF) =1122 new AF) . . = 53 ,370 m3 New AF = 53 ,370 + 1122 =In t h i s case, new AB a l so47 .567 m ~ t r e s= 47 567 mBG = AG - new AB = 50Tc = TM

MCTCand MCTC =

47.567 = 2.433 mW GMLlOOL or W BMLlOOL

BML

2 5I *COF i n t a c t WP =I *COF smal l r ec t I* CO F la r ge re c t ++ I*COF cargo a re a .P l an view a t water l ine

i12 ml- - - - - 4 7 5 6 7 . m - - -:>1 ; - - 4 7 4 3 3 m - 5 r i

I*COF l a rge r ec t1 2 8 0 3)+ 12 )80 7 .567 2 ) = 566 , 969 . 109 m 4

12I*COF small r e c t =

12 103) + 120 47 . 433 2 ) = 270,986 . 739 m 412I*COF cargo area =7 . 0 9 9 5 . 9 1 6 3)+42 37 .433 2 )=58 , 974.12812

I*COF i n t ac-t WP . . . . . 896 , 929 . 976m 4

m 4BML 896929 .9?6 . 7 200 = 124.57 4 m e t r e'; 'MCTC = 7380 x 124 . 574 = 91 .936 t m100 x 100Tc = TM = 7380 2 . 433) = 195 . 3 cm = 1 . 95 3 mMcTC 91 . 936Ta = AF Tc)L = 47 .567 x 1 .95 3100 = 0 . 929 m.


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New hydra f tTf or TaNew dra.f t s

Example - 3

26Fwd6.417 m+l . 0 247.441 m

Aft6.417 m-0 .9295.488 m

A box-shaped vesse l 180 m long and 20 mwide f loa t s in W a t a dra f t of 6.2 mfwd 8.2 m a f t . No:3 LH, beginning 40 mfrom the f o r w ~ r d end i s 20 m long andempty. Find the new dra f t s fwd and a f ti f th i s hold i s bi lged .Plan v ~ w a t water l ine

t20 m

- - - - - - - - - 120 m 20 m 40 mSide e l eva t i on

. . 0 . - - ----f 1. 0 m7. 2 m 7 . 2 m 6. 2'

m

v = 180 x 20 x 7.2 = 25920 m3 ) UnchangedW = 25920 x 1.025 = 26568 t by bi lg ingThe AG of the s h ip i s not known d i r ec t lyso i s to it

27Trim in cm= W.BG

MCTCor BG = MCTC x t r imw

BG MCTC x 200 = W.SML x 200 = BMLw lOOL W 90

= I*COF = 2 ~ 1 8 = 4.167 metresV 90) 12 25920)90Since t r im i s by s te rn AG < even keel ABOn even keel , AB wou ld be 90 metres .So AG of sh ip = 90 - 4.167 = 85.833 m.

Even keel cond i t i6nAG = 90 metresAB = 90 .me tr-esTrim = n i l

/iven cond i t ion Bilged cond i t ion

AG = 8 5. 8 33 m AG = 85.833 mAB = 90.000 m AB < 90 m due toV/ l t r ims un t i l b i lg ing . So t r imnew AB = AG has to be found.

Assuming even keel condi t ion :s = volume of l o s t buoyancyi n t a c t wate r -p lane a r ea

s = 20 x 20 x 7.2 =180 20) -20 20) 28803200 0 .9 metre .New hydraft = 7 200 0 900 = 8 100 mIn t h i s case , new AB new AF are equa l .


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2812 0 (20)60 + 40(20)160 = 3 200(new AF )New AF = 85 m. New AB a lso = 85 metres.New BG = AG - AB = 85.833 - 85 0 . 833 mSince AG > new AB, Tc i s by the head.TM = W BG 26568 x o.833 = 22131.144 tmTc = TM and

MCTCMCTC W GMLlOOL o r W BMLlOOL

BML = I*COF i n t a c t WP Volume of disp l .Plan view

- - - 75 m _ _ _ _ .

f20 m~ _ _ _ . . _ _ _ _ , _ _ _ _ ~ - f - - - . - - 1 J120 m 20 m 40 m

I*COF i n t a c t WP =I*CCF la rge r ec t + I*COF small r ec t .

I *COF la rge re c t =- 20(120 3)+ (20)120(25 2)=4380,000.000 m4.12I*COF small r e c t =20(40 3)+ ( 20)40(752 ) = 4 606,666. 667 m4.12I*COF in tac t WP = 8986 , 666 . 667 m 4

BML = 898 6 , 666. 667 . 25 , 9 20 = 346 . 708 m

29MCTC 26568 x 346.70 8100 x 18 0 511.741 tm.

Tc = TM = 22131.144 = 43 .2 cmMCTC 511.741 0.432 m.

Ta = Tc(AF)L

= 0.432(85)180

= 0.204 metre .

Tf =Tc - Ta = 0 .432 - 0 .2 04 = 0 . 2 28 m.Fina l hydra f tTf o r TaFina l dra f t s

Fwd8.100 m+0.,2288 . 3 28 m

Aft8 .100 m-0 .2047.896 m

Example 4I f in example 3, the permeab i l i ty of thehold was 60 , f ind the new d r a f t s fwda f t a f t e r b i l g i ng .The fo l lowing pa r t i c u l a r s a l ready ca l cu l a t ed in example 3, are unchanged byb i l g i ng : - V = 25,920 m3, W = 26,568 tAG = 85.833 metres .

s =

S = volume of l o s t buoyancyi n t a c t wate r -p lane a reaAssuming even keel cond i t ion :20 x 20 x 7 .2 x 60/100180(20) - (20)20(60/100) = 0 . 514 m.

New hydra f t = 7 0 . 514 = 7 . 714 m.In t h i s case new AF new AB a re equa l .To f i nd ne w AF, mome n ts of area a bout A:

30


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Plan view

---- --- ---- 120 m - - - - - - - - - 20 m 40 mj100 - 20,j0:4e = 1 12 = = /12.649 m.100

be b j100 - 12 = 2oj0:4 = 12.649 m.100Area of cargo = 20 x x 0.4 = 160 m2.In t ac t WP area i t s AF = 120 x 20)60 +40 x 20)160 + 160 x 130) = 292,800 m3 .New AF =292,800= 87.143 m a l so = new A .3,360New BG = New AB AG = 87.143 85.833

New BG = 1.310 m.Since AG < new AB, Tc w i l l be by s te rn .TM = W.BG = 26568 x 1.310 = 34,804.08 tmTc = - lli.. ci'na MCTC =MCTCBML = I*COF i n t ac t WPI*COF i n t a c t W area =+ I*COF small r ec t +

W.GML o r W.BMLlOOL lOOLVolume of d i s p l .I*COF l a rge r ec tI *COF cargo area

=

31Plan view

f 72 . 857 1

I*COF l a rge r ec t = [20 120 3 )-:- 12] +20)120 27.143 2 . = 4,648 ,181 .877 m 4 I*COF smai l r ec t = [20 40 3 12] +20)40 72 .857 2 = 4,353 ,180 .625 m4 I*COF c a r g o = [12 .649 12 .649 3 : 12]+

160 42 . 857 2 = 296,008.850 m4 I* COF in ta c t WP area = 9,297 ,371 .35 2 m4 BML = 9 297 371.352 7 25,920 = 358.695 mMCTC

Tc

Ta =

Tf =

26568 x 358.695 = 529.434 tm.100 x 180TM 34804.08 = 65.7 cm o. 6 57

MCTC 529.434F { T c ~ = 87.143{0.657) = 0.318L 180

Tc - Ta = 0.657 0.318 = 0 . 339

Fina l hydra f tTf or TaFina l dra f t s

Fwd7.714 m-0 .3397 . 375 m

Aft7.714 m+0 . 3188.03 2 m

m

m.

m.


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32Exe rc i se 32

Bi l ging an i ~ t e r m e d i a t e c o mp a r tment1 A bo x - shaped ves se l 180 m l ong & 24 mwi de f l oa t s i n W a t dra f ts o f 9 .5 mf wd & 8. 5 m a f t . An emp t y c o mp ar tment2 2 m l o ng 24 m wid e , who se a f t e rbu l k he a d i s 50 m f ro m the a f t e r endo f t he ve s s e l , g e t s b i l ged . Ca l cu l a t eth e ne w d ra f t s f wd and a f t .

2 A bo x - s haped ve s se l 200 m long & 20 mwid e i s a f l o a t in SW a t d r a f t s of 6 mfwd & 8 m a f t . No: 2 LH, 24 m lon g &20 m wi de, has p = 70 . I t s fwd bu lk head i s 30 m f r om the f w ~ end of theve s s e l . Find the new d ra :ts fwd & a f ti f t h i s LH ge t s b i lged .

3 A box - shaped 080 s h i p 160 m l o ng and18 m wi de i s in W drawing 6.4 m fwda nd 7. 2 m a f t . No : 2 ho l d , on thecen t re l i ne o f the sh ip , i s 5 m longand 1 2 m wide. Its f o r ward bulkhea di s 40 m f ro m the forwa r d end o f thes hip . Find the new dr a f t s if t h i sho ld , hav i ng p = 80 , ge t s b i l ged .4 I f t he s h i p i n q u e s t i o n 3 had a DBtan k 5 m long, 18 m wi de & 1 . 6 mh igh , b elow No 2 ho l d , c a l c u l a t e th e

new d r a f t s i f t he b i l g in g oc c ur re donly in the ho l d and no t in t he Dt ank. Note: Hold s on ly 1 2 m broad.

5 I f in ques t ion 4 , t he D tank a l s og o t b i lged , a long wi th the hold , f i ndthe new d r a f t s fwd and a f t .-o O o -

3 3P I 3 5

BI LGING OF AS I DE C O M P R T M ~ N T

Having s t u d i e d the e f f e c t s o f b i l g i n gan amidships compartment chap te r 32 ) ,a n end compartment chap te r 33 and anin t e r m e d i a t e compartment c h a p t e r 3 4 ) ,c a l c u l a t i o n s invo lv ing b i l g i ng o f a s i d ec ompartment i s j u s t one more s t ep .

In a dd i t i on to s inkage , change of KB,pos s i b l e change o f _ M and _poss ib l ec hange of t r im, b i l g i ng u f a s i d e compar tmen t causes list.

Before b i l g i n g , the CO and COG oft he s h i p were i n a v e r t i c a l l i r ie .Because o f b i l g i ng a s i d e compar tmen t ,t he CO of t he s h i p s h i f t s away from t heb i l g e d compartment and towards the_ v o l ume of buoyancy r ega ined . The C03 i s nowt r a n s v e r s e l y separa t ed from the COG b yt he d i s t a n c e BG. This G d iv ided by thenew GM f l u i d g ives the tan of the a n ? l eo f list, a s i l ~ u s t r a t e d in t he fo l l ow ingf i gu r e where in :W i s the w a t e r l i n e befo re b i l g i ng .B i s t he CO befo re b i l g i ng .G i s t he COG of the s h i p - i t s pos i t i oni s una f f e c t e d by b i l g i ng .Dia g o na l l y shaded a re a i s the volumel os t buoyancy and whose geometr ic

c e n t r e i s i n d i c a t e d by ' b .e r t i c a l l y shaded a re a i s t he v o l u ~ e ofb uoyancy r ega i ne d by pa r a l l e l s inka ge


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34and whose geomet r ic cen t re i s i nd i cated by b l in the f igure .WlLl i s the new wate r l i ne a f t e r bi lg ingand pa ra l l e l s inkage.

w1w

M1

8 1 '

K E E L

L2

L1L

B1 i s the COB a f t e r bi lg ing and pa ra l l e ls inkage. N o t e B ~ i s pa ra l l e l to b ~GN i s the t r ansve rse sepa ra t ion betweenand G a f t e r bi lg ing and para l le l

sinkage - the moment so formed causesthe sh ip to heel over to the s ide onwhich the b i lg ing occur red .b3 b4 a re the geomet r ic cen t re s of theemerged and immersed wedges .Bz i s the f i na l pos i t i on of COB.Note: B1-B2 i s pa ra l l e l to b3-b4 and

~ i s ve r t i c a l l y below G.W 2 ~ i s the f i na l wate r l ine .M1 i s th new metacentre a f t e r bi lg ing .NM1 i s the new metacentr ic heLght.0 i s the angle of l st due to bi lg ing .

35Fr om the f igure , t i s apparent t ha t :Tan - = NG NM1 NG i s gene ra l ly knownas BG and NM1 i s the new in i t i a l GM. Theformula then becomes Tan B = BG ; New GM

Calcu la t ion of l st ~ u to b i lg ing as ide compartment can be i l l u s t r a t e d bythe worked examples tha t fol low:Example 1A vesse l 200 m long 20 m wide i s boxshaped and a f lo a t in SW a t an even keeld ra f t of 8 m. A DB tank on th e s t a rboa rds i de i s r ec tangula r , 12 m long, 10 mwid e , 1 .2 m deep empty. Calcula te thel st i f t h i s tank i s now bi lged , givent ha t KG = 7.5 rn and FSM = 820 tm.v = 200 x 20 x 8w = 32000 x 1.0 25

c = volumei n t ac ts = 12 x 10 x 1 . 2200 x 20

= 32000 m 3) Unchanged= 32800 t ) by bi lg ingof l o s t buoyancywater-p lane area

= 1444000 = 0.036 metre.New hydra f t = 8.000 + 0.036 = 8.036 m.o f ind new KB, moments about keel :

32000 new KB) =32000 ( 4) - 144 0 .6 ) + 144 8.018)New KB = 4.033 metres .

Al terna t ive method - see f igure next pageBB1t = x v = 7.418 144) = 0.033 metre.v 32000


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36New KB Old KB BB1t = 4 . 0 33 metres.

- - - - - - - - - - 2 0 m- - - . - ->

L l

x

BMT = I*COF i n t a c t water -p lane volume= LB 3 = 200 ( 20 3 ) = 4.167 metres.l2V 12(32000)

KMT = New KB BMT = 4.033 4.167 = 8.2M = KM - KG = 8.2 - 7.5 = 0.700 metres

FSC = 820 32800 = 0. 0 25 met re .M f l u i d or GMF = 0.675 metres

BB1 yv = 5(144) = 0.0225 metre.v 32000Note: BB 1 i s the same as BG.Tan B = BG GMF =0.0225 : 0. 67 5= 0.0333

t = 1.91 o r lo 5 5 to s ta rboa rd.

37Example 2A box-shaped t anke r 180 m long and 16 mwide i s a f l o a t i n SW a t an even kee ld r a f t of 7.5 m. The wing tanks a re each4 m broad and the cent re tan ks, 8 m. TheKG i s 5 .5 m and FSM 1200 tm . Calcu la t et he list if No:3 s ta rboa rd tank , whichi s 15 m long and empty, i s now bi lged .v = 180 x 16 x 7.5w = 21600 x 1.025 = 21600 m 3 ) Unchanged= 22 140 t by b i lg fng

s = volume of l o s t b u o y ~ c yi n t a c t wate r -p lane a reas = 15 x 4 x 7.5 =2880 - 60 450 =2820 0.160 metre.

d ft 7 5 0 160 = 7.660 met res .ew hy ra = . .M1

1 - - 4 I

ww L

B1 2 m


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yx

Port side

Bl & Fl - '- - - - _,. - ;:: - -B F , ' 0.128 ; -_t_1_4 m s m

'-After bi lg ing & p a ra l l e l s inkage ,area of the water-plane i s cons tan ta l l d ra f t s . So F1 and Bi wi l l be insame ver t i ca l l i ne and:

yx

thea tthe

New KB = new hydraf t 2 = 3.830 metres .To f ind FF , the t ransverse sepa ra t ionbetween F and Fl , moments of area aboutthe s ta rboard s ide :Tota l area ( i t s d i s t ) area l o s t ( i t sdis tance) = balance area ( i t s dis tance) .2880 x 8 - 60 x 2 = 2820( i t s dis tance)Dist of Fl from s ta rboard s ide = 8.128 m

FF = 0.128 metres.Al t e rn a t iv e method

FF1 = ax =A

60 ( 6) =2820

0 .128 m

In th i s case, FF1 = BB .

39I*XX whole water -p lane= 180(16 3 ) = 61440 m 4.12I*XX l o s t area= 15(4 3 ) + 60 ( 6 2 ) = 2240 m4.12I*XX i n t a c t water -p lane . . = 59200 m 4.I*YY i n t ac t water -p laneI*XX i n t a c t water - r lane area - Ay2= 59200 - 2820(0.128 )= 59153.797 m4Note: Fl i s the cen t ro id of i n t a c t area .Hence the minus s ign in the formula.BMT = I*YY i n t a c t water -p lane volume

= 59153.797 + 21600 = 2.739 metres .KMT = K B + BMT = 3.830 + 2.739 = 6.569 mGM = KM KG = 6.569 5.5 = 1.069 mFSC = FSM w = 1200 22140 = 0.054 mGM f l u id or GMF = 1.015 m

Tan = BB1 = 0.128 = 0.12611GMF 1.015= 7.188 o r 7 11 ' t o s ta rboard .

Example 3A box-shaped vesse l , 150 m long and 17 mwide, i s a f l o a t in FW a t an even keeld r a f t of 8 m. A rec tangula r compartmenton the s ta rboard s ide i s 16 m long and8 .5 m wide. I f KG = 6.0 m and FSM = 900t m, and permeabi l i ty i s 60 f ind the


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40V 150 x 17 x 8 = 20400 m3)w = 20400 x 1.000 = 20400 t ) Unchangedby b i lg ing

S = volume of l o s t buoyancyi n t a c t wate r -p lane a re as = 16 x 8 .5 x 8 x 60/100

150(17) - 16 8 .5 )60 /100= 652.8

2550-81.6S = 652.8 = 0.265 m2468.4

e ~ hydra f t = 8 + 0 ..265 = 8.265 metres .Length of ca rgo = ' l e =

1 j 1 0 0 - p = 16/0. 4 = 10.119 metres .100

Breadth of cargo = be =b j100. -p = 8 .5 /Q .4 = 5.379 metres .' 100

Area o f cargo = 16 x 8 .5 x 0 .4 = 54.4 m2

Plan view a t w a t ~ ~ l i n eP0 r t s i de

yx

8. 5 m

S t a r boa r d s ide

yx

41A f t e r b i l g i ng p a r a l l e l s inkage , thea re a of the wate r -p l ane i s c o n s t a n t a tll d r a f t s . So F1 Bi w i l l be in thesame v e r t i c a l l i n e and the new K w i l lbe h a l f the new d r a f t . New K = 4.132 m.

To f i n d FF1, the t r a ns ve r s e s e pa r a t i onbetween F Fl , moments abou t s t bd s i de :Tota l a rea i t s d i s t ) - a re a l o s t i t sd i s t a nc e ) + c a rgo a re a i t s dis tance ) =ba lance a re a i t s d i s t a n c e ) .2550 8 .5 ) 136 4 .25 ) + 54 . 4 4 .25 ) =246 8 . 4 i t s d i s t a nc e from s t a r boa r d s i de )D i s t of Fl from s t bd s i de = 8 . 640 metres

FF1 = 8.640 - 8.500 = 0.140 metre .A l t e r na t i ve methodFFl = ax = 81 .6 4 .25 ) = 0.140 metre .A 2468.4

In t h i s case FF1 = BBl.I*XX whole wate rp lane= 150(173) - 61412.500 m4 12I*XX compartment a re a= 1 6 8 ~ 2 J + 136 4 . 25 2 = -327 5 .333 m4 1 2I*XX cargo a r e a= 10 .119 5 .379 3 ) + 54 .4 4 .25 2 )12

= +1113.838 m4 I*XX i n t a c t wate r -p l ane = 59251.005 m4

42


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I*YY i n t a c t wate r -p lane= I*XX i n t a c t wate r -p lane - Ay2.= 59251.005 2468. 40 (0 . 140 2 )= 59202. 624 m4.

BMT = I*YY i n t a c t water-plane volume= 59202.624 20400 = 2.902 metres .New KM = new KB new BM = 7.034 metres .GM = KM - KG = 7.034 6.000 1.034= m.FSC = FSM w = 900 20400 0.044 m.GM f lu id or GMF = 0.990

l

2

. . . . . . . . . .Tan = BB1 = 0.141414GMFB = 8.049 or 8 03 to s ta rboard .

Exerc ise 33Bilging a s ide compartment

m.

A . b o x ~ h a p e d vesse l 8 ~ m long 20 mwide i s a f l o a t in SW a t an even keelc r a f t of 7 m. A r ec t angu l a r DB tanko ~ the por t s id e i s 16 m long , 10 mwide and l m high i s empty. Calcu la t ethe list i f t h i s DB tank i s bi lged i fthe KG = 7.5 m and FSM = 1000 tm.

A . b o x ~ h a p e d tanker 200 m long 24 mw1je i s a f l o a t in FW a t an even keeld r a f t of 12 m ~ The wing tanks are 6 mbroad and the cen t r e tanks , 12 m. TheKG i s 8 .5 m and the FSM 1800 tm. Find

43 the list if n u m ~ r 4 s ta rboard wingtank, 18 m long empty ge ts b i lged .

3 A box-shaped vesse l 175 m long 18 mwide i s - -a f loa t in SW a t an even kee ld r a f t of 10 m. A rec tangula r compar t ment 15 _m long extends 6 m in breadthfrom the por t s h e l l p l a t i n g . Thiscompartment runs a l l the way from thekee l up to the upper deck and has p =40 . KG 6.8 m FSM 800 tm. Find thelist i f t h i s compartment ge t s b i l ged .

4 A box-shaped vesse l 150 m long 16 mwide f l oa t s in SW a t an even kee ld r a f t of 9 m. It has a l ong i tud ina lw a t e r - t i g h t bulkhead on i t s cen t rel i n e and DB t anks 1 .2 m high. KG i s6.0 m and FSM = 900 tm. A hold 12 mlong, on the p o r t s id e , having p = 30, ge t s b i l ged . Find the list

5 I f in ques t i on 4 the empty por t DBt ank , d i r e c t l y below the hold, a l so

~ t s bi lged i f ind the list6 A box-shaped vesse l 120 x 14 m i s inSW a t an even keel d ra f t of 8 m KG

5.8 m, rs 560 tm. A rec tangula r DBtank on the s tbd s id e , 15 x 7 x 1 m,h a l f f u l l of HFO RD 0.95 , i s b i l ged .Find the list Hint : Af.ter b i lg ing ,the HFO i s not in the sh i p anymore.C al cu l a t e the GG1 to p o r t , the new wthe new KG and new FSM, assuming t h a tthe HFO j s di scharged . Then bi lge thet ank , obta in BB1 to p o r t , compute thef i na l BG and thence the list Asusua l on sh ip s , assume KG of t h i s


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44CHAPTER 36

BILGING . - PRACTICAL .SHIPBOARD CALCULATIONS

Afte r s tudying chapte r s 32, 3 3 ~ 3435, the s t uden t would have a very c l ea ridea of the e f f e c t s on s t a b i l i t y ofbi lg ing a compartment . In t l ,ose chap-t e r s , the s h i p was considered to be box-shaped so t ha t the concept of thepr inc ip l e s involved could be absorbed

w i t o ~ t the d i ve r s i on tha t would haver e su l t ed from having too many var i ab leparamete rs . t would ndt be proper tol eave the s t uden t a t t ha t : withoutor i en t i ng his conceptual knowledgetowards p ra c t i c a l use on board sh ips ,wherein the d e t a i l s given in the hydro-s t a t i c t ab l e would have to be ~ o q i f i e dby the o f f i c e r concerned, befo re he canuse them f6r ca l cu l a t i ons involv ingb i l g i ng .

In t h i s chapte r , use has been made ofAppendix I , of t h i s book, wherein s u i t -ab le ex t r a c t s of the hydros ta t i cp a r t i c u l a r s of an imaginary genera lc argo sh i p m. v.VIJAY have been g i ven .The ca l c u l a t ions have been div ided i n t otwo pa r t s - p a r t A: bi lg ing of a DB t a nkand p a r t B: b i l g i ng of a cargo ho l d .P RT - ilging of a DB t nkExample 1M.v.VIJAY i s a f l o a t in SW a t an e v en

45kee l d r a f t of 4 . 8 m. n empty DB tank21 .5 m longf 20 m wide and 1 m deep ,

w h ~ s e G i s 60 . m, ge ts ' i l ged . Refe r r ingto . appendix I of t h i s book, c a l c u l a t ethe r e v i s e d hydros t a t i c p a r t i c u l a r s forthe b i l ge d cond i t i on .Vol of l o s t buoyancy= 21.5 x 20 x 1 w . . . . 1 - . . - - ~ ~ ~ ~ ~ - r . . . + - L 2- + - L - L . . i . . . L . , _ . _ . . _ _ , _ . _ _ , _ . . . . , . . _ . . ~ ~ ~ H _ 1= 430 cubic met res .efore bi lg ing

D raf t4 . 8 m W in SW9451 tDue to b i l g i ngAfte r b i l g i ng

Ku/w volume vol of d i s r9220.488 m3 9220.488 m+4309650.488 m3 9220.488 m3

Ente r ing hydros t a t i c t ab l e wih new u/wvolume of 9650.488 m3 i . e . W = 9891.8 t )new d r a f t = 5.000 m.Afte r b i l g i ng , though the d r a f t i s 5.000m, volume of d i sp l acemen t = 9 2 ~ 0 . 4 8 8 m3,and W = 9451 tonnes .In t he fo l lo wing t ab le , 1) and 2)i n d i c a t e the p a r t i c u la r s for the L1tac thu l l in sw , as take n from the hydro s t a t i c t ab l e , while 3 ) in d i ca t e s t her ev i s e d pa r t i c ula rs f or theco ndi t i l n .

Dra f t w TPC MCTC B( 1 ) 4. 8 9451 21.97 164.3 72.016( 2 ) 5 . 0 9891 22 . 06 165.7 7 2. 0147 2. 574


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46AF KB KMT KML1 71. 970 2.576 8.828 263.9( 2) 71. 913 2. 68 5 8.686 254.3( 3 ) 71.913 2.781 9.061 266.l

Calcula t ions i n support of ( 3 ) above areas fol lows:Draf t W: The method how the values in3) have been a r r i ved a t has a lreadybeen expla ined.TPC AF: These depend on the area andshape of the water -p lane a t the dra f t a twhich the sh ip i s f loa t ing . In thebi lged condi t ion , the dra f t i s 5 metres,corresponding to the wate r l ine W2 L2 inthe foregoing f igure .New KB: Taking moments about the keel ,

new KB) =ew volumeOld volumei t s KB) + i t s KB) volume l o s tvolume regained i t s KB9220.488 new KB =9220.488 2.576) - 430 0.5) + 430 4.9)

New KB= 2.781 metres.Note: The KB of the volume of buoyancyregained has been taken to be the meanof the two dra f t s 4.8 and 5.0 metres .

BB1t = dvvOR

= 4.9 - 0.5) 4309220.488 = 0.205 m. New KB = Old KB + BB1t = 2.576 + 0.205

47New KB = 2.781 m.

KMT KML: From hydros ta t i c t ab le , for5 m dra f t , i n t a c t hul l :KMT 8.686 mKB 2.685 mBMT 6.001 m

KML 2 5 4 ~ 3 mKB 2.685BML 251. 615

I*CL 57908.186 m4 I*COF 2428023.380 m4.Note: BM= I V or I = BM V). In thei n t ac t condi t ion , a t 5 m d ra f t , volumeof disp lacement= W + 1.025 = 9649.756m3 The value of I i s obtained by mul t i plying M by V.

The above values of the moment ofi n e r t i a , ca l cu l a t ed for 5 m d ra f t , withhul l i n t ac t , hold good in the bi lgedcondi t ion a l so , l::ecause they depend onlyon the water-p lane area - the water l inecorresponding to W2 L2 in the foregoingf igure .In the bi lged condi t ion , the volume ofdisplacement = 9220.488 m3.New BMT = 57908.186 + 9220.488 = 6.280 mNew KMT =new KB+ new BMT = 2.781+6.280

New KMT = 9.061 m.

New BML = 2 4 2 8 0 2 3 3 8 0 ~ 9 2 2 0 4 8 8 = 263.329 mNew KML = new KB+new BML 2.781+263.329

New KML = 266.110 m.


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48MCTC:New MCTC = W new BML) =lOOL 9451 263.329)1 0 0 1 ~ 0 )

New MCTC = 177.766 tm.New AB:

Aw2 \,.-i -11 - - - - - - - - - - 8 - - - - ~ r 7 ,

~ ~ c = J ~ - - - ~ /~ 6 0 m ----71I,..::------ 7 2 0 1 4

Moments o f volume abou t A,New volume new AB)d r a f t ( its AB )9220.488 new AB) =

= U./w volume a t 5 JTlvolume l o s t ( i t s AB)9650.488 72.014) - 430 60)

New AB = 72.S74 m.

Find the d r a f t s , fwd a f t , in example 1 .Sinc e the vesse l was on an even kee l a t4 .8 m d r a f t , AG = AB = 72.016 mb e f o r e b i l g i ng . Afte r b i l g i ng , AB =72.574 m.BG AB - AG = 72.574 - 72.016 = 0.558 mTc w i l l be by the s t e r n because AG < AB.

Tc

Ta

Tf

= W.BG = 9451 0.558)

=

MCTC 177.8Tc 0.297 metre .

AF Tc) 71.913 0 .297)L 140Tc Ta = 0.297 0 . 153

Fi n a l h y d ra f tTa o r TfFi n a l d r a f t s

Fwd5.000 m-0 .144 m4. 8 56 m

Example 3

= 29.7

= 0.153

= 0 . 144Aft5.000 m+0.153 m5.153 m.

cm

m.

m.

I f a t the end of example 1 , KG = 7.6 m,FSM = 1000 tm, and a heavy lift of 50 ti s s h i f t e d 10 m to s t a r boa r d , f ind ther e s u l t a n t listM = KMT K = 9 061 7 600 =

FSC = FSM 7 K = 1000 9451 =M f lu i =

1 . 461 m0 .106 m1 . 355 m

Tan B = dw =W.GM 10 x 50 = 0.039049451 ( l 355)= 2.236 or 2 14 to s ta rboa rd .

Example 4M.v.VIJAY i s a f l o a t in FW a t d r a f t s o f5.6 m fwd and 6.6 m a f t . A DB t ank 22 .mlong , 15 m wide and 1 .2 m deep has AG =100 m. Find the hydr os t a t i c p a r t i c u l a r s ,r e f e r r i n g to appendix I of t h i s book, i ft h i s empty t ank now ge t s b i lged .

50 51


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Fwd 5. 6 m a f t 6.6 mTrim 1.0 m by lhe s t ~ r nMean 6 . 1 m AF 71. 40 l m w2 T n - t - r H ~ . - . ~ ~ . . . . . . > L 2Corr = AF trim) = 0.510 W1 L 1

LHydraf t = 6.600 - 0.510I n i t i a l hydraf t 6.110 m I1) and 2) below have b e ~ n der ivedfrom the hydros t a t i c tab le and are forthe i n t a c t hul l condi t ion :

( l SW( 2 FWAB

Draf t6.1106.110AF

w12371.212069.4KB

TPC22.5021.95KMT

MCTC173.8169.6KML( l 71.948 71.393 3.262 8 . 204 214. l( 2) 71. 948 71. 393 3.262 8.204 214. l

Trim = W.BG o r BG = 100 169.6)MCTC 12069. 4I n i t1a l BG = 1.405 metres .

S ince the t r im i sAG of sh ip = AB

AG of the

Before bi lg ingDue to bi lg ing

by the s t e rn , AG < AB.- BG = 71.948 - 1.405sh ip = 70 . 543 m.u/w volume vol of disp12069.4 m3 12069.4 m3+ 396.0 m3

Enter ing the hydros ta t ic tab le with u/wvolume of 12465.4 m in FW i . e 12465.4x 1.025 = 12777.0 t in SW a t cons tan td r a f t ) , dr a f t = 6.2S9 m e t ~ e sLines 3) and 4) below are der ived fromthe hydros ta t ic t ab le , assuming the hul lto be i n t a c t . Line 5) i s for the bi lgedcondi t ion , for which re levant explana t ions are given subsequent ly .

Draf t w TPC MCTC( 3 ) W 6. 289 12777.0 22.584 17 5.4( 4) FW 6. 289 12465 .4 22.034 171.15 FW 6.289 .12069.4 22. 0 34 183.5AB AF KB KMT KML

3)71.928 71.259 3.355 8.160 209.4( 4 )71. 2 8 71.259 3.355 8 . 160 209.45)71.007 / l . 259 3.446 8.409 216.3Explanat ions for l ine ( 5)

TPC and AF: These depend on the area &shape of the water-plane a t the dr a f t a twhich the sh ip i s f loa t ing . In thebi lged condi t ior i , the dr a f t in FW i s6.289 m - W2 L2 in the foregoing f igure .New KB: Taking moments about the keel ,New volume new KB) = old vol i t s KB) -vol l o s t i t s KB) + vol regained i t s KB)12069.4 new KB)12069.4 3.262) - 396 6 ) + 396 6.2)

New KB = 3.446 metres .Note: KB of the volume of buoyancy

52 53


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regained has been t ake n to be the meanof the two dra f t s 6.110 and 6.289 m.Alte rna t ive method:In ta c t vo lume a t 6.289 m d r a f t i t s KB) -

v o ~ l o s t ( i t s KB) = new vol new KB)12465 . 4 3 .355 ) -396 0 .6 ) = 12069.4 new KB )

New KB = 3.445 metres .KMT and KML: From the hydros t a t i c t ab le ,fo r 6.289 m dra f t , cons ider ing t ~ hul lto be in tac t :

KMT 8.160 mKB 3.355 mBf -:1' 4. 805 m

KML 209.400 mKB 3.355 mBML 206.04S m

I*CL 59896.247 m4, I*COF 2568433.343 m4 .Note 1: BM= I V or I = BM V). I*CLI*COF have been obtained by mult iplyingBMT BML by the volume of d isp lacementa t 6.289 m d ra f t in FW with the hul lconsidered to be i n t a c t .Note 2: The above values of I hold goodfor the bi lged condi t ion a l so as theydepend only on the shape and area of thei n t a c t water -p lane , corresponding toW2 L2 in the foregoing f igure .In the bi lged condi t ion , volume of di s p l acem en t = 12069.4 m3.

New BMT = I*CL 7 V = 4.963 metres .KMT = ne w KB new BMT = 3.446 4.963

KMT = 8.409 metres .New BML = I*COF V = 212.805 metres .

KML = new KB new BML = 3.446 212.805KML = 216.251 metres .

New MCTC:MCTC = W.BMLlOOL = 12069.4 212.805)100 140)

New AB:A

New MCTC = 183.459 tm.

B7 1 . 9 2 8 m - -+1-------- 1 0 0 m

L 2

I n t a c t vol a t 6.289 m dra f t , in FW, i t sAB) - vol l o s t i t s AB) =n e w v o l i t s AB.)12465.4 71 .928) - 396 100) = 12069.4 AB)

New AB = 71.007 metres .Exerc ise 34

Bilging DB tank - ac tua l sh ipl M.v.VIJAY i s a f l o a t in W of RD 1.015drawing 4.6 m fwd and 5.8 m a f t . A DBtank 20 m long, 18 m wide and 1.2 mdeep has AG = 90 m and i s empty. I f

t h i s tank now ge ts bi lged , f ind thehydros ta t ic par t i cu l a r s r e fe r r in g toappendix I of t h i s book.54 55


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2

3

5

6

7

8

9

In que s t i on 1, f i nd the d r a f t s fwda f t a f t e r b i lg ing .I f i n que s t i on 2 KG = 8 m, FSM =1200 tm and a weight o f 50 t i ss h i f t e d 10 m t o s t b d , f ind the list.M.v.VIJAY i s in SW drawing 6.0 m fwd

5.0 m a f t . A DB tank of AG 20 m, 21m long , 16 m wide and 1 m deep hasHFO of RD 0.95 in it to a sounding of0.40 m. Calcu la t e the hydros t a t i cp a r t i c u l a r s , r e f e r r i ng to appendix Iof t h i s book i f t h i s tank now ge tsb i l ged . Note: As usual on sh ips ,assume t ha t the KG of t h i s tank = 5 m. regard l e s s of the sounding.In que s t i on 4, f i nd the new d r a f t sfwd and a f t a f t e r b i l g i ng .I f i n que s t i on 5before b i l g i ng waslist when a weight10 m to por t .

KG = 7.5 m and ~ S M8500 tm f ind theof 80 t i s sh i f t edM.v.VTJAY i s in SW a t an even keeld r a f t of 4 .6 m. A DB tank 20 m long10 m wide and 1.2 m deep i s fu l l ofSW b a l l a s t . The AG of the tank i s 100m. Find the new hydros t a t i c p a r t i -cu l a r s if t h i s tank ge ts b i lged ,r e f e r r i n g to appendix I of t h i s book.In que s t i on 7 , f i nd the new d r a f t sfwd and a f t .M.v.VIJAY a t 6 m even k e ~ l d r a f t insw, c o l l id e s with a barge . The forepeak t ank , f u l l with 102 t of FW AG

135 m, KG 4 m) ge t s b i l ged . S i n e ~ the ~ o p of t h i s tahk was below the o r i g -i n a l wate r l i n e , _the s h i p s wate r -plane a rea i s not adve rse ly a f f ec t ed .Refe r r ing to Appendix I of t h i s bookf i nd i ) the new hydros t a t i c p a r t i c -u la r s ii th e . new d r a f t s fwd a f t

PART B - Bi lg ing of a cargo 'hold .S t a b i l i t y ca l cu l a t i ons , a f t e r a cargohold ge ts bi lged , a re complex t ed ious .Such ea l cu l a t i ons a re not always

s o l va b l e on board owing to the l i m i t -a t i ons of the in fo rmat ion ava i l ab l e tothe sh ipmas te r . However; an idea of thep o s s i b l e ca l cu l a t i ons t ha t can be madeto a f a i r amount of accu racy , i s givenhere in the fo l lowing worked examples.

Though an empty hold has been bi lged ,in the worked examples t h a t fo l low, thes t u d e n t who has s t ud i ed the e a r l i e rchapte r s on b i lg ing , should be able toso lvb s i m i l a r such problems even whenthe bi : ged compartment has cargo in it.Example SM.v.Vijay i s a f l o a t in SW a t d r a f t s of4.11 m fwd 5.11 m a f t . No 2 LH AG 104m, i s 20 m long , 18 r wide empty. No 2DB t ank , one metre deep i s s i t u a t e dd i r e c t l y below No 2 LH. While corninga l ongs i de , a tug co l l i de s with t h i s s h i pcaus ing No 2 LH to ge t b i lged above thel eve l of the tank top . Ca lcu la te thehydros t a t i c p a r t i c u l a r s a f t e r theacc i den t , assuming t h a t No 2 LH i sr .ectangula r .

56. 57


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~ d 4.11 m a f t 5.11 mby s t e r n , Mean 4.61 m t r im one metreAF = 72.011 mCorrect ion = AF t r im) = 0.51 m.

LI n i t i a l hydraf t = 5.11 0.51 = 4.6 m.W = 9013 t , vol of d isp l = 8793.171 m3vol l o s t = 20(18)(4. - l ) = 1296 m3.Sum of the two volumes = 10089.171 m3

I f the hul l was i n t a c t and the vo l ume. of displacement was 10089.171 m3 wwould have been 10341.4 t and the draf t ,as per hydros ta t ic tab le , would havebeen 5.204 metres .

Actual ly , in No 2 LH the spacebetween the water l ine a t 4.6 m dra f t and5.204 m dra f t , having a volume of 217.44m3 i . e 20 x 18 x 0.604), i s dead space- t i s nei ther l o s t nor r ~ g i n e d asvolume of buoyancy. So the sh ip wil ls ink in excess of 5.204 m draf t in orderto rega in t h i s 217.44 m3. During th i ss inkage, only the i n t a c t water-planearea wi l l con t r ibu te to volume ofbuoyancy r ega ined . Since th i s s inkage in

excess of 5.204 m dra f t w i l lsmal l , the water-plane areaconsidered to be cons tan t .be verymay be

From hydros ta t ic tab le , consider ing theh u11 to be in tac t :TPC a t 5.204 m d r a f t = 22.142.TPC = l .025A100 or A = 22.142(100)1.025

-W a r ea a t 5.204 m d r a f tArea los t in No 2 LHI n t a c t water-plane area= 2160 .1 95 m2.= 360 rn 2.= 1800.195 rn2.

Sinkage in excess of 5.204 rn d r a f t= vol yet to regain =i n t a c t WP area 217.44 0.121 m1800.195Final hydraf t = 5.204 0.121 = 5:325 rn.Line (1) below contains the hydros ta t icpar t i cu la r s for the i n t a c t hu l l a t 5.325m dra f t in SW while l ine (2) i s for thebi lged condi t ion in SW a t the samedr a f t .

Draf t(1) 5.325(2) 5.325AB(1)72.006(2)66.353

w10610.59013AF71.78965. 364

KB2 . 8 532.799

TPC22.1918.50KMT8.5008.341

MCTC167.975146.403KML240.463230. 208

Calcula t ions in suppor t of l ine (2)above are g i ~ e n in the fol lowing pages.58


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w check:U/w vol a tDead space 5.325 m d r a f t = 10351.707 m320 18) 5.325-4.6)= 261 m3vol of l o s t buoyancy =Vol of disp by sub t rac t ion

10090.707 m31296 m38794. 707 m3Vol of di sp as per o r ig in a l hydraf t i s8793.171 m3. The di f fe rence i s i n s ig n i f i can t , i nd i ca t i ng tha t the ca l cu l a t i oni s correc t .

w = 8793.171 x 1.025 = 9013 tonnes .New TPC: TPC a t 5.325 m d r a f t , assumingt h a t the hul l i s i n t a c t - 22.19.

TPC = l .025A100 o r A = 100 22;19)1.025At 5.325 m d r a f t ,to ta l water-plane areaArea l o s t in No 2 LHIn t ac t water-plane area

===

2164.878 m2 360 m2 1804.878 m2 New TPC = 1.025 1804.878) = 18.500 t

100New AF: Moments of a rea about A,In t ac t WP area i t s AF) = e n t i r e WP areai t s AF) - area l o s t in No 2 LH i t s AF)1804.878 i t s AF) =2164.878 _ 71.789) - 360 104)

New AF = 65.364 metres .. , .New AB Mo.merlt.s of volume about A,

A

59I n t a c t volume i t s AB) = e n t i r e Vpl a t5.325 m i t s AB) - u/w vol No 2 i t s AB)8793.171 (New AB) =10351.707 72.006) - 1557 104)Note: U/w volume in No 2 LH =20 18) 5 .325 1) = 1557 m3.

New AB = 66.353 m.New KB :Moments of volume about the kee l ,I n t a c t v o l i t s KB) = en t i r e vol a t 5.325m d ra f t i t s KB) - u/w vol No 2 i t s KB)8793 . 171 i t s KB) =10351 . 707 2.853) - 1557 3.163)Note: KB of u / w vol No 2 = 4.325 1)

2New KB= 2.799 metres .

New KMT & KML:

k - - 71. 789 m - - ~ I 20 mFl F

- -65.364 m - - - - - . J 5 .42s

--------------- 104 m ------------------

From hydros ta t ic t a b le , a t 5.325 m d r a f t60 61


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KMKBBM

Transverse8.500 m2.853 m5.647 m

Longi tudinal240.463 m2.853 m237.610 mV 19351.707 m3.

I*CL 58456.089 m4 , I*COF 2459669.l m4.Note: BM= I V or I = M V).so ca l cu l a t ed for the i n t a c t WPmodif ied for the WP a rea in thecondi t ion and divided by the-volume of displacement to ge t thP.the bi lged cond i t ion .I*CL i n t a c t WP =

The Imay bebi lgedac t ua lM for

I*CL . e n t i r e WP - I*CL area l o s t in No 2= 58456.089 - 20 18 3 = 48736.089 m4.

11 1 I 12BMT = I = 48736.089 = 5.542 metres

v 8793.171KMT = new KB new BMT = 2.799 5.542

New KMT = 8.341 metres .

I*COF1 en t i r e WP = -I*COF Ah 2= 2 4 5 9 6 ~ 9 l 2164.878 6.4252 = 2549036.616 m 4

I *COF1No 2 a r e a = 18 203 + 360 38.636 212= 549386.579 m4.

I*COFl i n t a c t WP= I*COF1 en t i r e WP I*COF1 l o s t a reaI*COFl i n t a c t WP = 1999650.038 m4.

Double check of I*COFl above:I*COF i n t a c t WP= I*COF en t i r e WP - I*COF area l o s tI*COF a rea l o s t = 18 203) 360 32.211 2 12

= 385517.468 m4 I*COF i n t a c t WP = 2459669.l - 385517.468

= 2074151.633 m4

I*COE\1 i n t a c t WP = I*COF - Ah2= 2074151.633 - 1804.878 6.425 2= 1999645.142 4m

Resumption of ca l cu la t ionBML = I*COF1 = 1999650.038 = 227.409 mv 8793.171KML = n ew KB+ new BML = 2.799 + 227.409

New KML = 230.208 metres .New MCTC:New MCTC = W BML =lOOL 9013 227.409)100 140)

New MCTC = 146.403 tm.62 63


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Example 6In example 5, f ind the dra f t s fwd a f t .To f ind KG of sh ip :

Draf t4 .600

w9013

MCTC162.7

AB72 . 017

Trim = W.BGMCTC or BG = t r im (MCTC)w

BG= 100 162 . 7) = 1.805 metres .9013Trim s by the s t e rn , so AG < AB.

AG = AB - BG = 72.017 - 1.805 = 70.212 mFina l BG = AG - AB = 70.212 66.353Fina l BG = 3.859 metres .

Since AG > AB, Tc wi l l be by the head .Tc =W.BG= 9013 3 . 859)=237.6 cm = 2.376 mMCTC 146.403Ta = AF Tc) =

L65 .364 2 .376)140 = 1.109 m.

Tf = Tc - Ta = 2.376 - 1.109 = 1.267 m.

Fina l hydra f tTf o r TaFina l d r a f t sExample 7

Fwd5 .325 ml .267 m6 .592 m

Aft5. 3 25 m-1 .10 .9 m4 .216 m

in example 6, 150 t of FW i s

t r ansfer red from the fore peak tank tothe a f t e r peak tank through a d i s t anceof 130 m and 150 t from No 2 DBT to .No8 DBT, through a d i s t ance of 80 m f indthe new dra f t s fwd af t .Tc

Ta

Tf

=150 130)+150 80)= 215.2 cm = 2 .152146.403

= AF Tc) = 65.364 2.152)L 140

Tc Ta = 2.152 1.005=

I n i t i a l dra f t sTf or TaFina l dra f t s

Fwd6.952 m- 1.147 m +5.805 m

-oOo-

=

= 1.005

1.147Aft4.216 m1.005 m5.221 m

m

m.

m.


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CHAPTER 37

CALCULATION OFLIST BY GZ CURVE

Calcu la t ion of list has h i t he r t o beenadvocated by use of the formula Tan B =GG1/GM. Use of t h i s formula has beeni l l u s t r a t e d in chap te r 13 BhipS t a b i l i t y I . That method of ca l cu l a t i onsu f f e r s from l ack of accuracy because tassumes t ha t the i n i t i a l GM and hencBthe t r ansve rse KM remains cons tan tdes p i t e changes in the angle ofi nc l ina t ion . A more accura te method ofcomputing the angle of list would be byuse of the curve of s t a t i c a l s t a b i l i t y .For examinat ion purposes , t i s sugges t -ed t h a t wherever the list , ca l cu l a t ed bythe formula, exceeds f ive degrees , theGZ curve method should be used .The a t t e n t i o n of the s tuden t i sinv i t ed to chapter 22 Ship S t a b i l i t yLI) where in the curve of s t a t i c a ls t a b i l i ty of a sh ip with a list has beeni l l u s t r a ted. In such a case , the s h ip i sin s t ab l e equ i l ib r ium - GZ to one s id ei s cons idered pos i t i ve and to the o the rs i de , nega t ive .A t r ansve rse s h i f t of the COG of thesh ip GG1) to one s i de causes areduc t io n of GZ when the sh ip i nc l i nest o t h a t s i de and vice versa . Thisr educ t ion of GZ , or upse t t ing l eve r , canbe c a l c u l a t e d by the formula:Red of GZ o r u p s t i n l e v e r = GG1.Cos B-

Origin of the formulaMI

In the foregoing f igure , l e t us irstcons ide r the sh ip heel ing over only dueto ex t e r na l fo rces .KG i s the f i n a l f l u id KG of the sh ip .B i s the COB when the sh ip i s upr igh t .M i s the i n i t i a l metacen t r i c he igh t .

Bi l .S the COB a f t e r the s h ip hee ls .Mi i s the new metacentre a f t e r heel ing .GZ i s the r i gh t i ng l eve r caused by hee l .

i s the angle o f heel cons ide red .

66 67


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I f the COG of the ' sh ip is now Shif tedtransverse ly from G to G1, p a ra l l e l tothe keel , and the v e r t i c a J. through G,1cuts GZ a t Y: angle Y = 90 , ~ g l e YGG1= angle of heel = f:t GY = GGi Cos {1.).Since GY i s the reduct ion o f GZ due tot ransverse s h i f t of COG, t h e formula maybe s t a t ed as :Red of GZ or upse t t ing l e v e r GG1 .Cos ~

This upse t t ing l eve r may be allowedfor in e i th e r of two ways:Method lThe curve of s t a t i ca l s t a b i l i t y i sdrawn, as usua l , for the f i n a l f lu id KGof the ship, as shown in chap t e r s 23 and24 Ship Stab i l i t y I I ) . The ~ i sca lcu la t ed by the formula GG1 = dw/W.I f GG1 cos B) i s plo t ted fo r va lues o f ~from to 40, it w i l l be not iced t ha tt h i s curve very nea r ly co inc ides with as t r a i g h t l i ne drawn from GG1 a t 0 heelto 0.8 GG1) a t 40 hee l , a s i l l u s t r a t e di n the fol lowing f igure where in GG1 i sassumed to be 1.000 m. Hence GG1 Cos Ba t 0 = 1.000, a t 10 = 0.985, a t 20 =0.940, a t 30 = 0.866 a t 40 = 0.766

CD Xt yE

0 rJ E E E E0

E0 r\ 0 00 0 Q \ Q \ 00 r . 008 7 0 0 0 0 00 10 20 30 40

Hence the upse t t ing arm curve may bedrawn a s fol lows:GG1 i s . l a i d off upwards from the 0 hee lpos i t i on on the x -a x i s , a r r iv in g a t apo i n t c a l l ~ d X. 0.8 GG1) i s l a i d o f fupwards from the 40 hee l pos i t i on onthe x-ax i s , a r r i v ing a t Y. Poin t s X andY are jo ined by a s t r a i g h t l i n e whichr ep re sen t s the upse t t ing lever a t anyangle of heel upto about 30. The angleof hee l a t which the r i gh t i ng arm curve(GZ curve) and the upse t t ing arm curvel i n e XY .) i n t e r s e c t , i s the listMethodThe curve of s t a t i c a l s t a b i l i t y i s drawnonly a f t e r the value o f GZ, a t eachangle of hee l , i s reduced by GG1 Cos ~ .The GZ a t 0 heel wou ld nece s s a r i l y benegat ive . The angle of heel a t which themodif ied GZ c u r v ~ cut s the x-ax i s i s theangle of listxample l

M.v.VIJAY i s in SW d i sp l ac i ng 13000 t .KG = 7.788 m, FSM = 1372 tm. A t r a c t o ~weighing 50 t i s to be sh i f t e d 10 m tos t a rboa rd . Calcula te the r e su l t a n t list

FSC = 1372/13000 = 0.106 mSol id KG = 7.788 mFlu id KG = 7 .894 mKM appendix I ) = 8.139 mI n i t i a l f lu id M = 0.245 m

Note: Though the normal prac t i c e i s tosu b t ra c t FSC from the so l i d M in orde r68 69


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to ob ta in the f l u i d GM, the above changein procedure has been adopted forconvenience, in t h i s case because thef l u i d KG may be requ i red l a t e r on, incase the the list by formula e ~ c e e d s 5.GGi = dw/W = 50 10) /13000 = 0.03846 m.Tan = GGi/GM = 0 .03846/0 .245 = 0.15699

= 8.922 or 8 ~ 5 to s ta rboard .Since the list exceeds 5, it i s pr e f e r -rab le to ob ta in a more accura te answerby the GZ curve.From appendix I I

eo KN KG Sin G = GZ0 0.000 7 . 894 Sin oo 0.000 m5 0.798 7 .894 Sin 50 0.110 m10 1.595 7 .894 Sin 10 0.224 m

Note: It i s no t necessary to draw thee n t i r e curve of s t a t i c a l s t a b i l i t y j u s tto compute the value of list.ethod

Upset t ing l eve r a t 0 and a t 40 heel =GGi and 0 .8 GGi = 0.038 and 0.030 metre.The graph on the n ~ x t page shows boththe r ight ing arm and the upse t t ing armcurves on which the value of list i sr eadi ly apparen t .In order to cor r ec t l y ge t the shape ofthe curve in the v i c i n i t y of the or i g i n

a perpend icu la r i s e rec ted a t 57.3 heeland the i n i t i a l f l u i d Gr l a i d of f t : pwardN Init ial fluid GM

0 245m

>-

0::w>w_

t J2

xGZ in metres

r LO

0LO

0( ' )

0N

'lCl'-0\Cl0

c:

_

ww:::i:


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0

70to ar r ive a t poin t Z. Poin t Z and theor ig in are joined by a s t r a igh t l ine ,while drawing the curve of s t a t i c a ls t a b i l i t y , it i s ensured tha t the curvecoincides with th i s l ine for the f i r s tfew degrees . I t i s t h i s technique tha tmakes method 1 a littl more accura tethan method 2:Method

eo Ful l GZ - GG1 Cos S = Reduced GZ0 0.000 m 0 0 3 ~ C o s oo -0.038 m5 0.110 m 0. 038 , Cos 50 +0.072 m10 0 . 224 m 0.038 Cos 10 0.186 m

The graph below shows the values ofreduced GZ plo t ted aga ins t the co r re s -ponding values of heel .

0.2 metre

0.1

Z

5 10H E E L in degrees

71The l i s t , ca lcu la ted by the formula, was8.9 but , by the GZ curve, i s only 2.5by method 1 and 1.7 by method 2. In thecase of th i s sh ip , the change of KM, dueto change of angle of heel , is cons ider -able even for small angles o f heel . Ther esu l t s by use of the formula becomel e s s accura te i f the f l u id GM i s smal l .Note: The above curves are only for s tbdheel because GG1 i s to s ta rboard .

i ~ s tudent may, i f he so des i r e s ,draw the en t i r e curve for both, thes ta rboard and the por t s ide s , in orderto unders tand the topic be t t e r . Thec o r r e c t i o ~ to GZ, obta ined by theformula GG1 C o s e ) , would be addi t iveto the por t s ide because GG1 i s to s tbd .xample 2

M.v.VIJAY i s in SW a t a displacement of13700 t . KG i s 7.0 m and FSM 1400 tm.300 t of cargo i s loaded on the upperdeck, KG 12 m, 6 m to por t of the cent rel i ne of the sh ip . Calcula te the listcaused.GG j = dw/W = 5(300)/14000 = 0.107 m.Final KG so l id = 7.000 + 0.107 = 7.107 mFSC = FSM/W = 1400/14000 = 0.100 m.. .Final f lu id KG . . . . . . . . .. . . . = 7.207 mKM from appendix I . . . . .. . . = 8.073 mI n i t i a l f lu id GM . . .. . . . . = 0.866 mGG1 = dw/W = 6(300)/14000 = 0.12857 m.


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72Since the l st exceeds 5, t i s p r e f e r rab le to obta in a more acc u r a t e answerby the GZ curve. Use appendix I I .

,-Bo KN KG Sin ~ = GZ'O 0.000 7.207 Sin oo 0 .0 00 m5 0.793 7.207 Sin 50 0.165 m10 1 . 581 7.207 Sin 10 0.330 m20 3.130 7.207 Sin 20 0.665 m

Method 1Upset t i ng levers a t oo and a t 40 heel =GG1 and 0.8 GG1) = 0.129 and 0.103 metre0.9 metre z

Z

0.7ED

4. 5 DLIST to port CX

I00.5 :E(. )

=>4 -.......'0.3c:

>--5 0 .7 56 0 . 76 3 -0 . 007 . 010 l 5 0 1 l 519 -0 .0 18 > l / ). Q ,...-l15 2.22 7 2.265 -0 . 0 38 H0020 2.974 2.99 3 -0 .019 c25 3.699 3.698 +0.001 0.090 mrShip meets reqbirement 4 b)

Requirement 4 c ) : Area between 30 theangle of f looding to be a t l e a s t 0.03 mr.As ca lcu la ted above, area between 30the angle of f looding i s 0.143 mr. Thesh ip complies with requirement 4 c) .In example 5, the sh ip meets a l l thes t a b i l i t y requirements of the LoadlineRules .

Exercise 37Dynam s t a b i l i t y LL Rules requirements .

In the fol lowing cases , ver i fy which ofthe s t a b i l i t y requirements under theLoadl ine Rules have been met and whichhave no t . Use the .appendices of t h i sbook, where necessary . Where the angleof f looding i s not mentioned, assumet ha t it i s over 40.1 M.v.VICTORY, W = 70,000 t , KG = 9.41 mFSM = 6300 tm, M 13 . l m. Sta te a l sothe dynamical s t a b i l i t y a t 40 heel .

2 M.v.VICTORY, W = 85,000 t , KG =10.68 mFSM = 6761 tm. Sta te a l so thedynamical s t a b i l i t y a t 30 heel .3 M.v.VIJAY, W = 13,250 t , KG = 6.427 m.FSM = 1200 tm.

4 M.v.VIJAY, w = 10,777 t KG= 7.8 mFSM = 800 tm. S ta t e a l so the dynamicals t a b i l i t y a t 40 heel .

5 M.v.VIJAY, W = 19,943 t KG = 7.326 mFSM = 1342 tm, M 8.257 m angle off looding = 36.

-oOo-

94 95


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CHAPTER 41

EFFECT OF BEAM ANDFREEBOARD ON GZ CURVE

The beam and the f r eeboard i n f luenceGZ cons ide rab ly as i l l u s t r a t e d below. Inthe fol lowing f igu re , curves A, B and cbelong to t h r ee d i f f e r e n t box-shapedveEsels whose dra f t , KG f lu id and volumeof displacement a re the same.

Ship A s deck edge immerses a t 15 hee l .Ship B has the same bread th as sh ip Abu t has a g r e a t e r f reeboard . As seenfrom the fol lowing f igu re , the GZ valuesof sh ip B a re the same a s t ha t o f sh i p Au n t i l 15 - the hee l a t which A s deckedge immerses. On hee l ing f u r t he r , theGZ o f sh i p B i s gr ea t e r than t ha t ofsh ip A for the same ang le of hee l . Thisf ac t shows up c l ea r ly on i n sp e c t i n g

curves A and B in the foregoing f igu re .~ h range of s t a b i l i t y of sh ip B i sgr ea t e r , than t ha t of sh ip A, because ofits gr ea t e r f r eeboard .To sum up : - Grea te r f r eeboard means:

i )i i )iii)

i v )

No change in I n i t i a l Fluid GMDeck edge immerses a t a grea te rangle of hee l .GZ values unaf fec ted un t i l thedeck edge immerses but , t h e r e a f t e r , GZ values are gr ea t e r .Grea te r range of s t a b i l i t y .

f fect o be m on GZ:-Ship c, having the same f reeboard assh ip A, has g r e a t e r beam bread th ) . Forsh ip - shapes , BM = I /V for box-shapes,it i s . s im p l i f i e d as BM= B2 /12d, a sexp la ined in chap te r 19 of ShipS t a b i l i t y I I . Hence the GZ values ofsh i p C are grea te r than those of sh ip A,a t a l l ang les of heel .

96 97


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As apparer i t from the fo l lowing f igu re ,the deck edge of the broader sh ipimmerses a t a smal le r angle of hee l than

t h a t a ~ a nar rower sh ip . The range of~ t a b l t y a l s o i nc reases with i nc reasein beam. All o f the foregoing areapparen t when i n spec t ing curves A c inthe first f igu re of t h i s chap te r .

To sum up : - Grea te r beam bread th ) meansi )i i )iii)iv)

G r e a t e r I n i t i a l Fluid GM.Deck edge immerses a t a smal le rang le of hee l .Grea te r values of GZ a t any hee lGrea te r range of s t a b i l i t y .

The a c tu a l c a l c u l a t i o n s in suppor t o fthe f o r ego ing s ta temen ts are beyond thescope of t h i s book.Box-shaped vesse l s have been t aken fo ri l l u s t r a t i o n so t h a t the var i ab le para meters have been kept to a minimum forthe sake of easy comparison.

-oOo-

CHAPTER 42

CHANGE OF TRIMDUE TO CHANGE OF DENSITY

When a sh ip i s in equ i l ib r ium, herCOG COB would be in the same ver t i ca ll i ne . When she proceeds to water ofano ther dens i ty , her volume of d i sp l a c e ment, and hence her dra f t , would change.I f the AB of the sh ip a t the new dra f t ,i s d i f f e r en t from the AB a t the e a r l i e rd r a f t , the COB would now be l ong i tud in a l l y separa ted from the COG by the thed i s t ance re fe r red to , in e a r l i e rchap te r s , as BG. The trimming moment, soformed, would be W.BG. The sh ip wouldthen a l t e r her t r im un t i l the COB comesd i r e c t l y under the COG. The t r im, socaused, can be ca lcu la t ed by using thehydros ta t i c par t i cu l a r s of the sh ip fo r

t ~ new d r a f t a t the new dens i ty . Thiss imple phenomenon can be i l l u s t r a t e d bythe fo l lowing example.Example 1M.v.VIJAY i s6.40 m a f t .a f t if sheappendix I ) .

in SW drawing 3 .60 m fwdFind the hew d r a f t s fwd andnow proceeds to FW. Use

Fwd 3.600 m, a f t 6.400 mt r im 2.800 m or 280 cmFrom appendix I , AF =

mean 5.000 m,by the s t e r n .71 .913 metres .

98 99


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Correc t ion to a f t d r a f t = AF t r im) /L71.913 2 .8 ) /140 ~ 1.438

I n i t i a l h y d ra f t = 6.40 - 1 . 438 = 4.962 mSW d r a f t

4.962w t9807.4

MCTC tm165. 4 34

AB m)72.014

Trim in cm = W BGMCTC

o r BG = trim MCTC)wI n i t i a l BG= 280 165 . 434) = 4.723 met res9807.4Since t r i m was by s t e r n , AG was < AB.AG of vesse l = 72.014 - 4.723 = 67.291 mSo f a r , the ca l cu l a t i on has been e x a c t l ythe same a s in example SB of chap te r 27in Ship S t a b i l i t y IIFrom appendi x I ,

w tSW 10052.6FW 9807.4

d r a f t5.0735.073MCTC tm AB m AF m166 . 212 72.013 71 .887162.158 72.013 71.887

BG= AB - AG= 72.013 - 67 . 291 = 4.722 mSinc e AB > AG, f i n a l t r im i s by s t e r n .Tr i m = W BG = 9807. 4 4.722) = 28 5 . 6 cm.

MCTC 162.158Ta = AF t r im)

L= 71.887 2.856) = 1.466 m140

Tf Tc Ta = 2 .8 56 - 1 . 466 = 1.390 m

Fina l hyd ra f tTf or TaFina l d r a f t s

Fwd5..J)73 m-1 . 390 m3.683 m

Aft5.073 m+ l . 466 m6.539 mThe change of t r i m ac t ua l l y caused byt h i s change in dens i ty of wate rd i sp l aced = 285.6 280 5 .6 cm only .This chap te r i s only meant to i l l u s t r a t e , in theory , t ha t the t r im o f as h i p i s l i a b l e to s l i g h t change when thedens i ty of water d isp laced changes . Iti s no t of much impor tance in prac t i c-a lopera t ion of sh ips . Since thec a l c u l a t i on i s not only s i m i l a r to , buts imp le r than , o t he r p r o b l e m ~ on t ~ i m n oe x e rc i s e has been s e t on t h i s top ic .

-oOo-

100 101


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CHAPTER 43

CENTRE OFPRESSURE

Cent re of pressu re i s t ha t pointthrough which the t h r us t , o r t o t a lpressu re , may be cons idered to ac t .Pres sure and th rus t were explained inChapter , 2 in SHIP STABILITY I .Calcula t ion of pressu re and th rus t oncurv i - l i nea r , immersed areas was done inChapter 20 in SHIP STABILITY I I . It i spresumed t ha t the s tuden t has gonethrough those chapters and, therefo r e ,needless r epe t i t i on i s avoided here .The depth z of the cen t re ofpressu re (COP) below the water sur face(WS) may be ca lcu la ted by the formula:

z I wsd

Where:I ws i s the moment of i ner t i a orsecond moment of the immersed area aboutthe water sur face expressed in quadrome t re s .

A i s the area of the i mmersed por t ioni n square metres .d i s the depth , in metres , of the

geometric cen t r e of the immersed por t ionbelow the wa t e r su r f ace .

Standard formulae for the moment ofi n e r t i a I about an ax i s pass ingthrough the geometr ic cen t re of someregu la r shapes a re given below. Studentsshould l ea rn these by hear t .yI~

1 IIIcL- fI*YY: ; : LB3TI

x- -xl IIII

I XX : ; : BL3T I y

~I*XX : ; : Bn X ;v-X6

IE D-- 1

I XX : ; : TTD 4 G4 x --xonce I of an a rea , about an ax is

pass ing through i t s geometr ic cen t re ,known, I about any o ther pa r a l l e l ax iscan be ca lcu la ted by the theorem ofpa r a l l e l axes .

102 103


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Theorem of p a r a l l e l axes: I f I*GG i s themom nt of i n e r t i a of an area about anax i s pass ing through i t s geometr iccen t r e , and ZZ i s an ax i s p a r a l l e l toGG , then

I*ZZ = I*GG Ad 2where d i s the d i s t ance betweenGG and a x i s ZZ.Example 1 :I*ZZ = I*GG Ad 2

= LB 3 LB B/2) 212= LB 33

Example 2:z~ ~ i

D

x _ _LI*ZZ = I*GG Ad 2

BrD/3- . - It - - G

2D/3- -- x

= BD 3 BD D/3 )236 2

axis

= BD3TI

~ X X = I*GG Ad 2 = BD3 BD( 2 D 3 )2 = BD336 2 4

Example 3:I*XX = I*GG Ad 2

i D4 fTD 464 16

= fTD 464= 5 r D464

fTD 2 ( D / 2 24

X XD IIG I___ G1y

y yI*YY = I*GG Ad 2 = ITD 464

CALCULATION OF POSITION OF COPWorked example 1One s i de of a tank i s a v e r t i c a l , r e c t -angula r bulkhead 15 m long 10 m high .Find KP ( the he i gh t of the COP above thebot tom of the tank) when the tank has SWin it to a sounding of 9 metres .

z = I*ws = I*GG Ad 2Ad Ad= 15 9 3 )+15(9)(4 .5 2 )12 15( 9 ) ( 4 . 5 )


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z = Depth of COP below WS = 6.000 m9.000 m3 ~ 0 0 0 mounding ... ... ... ... =KP height of COP above bottom)

~ o r k e d example 2Calcu la t e KP height of the COP abovethe bottom of the tank) in workedexample 1 , when the sounding i s 10 m

z = I*ws = I*GG + Ad zAd Ad= 15 10 3 )+15 10)5 Z12

15 10) 5)

W +----- 15 m - - - - - - ; > S'I'5 m- ..[,- - - . - - - G5 rnJ,K

z = Depth of COP below ws = 6.667 m10.000 m3.333 mounding . . . . . . . . . . . . . . . =KP height of COP above bottom)Worked example 3

C a l c u l a t ~ KP he igh t of the COP abovethe bottom of the tank) in workedexample 1 , when the sounding i s 12 mW

z = I*ws = ~ G G + Ad zAd Ad 7 rnG - :._ ___ _1 G; 15 10 3 )+15 10)7 z12

15 10 ) 7 )z = Depth of COP below ws =ounding =

5 mK

KP height of COP above bottom)8.190 m12.000 m3.810 m

Worked example 4One bulkhead ~ f a tank cons i s t s o f at r i ang l e , apex downwards, which i s 14 mbroad and 12 m high . Calcu la t e KP

he igh t of the COP above the bottom ofthe tank) when the saunaing i s 9 metres .By s i m i l a r t r i ang l e s ,breadth of water su r -face = 10 .5 metres .A = 10.5 9) = 47.25 m2,2z = I*ws = I*GG + Ad zAd Ad

= 10 5 ( 9 3 ) +4 7 2 5 ( 3z )3647 .25 3)

- -- - 14 rn _ _ ... ..

_ _ -Kz = Depth of COP below ws = 4.500 mSounding . . . . . 9.000 mKP h ~ i g h t of COP above bottom) 4.500 mWorked example 5Calcula te KP he igh t of the COP abovethe bottom of the tank) when thesounding of the tank in worked example 4i s 14 rn.

A = 14 12)/2

I*GG = 14 12 3 ) = 672 m436

- - - s

8 rn

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z = I*ws = I*GG + Ad 2Ad Ad

6 7 2 + 84 ( 6 284 6)z = depth of COP below ws = 7.333 m14.000 m6.667 mSounding . . . . . . . . . . . . . . . =KP he ight of COP above bottom)Worked exampleOne bulkhead of a tank cons i s t s of at r i a n g le , apex upwards, 14 m broad and12 m high . Calcula te KP he ight of theCOP above the bottom) when the soundingi s 9 m.

AFw - - N c - - s

12 m Jt9 m 4mJ lD K E14 m

By s imi l a r t r i ang l e s , BC = 3.5 metres .Let F J be the geometr ic cen t r e s oft r i angle s ABC ADE . To ca l cu l a t e thep s t of G, the geometr ic c e n t r e oft r apezoid BCED immersed a rea ) :Taking moments about A,Area ADE AJ)-Area ABC AF)= Area BCED(AG)

84 8) 5.25 2) = 78.75 AG)AG = 8 . 4 metres . NG = d = depth ofgeometr ic cen t re below ws = 5.4 metres .To f ind I ws o f immersed area

I*ws of ADE = I* JJ + A JN 2 )- 14 123) + 84 5 2 ) = 672 + 210036

= 2772 m4 I*,ws of ABC = I*FF + A(FN 2 )

3.5 3 + 5 .25 1 2 =36= 7.875 m4

2.625 + 5.25

I*ws BCED = I*ws ADE - I*ws ABC= 2772 - 7 . 875 = 2764.125 m4

Alternat ive method:

w6 m

l3 mJ

A

4.5 mi

- - - s

t3 mD 5.25 m R 3.5 m Q 5. 25 m E

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I*ws BCQR =I*ws CQE =

I*ws BRD =I*ws BCED . .

3.5(9 3 )3 =

5.25 9 3 ) =45 .25 93) =

4.. . ... =

850.500 4m

956.8125 m4

956.8125 m4

2764.125 4 .m Note: In case the s tudent i s not able tor ead i ly fol low the above s t eps , he i sreques ted to re fe r to examples 1 2 on

~ l c u l t i o n of moments of i n e r t i a on thet h i rd page of t h i s chapter .To f ind d depth of geometr ic centre ofimmersed area BCED):Taking moments about ws, Area BCED(d) =Area BCQR 4.5) +Area CQE 6) +Area BRD 6)

so t ha t78 .75 d) = 31 .5 4 .5 )+23 .625 6)+23 .625 6)

d = 5.4 metres .ompletion o f ca l cu l a t i on

z = I*ws =dKP = 9 .000

2764.125 =78.75(5 .4)-6.500 metres .

6.500 = 2.500 metres .Compare the KP here wi th t ha t of workedexample 4 wherein the same tank had thesame sounding but was apex downwards.

Exercise 38COP - regular shapesFind theCOP abovebulkhead in

t h ru s t and KP (he ight of thethe bottom) of a ver t ica lthe fol lowing cases :Shape Breadth Height Sound- RDing

1 Rectangle 16 m2 Triangle 15apex down3 Triangle 14apex up4 Rectangle 125 Triangleapex down6 Triangleapex up

11

10

7 Rectangle 128 Triangle 16apex down9 Triangleapex up10 Circ le

12

4

10 m 10 m 1.02510 10 1.000

10 10 1.010

9 6 1.0159 6 1.012

9 6 1.012

8.4 11.4 1.0008.4 11.4 1.015

8.4 11.4 1.0

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Ship Stability III by Capt. Subramaniam