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4.9 Calculation of the mooring system

In the following pages, the calculation of a mooring system for an oil tanker will be

presented. For ship dimensions will be used LS Annes dimensions.

Chapter 1

Primary dimensions determination

1.1 Length of the vessel [m]

L ! "#$.$ m

1.2 Service speed [ms]

%! %&'#.$"(! ).")* m+s "."-

%& ships ser%ice speed/ %&! "0 &d

1.! "roude#s num$er

Fr!v

gL !6.168

9.81105.5 ! #."1"2 ".3-

% ser%ice speed

g gra%itational acceleration 1.*"m+s3-

L length of the %essel

1.4 "inesse $loc% coefficient&

4he block coefficient will be calculated using Logace%s formula for tankers and bulk

carriers.

56! "."#$ 7 ".)* ' Fr ! #.2*3 ".0-

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1.' Length over $readth fraction&

4here are used the ships dimensions.

L

B !

105.5

16.8 ! ).3* ".(-

6 7 breadth of the ship/ 6!").* m

1.( )raft of the vessel&

Also, taken from ships particulars, 4 ! ).0 m

1.* )isplacement [m!]

8 ! L'6'4'56 ".$-

8 ! "#$.$'").*').0'#.2*3 ! *20".1#$ m0

1.+ ,eight [m]

9 ! 2.( m

1.9 "ree$oard [mm]

F ! 94-'"#0 ".)-

F! 2.().0-'"#0! ""## mm

6ecause the block coefficient is greater than #.)*, the freeboard will be multiplied with m,

where:

m!Cb+0.68

1.36 !".#2$ ".2-

4herefore,

F"! F'm ".*-

F"! ""##'".#2$ ! ""*3.$ mm

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Also, because the height of the ship is greater thanL

15 !2.#0, the freeboard will be

increased with m , where:

m! 9L15

'L0.48 ".1-

m ! 2.( 7105.5

15 -'105.5

0.48 ! *#.)

4herefore,

F3! F"; m "."#-

F3! ""*3.$;*#.) ! "3)0." mm

1.1- echnical characteristics

Sail area

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hi! "$m / h!").3m

a

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Chapter 2

/nchors

4he Bruson anchor is chosen, a con%entional type stockless anchor.

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Chapter !

/nchor ca$le

It is composed of a cable shackle, common links Coined by Coining links and an end link.Length of a shackle is 32.$ m and the cable has "* shackles.

L6b! 3($ m L4b! 3$# m

L6b7 length of the port side anchor chain

L4b7 length of the starboard side anchor chain

Specific weight:

D! #.#"1*'d3! (1.$ kg+m 0."-

!.1 Stoppers

5able stoppers, usually known as slips, are pro%ided to hold the cable prior to letting go an

anchor, or to act as pre%enters when the ship is riding on the brake of the cable holder, or to hold

the cable temporarily so that the inboard part of the cable can be handled, or to house the anchor

securely in the hawsepipe.

5hoosing the stoppers for our installation, will reDuire finding the %alue of strength test of the

stopper, gi%en by the relation:

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i!;D'l4b-'a 0.3-

a 7 %alue depending on the running speed of the cable, which will be considered "# m+min, and

the cable caliber

a! v2/d ! "##+$# ! 3

i!0#)#;(1.$'3$#-'3 ! 0#*2# E&

And for the port side:

i! ;D'l6b-'a 0.0-

i! 0#)#;(1.$'3($-'3 ! 0#02$ E&

!.2 he anchor ha0se

According to the ?omanian &a%al ?egister, the thickness of wall of the anchor hawse must

respect the condition:

s 0.4d therefore,

s! 3# mm

4he thickness of the inferior part of the hawse will be:

sinf!3$ mm

4he diameter of the hawse, according to the ?.&.?, is:

n!003

Q ! (21." mm 0.(-

! 0#)# E&

Ge calculate a larger diameter:

na! n; 3#.1 ! $## mm 0.$-

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!.! he chain 0ell

4here will be two circular wells, ha%ing the diameter:

p! E'd ! 0#'$# ! "$## mm 0.)-

k 7 structural coefficient : 0#H01

4he %olume of the chain well is gi%en by the relation:

4b!l4b'ku'd3'"#) 0.2-

ku7 filling coefficient/ ku! *.$

4b!3$#'*.$'3$##'"#)! $.0 m0

4he height of taluJare Jone:

93!Dp

2 ' tg K 0.*-

where, K!($

93!1500

21=750mm

4he %olume of taluJare Jone:

V2=

1

3Dp

2H2

4 0.1-

3!

1

31.520.75

4! #.(( m0

5omplete filling %olume:

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"!3! (.*) m0 0."#-

9eight of the complete filling Jone:

H1=

4V1

Dp2 0.""-

H1=

44.86

1.52=2.75m

Free Jone height

4his Jone is designed for maintenance and obser%ation of the cable

H3

15d/

So, 90! 2$# mm

6ut, 9!9";93;90 0."3-

must be an integer, therefore, 90! ".$ m

9!3.2$;#.2$;".$ ! $ m

4he %olume of this Jone is gi%en by the relation:

V3=H3Dp

2

4! 3.)$ m0 0."0-

4he %olume of the chain well will be:

t!;0! $.0;3.)$ ! 2.1$ m0 0."(-

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Chapter 4

ooring euipment

3indlasses

4he force of the cable holder must be at least:

P1=9.8a d2=9.84.25502=104.125KN (."-

4hrust force on a cable holder:

Ft=1.5P1=156.2KN (.3-

4.1 Calculation of the gipsy 0heel

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Mitch of the cable is:

p!*d / p!(## mm

iameter of the cable holder:

Dcb= 8dsin36

=680.5mm (.0-

4.2 he nominal P of the gipsy 0heel, measured in ?ot+min, is:

nb= v

40d103 (.(-

nb= 10

4050103=5m /min

% 7 speed of lifting the cable/ %!"# m+min

4.! omentum of the gipsy 0heel

Mb=P

1DCb2 (.$-

Mb=104.125680.5

2=35428Nm

4.4 Po0er of gipsy 0heel

Pb=Mbnb

30 (.)-

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Pb=354285

30=18550W

Chapter '

Calculation of anchoring depth

4he relation is:

hmax=(Ks

max

Q

q +2 !)+(Ks

max

Q

q +2 !)2

+ 02

$."-

where: ksmaN7 break out coefficient of the anchor/ EsmaN!0

D 7 specific mass of the cable

7 total mass of the anchor

f 7 friction coefficient of the cable/ f!#.#3$

l37 shackle length/ l3!32.$ m

l#7 length of the free cable

0="b22 $.3-

0=250227.5=195m

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hmax=( 3306049.5 +27.50.025)+(3306049.5

+27.50.025)2

+1952=83.438m

'.1 "i5ation limits:

#max=0.87qg( 0

2hmax2 )

2hmax $.0-

#max=0.8749.59.81(195283.4382)

283.438=73364N

'.2 "riction $et0een ca$le and ground coefficient:

!$=

q02qhmax

2 2hmaxKsmaxQ

2hmaxq2 $.(-

! $=49.5195249.583.4382283.43833060

283.43849.527.5 =0.0251

5ondition:

|!$!|

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)etermination of the tension variation in the gipsy 0heel 0hen heaving the

anchor at depth h

4his determination consists in going through ( steps:

a- Mulling the ship on the anchoring line in the same time with hea%ing the shackles:

4ension at the hawse pipe will be:

"=0.87g (KsmaxQ+q2 !+qh )+qg h )."-

where,

h 7 anchorage depth will be 0# meters/

4he tension at the gipsy wheel is calculated with:

"b1="N1

%ha&s' ).3-

where, Ohawse #.1*

4hese tensions will be calculated for l3!32.$ and l3!# meters.

For l3!32.$ :

"N(

(1)=0.879.81(33060+49.527.50.025+49.530)+49.59.8130

4&I"-! "#$**# &

"b((1)=

105880

0.98=108041N

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For l3!#:

4&I3-! "#$$1# & and 4bI

3-! "#22(( &

b- Mulling the ship in anchoring line, while dumping the shank of the anchor, respecting

the neNt conditions:

4&IIP"Q!4&I

3-!"#$$1# &/

4&II3-!4&III/

4bII"-!4bI

3-!"#22(( &/

4bII3-!4bIII/

c- 6reaking ground, using the neNt conditions:

"N(((=0.87g ( )sminQ+qh )+0.87)dg (Q+qh) ).0-

"b(((= "N(((%h a&s'

).(-

where: kd dynamic coefficient/ kd!3

ksmin minimum breaking coefficient/ ksmin!"

"N(((=0.879.81(13060+49.530 )+0.8729.81(3060+49.530)

N(((=" "")02# &

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"b(((= ""*2($ &

4he neNt condition must be respected:

M" R 4bi"-R! "#("3$ & R "#*#(" &

Ft R 4bIII"-R! "$)"*2 & R ""*2($ &

4he second condition is not fulfilled, so @band Mbfrom the gipsy wheel calculation will be

modified:

Mb=

P1Dcb2

Mb=1125410.68

2=38263Nm

4herefore, we recalculate the power on the gipsy wheel:

Pb=Mbnb

30 ).$-

Pb=382635

30=20035W

d- 9ousing the anchor by respecting these conditions:

"N(V=0.87g(Q+qh$)

).)-

where # *h $*h

"b(V="N(V

%ha&s' ).2-

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For h!h!0# m

"N(V

(1)=0.879.81(3060+49.530)=38790N

"b(V(1)=

38790

0.98=39582N

For h!#

"N(V(2)

!3)"") & and "b(V(2)

!3))(1 &

Chapter *

)etermination of e5ternal forces that act upon the ship and verification of the

heaving line

*.1 3ater forceis composed of the friction force Faf- and the waste force Far-

Fa!Faf;Far 2."-

4he friction force is defined by:

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Fa!=1

2+a( c !+,c!)-.va

2

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%a7 speed of water/ 3 m+s

,c!7 correction term for the hull co%ered with clamshells/

,c!=1.9103

/

So,

Fa!=1

21.026(1.92103+1.9103 )25164=19.7KN

?esidual force is gi%en by the relation:

Fa/=1

2+ac/(LB")

2/3v a2 [KN]

2.)-

where, cr7 usual residual coefficient/ cf! #.2'"#0

cr 7 coefficient depending on the breadth, the draft of the ship and the residual

coefficient

c/$=c/+0.16(B"2.5)103 2.2-

c/$=0.7103+0.16( 16.86.3 2.5)103=7.266104

4herefore,

Fa/=1

21.0260.7103(105.516.86.3)2 /34=0.717KN

4he friction force is added to the residual force so that the water force will be found.

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Fa=19.7+0.717=20.415KN

*.2 he 0ind force

Fv=0.001pv0 v[KN] 2.*-

where, A%7 reef surface/

p%7 wind pressure/ p%! $2 &+m3

4he reef surface will be calculated with the following relation:

0v=1.6103F

2L(sin1+BLc2s1)[m2] 2.1-

4he calculations of F%and A%will be made for >!#/

2 and arctgL

B .

For >!#,

0v=1.61031263.1105.5( 16.8105.51)=33.95m2

Fv=0.0015733.95=1.935KN

For >!

2 ,

A%! 3"0.3" m3 F%! "3."$ E&

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For1=a/ctg

L

B ,

A%! 3"$.*1 m3 F%! "3.0 E&

*.! 3ave forces

F&=p&0 i

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F&=0.39972.765=29.03KN

For 1=

2 ,

Ai! ()$.3$$ m3 Fw! "*$.)0 E&

For 1=82.1 ,

Ai! (2".#33 m3 Fw! "*2.10 E&

*.4 6erifying the mooring line

4here are 3 possibilities to moor. U is chosen 3$

.

a- mooring on both lines, where:

"N=Fa+Fv+F&2cos5 2."0-

"N=20.415+1.935+29.03

2cos25 =28.34KN

b- mooring on a single line, taking into account the condition#*"N

1 /

for1=

2 /

#2=Fa

2+(Fv 2

+F&2

)22Fa(Fv

2

+F& 2

)cos(1802 ) 2."(-

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#2=20.4152+(12.15+185.63 )2220.415(12.15+185.63 )0.87862=180.106KN

4he condition# *"N( is not fulfilled so mooring on two mooring lines will be choosen.