C L DUBEY – EXERCISE – 08 (DRY DOCKING) |
Miscelleneous

# C L DUBEY – EXERCISE – 08 (DRY DOCKING)

###### Solution –

Given W = 19200, LBP = 235m, KM = 10.2m, KG = 9.4m, TPC = 32, MCTC = 254tm, draft fwd = 9.42m, draft aft = 10.88m in RD 1.005

In this Question first change of trim has to be considered due to change of density.
We have BB1 =
(change in v/w vol.) ⨯ (Distance between LCB + LCF) /Final volume

We can calculate, v/w vol. in 1.005
= 19200/1.005
= 19104.477m2

v/w vol. in 1.020
= 19200/1.020
= 18823.529m3

BB1 = 280.95 ⨯ 8/18823.529
= 0.119m

Since COF is abaft midship vessel will him by stern

Change in trim = (W⨯BB1/ MCTC)
= 19200 ⨯ 0.119/252.76 ⨯ 100

MCTC in 1.025 = 254
MCTC in 1.020 = 252.76
Trim caused ( TC) = 0.090m

Change in trim aft = (0.090 ⨯11.5)/235 = 0.043m
Change in trim fwd = (0.090 – 0.043) = 0.047m

Also,

FWA = (W40 ⨯ TPC)
= 19200 /40 ⨯ 32
= 15cm,

DWA = 0.015 ⨯ 15/0.025
= 9cm

 Fwd Aft Drafts 9.42m 10.88m DWA -0.09m -0.09m 9.33m 10.79m -0.047m +0.043m 9.283m 10.833

Therefore, Trim now = 1.55m
Now P = (trim ⨯ MCTC)/d
= 155 ⨯ 252.76/112.5
= 348.25

MCTC in 1.020 = 252.76

We know that –
GG1 = P ⨯ KM/W
= 0.185m,

Old GM = (KM – KG)
= (10.2 – 9.4)m
= 0.8m

###### Solution –

Draft fwd = 4.10m, draft aft = 6.20m,
Trim = 2.10m

Corn to aft draft = 2.10 ⨯ 70/144
= 1.021m

Hydrostatic draft = (6.20 – 1.021)m
= 5.179m

• When level is lowered by 1.5m the water level = 5.5m above. It means stern has already taken to the block while forward end is still afloat.

Change in draft aft = (5.5 – 6.20)m = 0.70m
70 =   P/17 +( P⨯70/150) ⨯70/144
P = 245.03 tonnes

GG1 = P ⨯ KM/W
= 245.03 ⨯ 6.9/6220
= 0.272m

Virtual  GM = (Old GM – GG1)
= (0.7 – 0.272)m
= 0.428m

• When level is further lowered by 1.0m water level = 4.5m. Since water level is lower than hydrostatic draft ship has taken over block overall.

Then P = (TPC ⨯ mean rise)
= 0.679 ⨯ 100 ⨯ 17
P = 1154.3 tonnes.

GG1 = (P⨯KM)/W
= ( 1154.3 ⨯ 6.9 )/ 6220
= 1.280m

###### Solution –
• Trim = 3.60m, d = 74m

P = (trim ⨯ MCTC)/d
= (360 ⨯ 290)/74
= 1410.8 tonnes

GG1 = (P ⨯ KM)/W
= (1410.8 ⨯ 9.4)/19500
= 0.680m

Old GM = 0.5m,
GG1 = 0.680m

###### New GM = -0.18mSince GM is –ve it is not safe to dry dock vessel in this condition.
• Trim caused = Tc = (1410.8 ⨯ 74/290 ⨯ 100)
= 3.600m

Change in draft aft = 3.60 ⨯ 76/165
= 1.658m

Change in draft fwd = 1.942m
Mean rise = (1410.8/21.3 ⨯ 100)
= 0.662m

 Fwd Aft Draft 12.60m 9.00m -0.662m -0.662m 11.938m 8.338m -1.942m +1.658m 9.996m 9.996m

Method 2 =

Corn to aft draft = (3.6 ⨯ 76)/165
= 1.658m

Hydrostatic draft  = 10.658m,
Mean rise = 0.662m

###### Solution –

Clearly stern will take block once level of water reaches 5.8m. i.e equal to aft draft present level = 4.8m

Corn to aft draft = Trim ⨯ LCF/LBP
= 2.4 ⨯ 72.013/145
= 1.192m

Hydrostatic draft  = 4.608 m
Since present level of water is greater than hd ship has not taken block overall.

Reduction in aft draft = 100cm
Or, 100 = P/21.89 + (P ⨯ 72.013/162.7) ⨯ 72.013/145
Or, P1 = 376.64 tonnes

GG1 = P ⨯ KM/W
= 0.382m

Old GM = 1.151m, FSC = 900/(2013 – 376.64)
= 0.104m,

###### Solution –

Draft Fwd = 3.9 m,
Draft aft = 5.2m
Corn to aft draft = 0.631m
Hydrostatic draft = 4.568m
Now, Water  level = 4.5m.

Since water level is almost same to hydrostatic draft we will consider that ship has taken block aft but not overall.

Reduction in aft draft = 70cm
Or, 70 = P/14 + P ⨯ 68/130 ⨯ 68/140

Or, P = 215.06 tonnes
GG1 = P ⨯ KM/W
= 215.06 ⨯6.1/5210
= 0.252m

GM = (0.3 – 0.252)m
= 0.048m

Now Trim caused = 215.06 ⨯ 68/130 ⨯ 100
= 1.125m,
Ta = 1.125 ⨯ 68/140
= 0.546m, Tf = 0.578m

 Fwd Aft Draft 3.9m 5.2m -0.154m -0.154m +0.578m -0.546 4.324m 4.50m
###### Solution –

Once vessel has taken block allover change in trim = 2.8 m

P = 280 ⨯ 161/72.1
= 625.24 tonnes

GG1 = P ⨯ KM/W
= 625.24 ⨯ 9.57/8487
= 0.705m

GM = (0.868 – 0.705)
= 0.163m

Moment of statical stability = (W ⨯ GZ)
= W ⨯ GM SinӨ
= 8487 ⨯ 0.163 ⨯ Sin 10

###### Solution –

Clearly stern will take water level equal aft draft i.e 5.8m

Present level = 4.8m
Corn to aft draft = 1.238m, Hd( Hydrostatic draft ) = 4.562m
Since level ˃ Hd means stern has taken to block while forward end is shall afloat
Reduction in aft draft = 1.0m = 100cm
or, 100 =   P/21.9 + P ⨯ 72.2/162.5 ⨯ 72.2/140
Or, P = 363.90 t

GG1 = P ⨯ KM/W
= 0.374m

Virtual GM  = ( old GM – Virtual loss of GM )
= (old GM – 0.374)
= (1.18 – 0.374)
= 0.806m

Trim caused = 1.617m,
Rise = 0.116m,
Ta = 0.834, Tf = 0.783

 Fwd Aft Draft 3.400m 5.800m -0.166m -0.166m +0.783m -0.834m 4.017m 4.800m
###### Solution –

P = 60 ⨯ 155/80
= 116.25 tonnes
Old GM = 0.4m

GG1 = P ⨯ KM/W
= 116.25 ⨯ 7.2/11000
= 0.076m

New GM = (old GM – GG1)
New GM = 0.324m

Corn to aft draft = 0.6 ⨯ 80/180
= 0.267m

Hd (Hydrostatic draft )= 6.433m
Rise = 116.25/2200
= 0.053m

###### Solution –

GM should be 0.3m, then GG1 = 0.20m
We know that –
0.20m = P ⨯ 6.0/4740
P = 158 tonnes
We have P ⨯ d = (trim ⨯ MCTC)
= 158 ⨯ 60/79

Trim = 120cm
= 1.2m
Therefore, maximum permissible trim = 1.2m by stern

For unstable GG1 = 0.50m,
P = 305 tonnes
Trim = 3.00m
Ta + Tf = 1.5

Rise = P/ TPC ⨯ 100
= 305/1000
= 0.305m

Draft should be fwd = (4.22 – 1.5)m
= (2.72 – 0.305)m
= 2.415m

Aft = (4.22 + 1.5)m
= (5.72 – 0.305)m

###### Solution –

Question same as Q. No. – 8

###### Solution –

After level dropped by 1.25m,
New water Level = (10 – 1.25) m
= 8.75m

Correction to aft draft = (1.38 ⨯ 89.5)/ 185
= 0.668m

(hydrostatic draft ) Hd = 8.712m
Clearly since level of water 8.75m ˃ Hd 8.712m

It means stern has taken to the block and fwd end is still afloat.
Reduction in aft draft = 0.63m
Or, 63 =    P/29.6 + P ⨯ 89.5/4100 ⨯ 89.5/185
Or, P = 451.96 tonnes

We know that –
GG1 = P ⨯ KM/ W
= 451.96 ⨯ 11.50/29000
= 0.179m

Therefore, GM = 0.371m
Trim caused = Tc = 451.96 ⨯ 89.5/410 ⨯ 100
= 0.986m

###### Rise = 451.96/29.6 ⨯ 100= 0.153m

Ta = 0.477m, Tf = 0.509m

 Fwd Aft Draft 8.00m 9.38m -0.153m -0.153m +0.509m -0.477 8.356m 8.750m
###### Solution –
• When level lowered by 1m

Present level = 5.5m
Correction to aft draft = 2.2 ⨯ 62/120
= 1.137m
Hd ( hydrostatic draft )  = 5.063m
Since present level is 5.5m it means stern has taken to block and forward end is still afloat.

Reduction in aft draft = 0.7m
Or, 70 = P/15 + P ⨯ 62/110 ⨯ 62/12
Or, P = 195.6 tonnes

GG1 = P ⨯ KM/W
= 0.202m

GM = 0.398m

• When water level is further lowered by 1.0m

Level = 4.5m
Since water level < hydrostatic draft ship has taken over block over all.
Reduction in Hydrostatic draft = (5.063 – 4.5)m
= 0.563m

Or, 56.3 = P/15
Or, P = 844.5 tonnes

GG1 = P ⨯ KM/W
= 0.870m

###### Solution –

When level is lowered by 1.5m then water level = 5.0m
Correction to aft draft =( trim ⨯ LCF)/ LBP
= 1.031m
Hydrostatic draft = 4.669m

Since water level is greater than 4.669m it means stern has taken to block while forward end is still afloat.

Reduction in aft draft = 70cm
Or, 70 = P/18 + P ⨯ 75.5/150 ⨯ 75.5/145
Or, P = 220.4 tonnes

GG1 = P ⨯ KM/W
= 0.147m

Virtual GM = (0.6 – 0.147)m
= 0.147m

Trim caused = Tc = 220.4 ⨯ 75.5/150
= 110.93
Or = 1.109m

Rise = 220.4/18 ⨯ 100
= 0.122m

Ta = 1.109 ⨯ 75.5/145
= 0.577m

###### Tf = 0.532m
 Fwd Aft Draft 3.72m 5.70m -0.122m -0.122m + 0.532 -0.577m 4.130m 5.000m
###### Solution –

In this question KMT,MCTC & TPC is not given
Since this is a box shaped vessel both can be found.
MCTC = W ⨯ GML/100 ⨯ LBP
TPC = A/100 ⨯ density of water displaced

KMT = KB + BM
BMT = b2/12D
= 16 ⨯ 16/12 ⨯ 3.5
= 6.095
KM = (6.095 + 1.75)m
= 7.845m

For TPC
Area of water plane = (80 ⨯ 16)m2
= 1280m2
TPC = 1280/100 ⨯ 1.025
= 13.12 tonnes / cm

For MCTC
GML = (KML – KG)
& KML = KB + BML
BML=   I/V
I = lb3/12 for long   bl3/12

V = volume = l ⨯ b ⨯ d
= b ⨯ l3/12 ⨯ (l ⨯ b ⨯ d)
= l2/12d
= 80 ⨯ 80/12 ⨯ 3.5
= 152.38m

KB = depth/2
= 3.5/2
= 1.75m

KML = (1.75 + 152.38)m
= 154.13m

###### Therefore, GML = 149.93 = 150

W = (80 ⨯ 16 ⨯ 3.5 ⨯ 1.025)
= 4592 tonnes

MCTC = 4592 ⨯ 150/80 ⨯ 100
= 86.1 tm

P = 100 ⨯ 86.1/40
= 215.25 tonnes

GG1 = P ⨯ KM/W
= 0.367m

GM old = (KM – KG)
= (7.845 – 4.2)m
= 3.645m

###### Solution –

GM now = 0.5m
GM required = 0.3m
GG1 = 0.2m

We know that –
GG1 = P ⨯ KM/W
0.2 = P ⨯ 6.7/19200
P = 573.13 tonnes

Therefore, permissible P = 573.13 tonnes
We know P ⨯ d = trim ⨯ MCTC
= 573.13 ⨯ 74/145 ⨯ 100 = trim in metres

###### Solution –
• After taking block allover change in trim = 50cm

P = 50 ⨯ 80/40
= 100 tonnes
GG1 = P ⨯ KM/W
= 100 ⨯ 4.5/3000
= 0.15m

###### Residual GM = 0.3m
• Original hd (Hydrostatic draft)  = Aft draft minus correction.

Correction to aft draft = trim ⨯ LCF/LBP
= 0.5 ⨯ 40/84
= 0.238m

Hd = (4.5 – 0.238)m
= 4.262m
Rise = P/ TPC ⨯ 100
= 100/1200
= 0.083m

###### Therefore, Fwd & Aft = 4.178
• When unstable GG1 = 0.45

0.45 = P ⨯ 4.5/3000
P = 300 tonnes
Rise = P/ TPC ⨯ 100
= 0.25

###### Solution –

Correction to aft draft = ( 2 ⨯ 115)/240
= 0.958m

Hd (Hydrostatic  draft ) = 7.042m
After landing flat = draft is 6.2m
Change in Hydrostatic draft = (7.042 – 6.2)m
= 0.842m

Rise = P/ TPC ⨯ 100
= 0.842m

GG1 = P ⨯ KM/W
= P ⨯ KM/(W – P)
= 1.301m

###### Solution –
• When flat on blocks change in trim = 1.800m

P = 1.8 ⨯ 360/105
= 6.171m ⨯ 100
= 617.1 tonnes
GG1 = P ⨯ KM/W
= 617.1 ⨯ 8.20/38700
= 0.131m

###### New GM = 0.609m
• Correction to aft draft = 1.800 ⨯ 105/200

= 0.945m
Hd = 8.205m
Rise = P/TPC ⨯ 100
= 0.128m

###### Therefore, New GM = (8.205 – 0.128)m = 8.076m
• When draft is 7.8m

Reduction in hydrostatic draft = rise = (8.205 – 7.8)m
= 0.405m
0.405 = P/4800
P = 1944 tonnes

GG1 = P ⨯ KM/W
= 0.417m

###### Therefore, New GM = 0.423m
• When she become unstable GG1 = old GM

= (8.2 – 7.46)/Old GM
= P ⨯ 8.2/38700
P = 3492.43 tonnes

Rise = P/TPC ⨯ 100
= 0.727m

###### Solution –

Old GM = KM – KG
= (7.8 – 7.4)m
= 0.4m

New GM = 0.1m ( As per question )
We know that –
GG1 = P ⨯ KM/W
0.3 = P ⨯ 7.8/26240
P = 1009.23

###### Solution –

Old GM  =  (7.12 – 6.96)m
= 0.16m

GG1 permissible = 0.16m
0.16 = P ⨯ 7.12/21480
P = 482.696 t

###### Solution –

Correction to aft draft = 1.3 ⨯ 96/200
= 0.624m

Hd( hydrostatic draft)= 6.976m
Change in draft = rise = 0.176m
0.176 =      P/18 ⨯ 100
P = 316.8 tonnes

###### Solution –

Change in trim = 3.3m
P = 3.3 ⨯ 350 ⨯ 100/110
= 1050 tonnes

GG1 = P ⨯ KM/W
= 1050 ⨯ 9/43200
= 0.218m

###### Solution –

Present GM = 0.36
Required = 0.15m

GG1 = 0.21m
0.21 = P ⨯ KM/W
P = 500 tonnes

Now, P ⨯ d = trim ⨯ MCTC
Trim = 500 ⨯ 98/210 ⨯ 100
Run down = 0.1m per 100m will provide 0.18m trip