Miscelleneous

C L DUBEY – EXERCISE – 08 (DRY DOCKING)

  1.    A ship of 19200 M.T. displacement, 235m long has KM 10.2m, KG 9.4m, TPC 32 MCTC 254tm is floating at draft forward of 9.42m and aft 10.88m in water of RD 1.005. The centre of floatation is 5m abaft amidships, centre of buo yancy is 3m forward of amidships. Vessel is drydocked in water of density 1.020. Calculate her virtual GM on taking blocks all over.
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

About the author

Manish Mayank

Graduated from M.E.R.I. (Mumbai). A cool, calm, composed and the brain behind the development of the database. The strong will to contribute in maritime education and to present it in completely different and innovative way is his source of inspiration.

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