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 