89.NA February 2020 Q.8(b)

A ship of 130 m long displaces 14000 tonnes when floating at draught of 7.5 m forward and 8.10 m aft. GML  - 125 m, TPC - 18, LCF - 3 m aft of midships. 
Calculate the final draught when a mass of 180 tonnes lying 40 m aft of midship is removed from the ship.

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GIVEN

L = 130 m
W = 14000 t
df = 7.5 m
da = 8.10 m
GML = 125 m
TPC = 18
LCF = 3 m aft of midship
m = 180 t
d = 40 - 3 = 37 m

TO FIND

a)Draught Forward final
b)Draught Aft final

SOLUTION

As it is given that 180 t mass is removed from the ship. So the ship draught will reduce.So ship will rise in water.

We know that,
Bodily rise = Mass removed / TPC
                   = 180 / 18
                   = 10 cm

As we know that,
MCT1 cm = (W x GML ) /  (100 x L)
                  = (14000 x 125) / (100 x 130)
                  = 134.6 tonne meter

Change in trim, t = Trimming moment / MCT1 cm
                            = (m x d) / MCT1 cm
                            = (180 x 37) / 134.6
                            = -49.48 cm by stern

Change in draught  forward = -(t/L) x [(L/2) + LCF]
                                             = -(-49.48 / 130) x [(130 / 2) + 3]
                                             = +25.88 cm

Change in draught aft = (t/L) x [(L/2) - LCF]
                                    = (-49.48 / 130) x [(130 / 2) - 3]
                                    = -23.60 cm

a) New draught Forward = Initial draught - Bodily rise + Change in draught Forward
                                        = 7.5 - 0.1 + 0.2588
                                        = 7.659 m

b) New draught Aft = Initial draught - Bodily rise + Change in draught Aft
                                 = 8.10 - 0.1 -  0.2360
                                 = 7.764 m