3.NA September 2018 Q.8 (B)

A ship 120 m long floats at draught of 5.50 m forward and 5.80 m aft ; MCT1 cm 80 tonne m , TPC 13 , LCF 2.5 m forward of midships. Calculate new draught when a mass of 110 tonne is added 24 m aft of midship.

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Given

L = 120 m
df = 5.5 m
da = 5.8 m
MCT1 = 80 t m
TPC = 13
LCF = 2.5 m fwd of midship
m = 110 t
distance = 24 m aft of midship

To find

1. New draught

Solution

Bodily sinkage = mass added / TPC
                         = m / TPC
                         = 110 / 13
                         = 8.461 cm

d = distance the new mass added + LCF
   = 24 + 2.5
   = 26.5 m

We know that Change in trim, t = Trimming moment / MCT1 cm
                                                   = (m x d) / MCT1 cm
                                                   = (110 x 26.5) / 80
                                                   = 36.44 cm by stern (As weight is added aft)

Distance from LCF to forward, FL = (L/2)  - 2.5 (As LCF is fwd)
                                                       = (120/2) - 2.5
                                                       = 57.5 m

Distance from LCF to aft , WF = (L/2)  + 2.5 (As LCF is fwd)
                                                 = (120/2) + 2.5
                                                 = 62.5 m

Change in trim forward = - (t/L) x FL
                                      = - (36.44/120) x 57.5
                                      = - 17.46 cm

Change in trim aft = + (t/L) x WL
                             = + (36.44/120) x 62.5
                             = + 18.98 cm

New draught forward = Initial draught fwd + Bodily sinkage + Change in trim fwd
                                   = 5.5 + 0.0846 - 0.1746
                                   = 5.41 m

New draught aft = Initial draught aft + Bodily sinkage + Change in trim aft
                           = 5.8 + 0.0846 + 0.1898
                           = 6.074 m