Stability – I : Chapter 11

17. A ship of W 16000t, KM 7.5m, KG 6.0m , TPC 25, is listed 3 degree to port .Her present mean draft is 8.6m and she is to finish loading at 8.8m mean draft . Space is available 5m 0f the centre line, on either side. State how much cargo must be stowed on either side to finish upright

Solution :

W = 16000t,
KM, 7.5m KG = 6.0m,
Initial  GM = 1.5 m
TPC = 25,
List  = 3 degree P
Present draft = 8.6m
Final draft = 8.8

Sinkage available = (8.8 – 8.6)m
= 0.2m
= 20cm

Now , Cargo to load = (Sinkage x TPC)
= (20 x 25)
=  500 t

We know that :
LM  =(W x GM tanθ)

Again , Initial LM = (16000 X 1.5 X Tan 3⁰)
= 1257.78 tm

Let  ‘X’  t 0f cargo will load on port side
Cargo to be load on stbd side = (500 – X) t
So, final LM caused due to loading of cargo would be

LM(1)   =  (Weight x distance)
= (X x 5)
=   5X (P)

LM(2)   =  (Weight x distance)
=   (500 –X) x  5
=   (2500 – 5X ) (S)

As per above calculation ,
Initial LM =    1257.78 ( P)

Hence  to keep vessel upright ,
LM(S)   =    LM(P)
(2500 – 5X)  = (1257.7 + 5X)
2500 – 5X – 5X   = 1257.7
(- 10X)  = (1257.7 – 2500)
X  = (1242.3/10)
= 124 .3 t

Hence cargo  loaded is  124.23 t on  port side
So, cargo loaded in stbd is (500 – 124.23)
= 375.7 t