Stability – I : Chapter 6 |

# Stability – I : Chapter 6

###### Solution:

Given:-

Weight = 8800t,
Original KG =6.2m,
KG = 1.7m

###### VM
8800 t 6.2m 54560tm
(+) 200 load 1.7m Final VM  54900 tm

Final W – 9000 t                                               Final VM  54900 tm

So, we can calculate

Final KG = (Final VM/ Final W)
Final KG = (54900 / 9000)
= 6.1m

Here ‘VM’ stands for vertical moments.

###### Solution:-

Given:-
Weight =12600 t
Original KG =6m
Cargo discharged  = 600 t,

Cargo discharge 11 m above the keel

###### VM
12,600t   6m 75,600tm
( -) 600t ( discharge) 11m  (-) 6600tm

Final Weight =12000 t                               Final VM =69000tm

We know that :

Final KG         =        (Final VM / Final Weight)

Final  KG         =          ( 69,000 / 12000)

= 5.75m

###### VM
12,600t    4m 39,600tm
(+)100 t  load  15m  ( +)1500tm

Final W 10000t                                             Final VM =41000tm

We know that:

Final KG         =        (Final VM / Final Weight)
Final KG        =   ( 41000 / 10000 )
=   4.11m.

###### Solution:

Given:-

Weight = 11500t
Initial KG = 6.3m
Cargo discharged = 500t, KG =3m

###### VM
11500t  4m 39,600tm
(+)100 t  load  15m  ( +)1500tm

Final W =11000t                                          Final VM =70950tm

We know that:

Final KG         =        (Final VM / Final Weight)
Final KG         =    (70,950/11000)
=6.45m

###### Solution:

Given:-

Weight =10000t,
Cargo shifted downward =500t,

We know that:-

GG1()  = (w x d)/( total weight)
= (500 x 15)/ 10000
=0.75m

Since, The weight is shifted vertically   downward,  the KG of ship will also go down parallel to the KG of the weight

Hence, KG of the ship decrease by 0.75m

###### Solution :

Weight =9000 t
Original KG = 10.5m ,
Cargo shifted  = 300 t

Vertical Distance of cargo shifted
= (height of UD – height of LH)
= (11.5 -2.5)
= 9m upwards.

###### VM
9000t   10.5m    94500tm
(+)300t  9m (+)2700tm

Final W = 9000                                             Final VM =200t

We know that :

Final KG         =        (Final VM / Final Weight)
Final  KG           = (97200 / 9,000)
= 10.8m.

###### Solution:-

Given:-
Weight = 9009 t,
Original KG =8.7m,

Let cargo loaded be   X   tonnes
Final KG =9m

###### VM
9009 t     8.7m    94500tm
(+) X t (load)  15m (+) (15 X) tm

Final W = (9009 + X) t                 Final VM =(78378.3 + 15X) tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final  KG     = (78378.3 +15X ) /(9009 + X)
9.0         = (78378.3 + 15X)/ (9009+X)
9 x (9009+X)  = (78378.3 + 15X)
(81081 + 9X) =   (15X  + 78378.3)
Hence, 6X    = 78378 .3 – 81081
6X   = 2702.7t
X  = 450.45 t

###### Solution:

Given:
Weight = 12000t,
Initial KG =7.6m,
Cargo shifted = 300t,
Change of KG = (8 – 2)
= 6m downwards

Derrick Height = 20m,
KG of the Weight = 8m

(since the weight is on the UD)

###### VM
12000 t      7.6m    91200tm
300t (shift)  12m  (+) 3600 tm

Final W =12000 t                                          Final VM =94800tm

We know that:

Final KG        =        (Final VM / Final Weight)
FKG              =          (94800 / 12000)
= 7.9m

###### VM
12000 t      7.6m    91200tm
300t (shift)  6m  (-)1800 tm

Final W =12000                               Final VM = 89400tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final  KG         =   (89400 / 12000)
= 7.45m.

###### Solution:

Given:-
Weight = 4950t,
Initial KG = 9.2m
UD KG =8m
Height of the derrick = 25m

###### VM
4950 t           9.2m      45540 tm
(+) 50t (load) 25m    (+) 1250 tm

Final W = 5000t                                    FVM = 46790tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final KG   = (46790 / 5000)
= 9.358m

###### VM
4950 t           9.2m      45540 tm
(+) 50t (load) 8m    (+) 400 tm

Final W = 5000t                                      Final VM = 45940tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final KG = (45940 / 5000)
= 9.188m

###### Solution:-

Given:-
Weight = 6000t,
Original KG = 8m,
Cargo discharged = 20t,
KG of cargo discharged = 5m
Height of the derrick = 22m

###### VM
6000t   8m       48 000 tm
20t 17m   (+) 340 tm

Final W = 6000                                  Final VM    48340 tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final KG = (48340 / 6,000)
= 8.056m

###### VM
6000t   8m       48 000 tm
(-)20t (discharge)    5m   (-) 100 tm

Final W =5980t                                          Final VM = 47900 tm

We know that:

Final KG        =        (Final VM / Final Weight)
Final KG      = (47900/5980)
= 8.010m. 