1. In a vessel of 8800 tonnes displacement and KG 6.2m, 200 tonnes of cargo was loaded in the lower hold 1.7m above the keel. Find the final KG.
Solution:
Given:-
Weight = 8800t, Original KG =6.2m, Cargo loaded = 200t, KG = 1.7m
Final W – 9000 t Final VM 54900 tm
Final KG = (Final VM/ Final W) Final KG = (54900 / 9000) = 6.1m
Here ‘VM’ stands for vertical moments.
2. 600 tonnes of cargo were discharged from a vessel from the upper 11m above the keel. In the original KG displacement were 6m and 12600 tonnes, calculate the KG.
Solution:-
Given:- Weight =12600 t Original KG =6m Cargo discharged = 600 t,
Cargo discharge 11 m above the keel
Final Weight =12000 t Final VM =69000tm
Final KG = (Final VM / Final Weight)
Final KG = ( 69,000 / 12000)
= 5.75m
3. In a vessel of 9900 tonnes displacement and KG 4m, a heavy lift of 100 tonnes is loaded on the UD (KG 15m ). Find the final KG
Solution:
Final W 10000t Final VM =41000tm
Final KG = (Final VM / Final Weight) Final KG = ( 41000 / 10000 ) = 4.11m.
4. 500 tonnes of cargo of discharge from the lower the lower hold (KG 3m ) of a vessel whose displacement and KG before discharging were 11500 tonnes and 6.3m . Find the final KG.
Solution:
Given:-
Weight = 11500t Initial KG = 6.3m Cargo discharged = 500t, KG =3m
Final W =11000t Final VM =70950tm
Final KG = (Final VM / Final Weight) Final KG = (70,950/11000) =6.45m
5. 500 tonnes of cargo was shifted 15 meters vertically downwards in a vessel of 10000 tonnes displacement. Find the effect it has on the KG of the vessel and whether KG increases or decreases.
Solution:
Given:-
Weight =10000t, Cargo shifted downward =500t,
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
6. In a vessel of 9000 tonnes displacement , KG 10.5m, 300 tonnes of cargo was shifted from the LH (KG 2.5m)to the UDC (KG 11.5m). Find the resultant KG of teh vessel.
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.
Final W = 9000 Final VM =200t
Final KG = (Final VM / Final Weight) Final KG = (97200 / 9,000) = 10.8m.
7. In a vessel of 9009 tones displacement, KG 8.7m, how many tonnes of cargo can be loaded on the upper deck (KG 15m ) so that the final KG would be comes 9m.
Solution:-
Given:- Weight = 9009 t, Original KG =8.7m,
Let cargo loaded be X tonnes Final KG =9m
Final W = (9009 + X) t Final VM =(78378.3 + 15X) tm
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
8. A heavy lift derrick , whose head is 20m above the keel , is to shift a locomotive weighing 300 tonnes from the UD (KG 8m ) to the LH (KG 2m). If the displacement and initial KG of the vessel were 12000 tonnes and 7.6m, find the KG of the vessels
a. When the derrick has taken the weight off the UD and
b. After shifting is over.
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)
Case – 1
Final W =12000 t Final VM =94800tm
Final KG = (Final VM / Final Weight) FKG = (94800 / 12000) = 7.9m
Case – 2
Final W =12000 Final VM = 89400tm
Final KG = (Final VM / Final Weight) Final KG = (89400 / 12000) = 7.45m.
9. On a vessel of 4,950 tonnes displacement, KG 9.2m, the ship’s jumbo derrick is to be load a weight of 50 tonnes from the wharf , on to the UD( KG 8m) . if the head of the derrick is 25m above the keel , calculate the KG of the vessel
a. when the weight is hanging by the derrick on the centre line but 2m above the UD , and
b. after loading .
Solution:
Given:- Weight = 4950t, Initial KG = 9.2m Cargo loaded = 50t , UD KG =8m Height of the derrick = 25m
Case – 1
Final W = 5000t FVM = 46790tm
Final KG = (Final VM / Final Weight) Final KG = (46790 / 5000) = 9.358m
Case – 2
Final W = 5000t Final VM = 45940tm
Final KG = (Final VM / Final Weight) Final KG = (45940 / 5000) = 9.188m
10. A ship’s derrick, whose head is 22m above the keel, is used to be discharge a weight of 20 tonnes (KG 5m), lying on the centre line. If the vessel’s displacement and KG before discharging were 6000 tonnes and 8m. Calculate the KG
a. As soon as the derrick lifts the weight and
b. After discharging .
Solution:-
Given:- Weight = 6000t, Original KG = 8m, Cargo discharged = 20t, KG of cargo discharged = 5m Height of the derrick = 22m
Case – 1
Final W = 6000 Final VM 48340 tm
Final KG = (Final VM / Final Weight) Final KG = (48340 / 6,000) = 8.056m
Case – 2
Final W =5980t Final VM = 47900 tm
Final KG = (Final VM / Final Weight) Final KG = (47900/5980) = 8.010m.