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# Chapter 11- Stability Calculation

Updated: Jun 5, 2020

### 1. On a ship of W 5000t , GM 0.3m, 20t was shifted transversely by 5m. Find the list.

Solution:

W = 5000t, GM = 0.3m, Weight shifted (w) = 20t & d= 5m

Listing moment = (20 x 5 ) = 100tm Tanθ = ( Listing Moment) /(W x GM) = 100/(5000 x 0.3) = 3.8 degree.

### 2. On a ship of W 8000t, GM 2.0m, If the following transverse shifting were done , find the list:

• 200cargo shifted 4m to stbd

• 100t cargo shifted 2 m to port

• 100t cargo shifted 4m to port

• 50t stores shifted 20m to stbd

Solution:-

Listing moment due to shifting of cargo LM(1) = ( weight x distance ) = (200 x 4) = 800 tm (S)

Again, LM (2) = (Weight x distance ) = (100 x 2) = 200 tm (P)

LM(3) = ( weight x distance ) = (100 x 4) = 400 tm (P)

LM(4) = ( weight x distance ) = (50 x 20) = 1000 tm (S)

Final listing moment (LM) = LM(1) + LM(2) +LM(3)+ LM(4) = 800(S) + 200(P) + 400(P) + 1000(S) =1800(S) + 600(P)

So, final listing moments = (1800 – 600) = 1200 tm (S)

Now, Tanθ = (Final LM )/(W x GM) = (1200/(8000 x 2 ) = 4.29 degree (S)

### 3. If 200t cargo was shifted downwards by 10m and to starboard by 5m on a ship of W 10000t , KG 7.0m, KM 7.4m, find the list.

Solution :-

Given : Cargo shifted (w) = 200t, Distance = 10m, d = 5m from CL to stbd W = 10000 t , KG = 7.0m & KM = 7.4m Final W =10000t Final VM= 68000 tm Final KG = (final VM)/ (Final W) Final KG = (68000/10000) = 6.8m

Final GM= (KM – KG) = (7.4 – 6.8) = 0.6m

Listing moment (LM) = (weight x distance ) = (200 x 5) = 1000tm

tanθ = LM/(W x GM) = 1000 /(10000 x 0.6) = 9.46 degree

### 4. A quantity of grain estimated to be 100t shifts transeversly by 12m and upwards by 1.5m, on a ship of W 12000t , GM 1.2m. Find the list caused.

Solution:

Grain shifted (w) = 100t, Distance = 12m & transversely, d = 1.5( î) W = 12000 t, GM = 1.2m

LM caused = (weight x distance) = (100 x 12) = 1200 tm

GG1 (î) = (distance x weight) /W = (1.5 x 100) / 12000 = 0.0125 m

Final GM = (GM – GG1) = (1.2 – 0.0125)m = 1.1875m

Tanθ = LM /(W x GM ) = 1200/(12000 x 1.1875) = 4 degree 8’.

### 5. A ship displaces 4950t and has KG 4.85m, KM 5.79m. cargo weighing 50t is loaded 1.25m above the keel and 4m port of the centre line . Find the list

Solution:

W = 4950 t , KG = 4.85m KM = 5.79m Weight loaded (w) = 50 t KG = 1.25m d = 4 m to CL to port

LM caused = (weight x distance ) = (50 x 4) = 200 tm to port . Final W = 5000 t Final VM = 24070 tm Final KG = (Final VM)/( Final W) Final KG = (24070/5000) = 4.814m

Final GM = (KM –KG) = (5.79 – 4.814) m = 0.976m.

Tanθ = LM/(Weight x GM) = 200/ (5000 x 0.976) θ = 2.34 degree

### 6. A weight of 100t is discharge from a position 2.45m above the Keel and 6m to port of the centre line of a ship of w 10000t, KM 8.25m, KG 7.45m. Find the list

Solution:

Weight discharged (w) = 100t, & KG = 2.45m Transverse distance = 6m to part from CL W = 10000 t, KM = 8.25 m, KG = 7.45m LM caused = (weight x distance) = (100 x 6) tm = 600 tm (P) Final W = 9900 t Final VM = 74255 tm Final KG = (Final VM)/ (Final W) Final KG = (74255 / 9900) = 7.5m

Final GM = (KM –KG) = (8.25 – 7.50) m = 0.75m

Tanθ = LM /( W x GM) = 600/(9900 x 0.75) = 4.62 degree.

### 7. A ship of 10000 t displacement, KG 8.3m carries out the following cargo operation: If the final KM is 9.6m, find the list.

Solution: Final W = 9000 t Final VM =81560 tm Final LM = 700(S)

Final KG = (Final VM)/ (Final W) Final KG = (81560 / 9000) = 9.062 m GM = (KM – KG) = (9.6 – 9.062) = 0.538m

Tanθ = (Final LM)/(final W x GM) = (700)/(9000 x 0.538) = 8.2 degree

### 8. A ship of W 9000t, KG 8.3m loads 600t of cargo (KG 4.0m, 3m to port of the centre line ) and discharge 400t of cargo ( KG 9.0m, , from 5m to port of the centre line ). 200t of cargo is then shifted upwards by 5m and to starboard by 8m. 300t of cargo is then is then shifted 1m downwards and 4m port .Find the list if the final KM is 8.95m.

Solution : Final W = 9200 t Final VM = 74200 tm FLM = 600(S)

Final KG = (Final VM)/(Final W) Final KG = (74200)/(9200) = 8.065m

Final GM = (KM – KG) = (8.95 – 8.065) = 0.885m

tanθ = (600 )/(9200 x 0.885) = 4 degree ( S)

### 9. A ship of W 18000 t, KG 7.75m discharge 1500t (6.0m above the keel and 3m port of the centre line ) and loads 500t (10m above the keel and 4m port of the centre line ).

• Cargo was then shifted as follows ;

• 500t upwards 2m and to starboard 4m

• 800 t downwards 2m and to port 3m.

If the final KM is 8.935 m, find the list .

Solution: