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:
Final W = 17000 Final VM = 134900 tm FLM =2100(S) Final KG = (Final VM) / (Final W) Final KG = (134900/17000) = 7.935m
Final GM = (KM –KG) = (8.935 – 7.935) = 1.0 m
We know that : Tan θ = Final LM/(Final W x GM) = 2100 /(17000 x 1) = 7 degree 2.6min
10. A ship listed 8degree to port , displaces 12000t and has KM 7.54m and KG 6.8m . Find how many tonnes of SW ballast must be transfered from NO 2 port DB tank to NO2 stbd DB tank , to up right and vessel , if the tank- centre are 10m apart .
Solution :
List = 8 degree (P) , Displacement (W) = 12000t, KM = 7.54m & KG = 6.8m GM = (KM – KG) = (7.54 – 6.80) = 0.74m
listing moment (LM)= (W x GM x tanθ) = (12000 x 0.74 x tan80) = 1248 tm Listing moment (LM) = (W x D) 1248 tm = (W x 10m) W =124.8 t
Here to upright the vessel the lighting moment caused on particle must be same at stbd side. So we have to transfer 124.8t of sea water to the stbd side of DB tank to upright the vessel.
11. A ship displacing 4000t has GM 1.0m (KM 10.0 m and KG 9.0 m) and is listed 10 degree to port .If 16000t of cargo is now loaded on the centre line and the final GM is 1.0m KM 9.0m and KG 8.0m state weather the list would change .if yes , find the new list .
Solution :
Displacement (W) = 4000t, GM = 1.0m (initial) , list = 10 degree (P) W = 16000t, G M = 1.0m (final) Listing moment (LM) = (W x GM tanθ) = (4000 x 1 x tan 100) = 705.3 tm
According to question , 16000 t of cargo loaded So Final W = (16000 + 4000) = 20,000t
But, Initial GM = Final GM Initial LM = Final LM Tanθ = (FLM/(W x GM ) = 705.3/(20000 x 1) = 2 degree 1 minute ( P).
Hence we can say that,(Yes) list would change due to the loading of cargo.
12. A ship of W 10000t. GM 1.5m, is listed 5 degree to stbd . If cargo is shifted vertically until her final GM is 0.5m, state whether the list would change . if yes find the new list .
Solution :
Displacement (W) = 10000t, GM = 1.5m, list = 5 degree Final GM = 0.5m Listing moment (LM)= (W x GM tanθ) = (10000 x 1.5 x tan50 ) = 1312.33 tm (S)
According to question, cargo shift vertically and now GM becomes 0.5 m Hence, Due to change of GM , list must be change . Tanθ = LM / (W x GM) = 1312.33/(10000 x 0.5) = 14.7dgree stbd
13. A ship of W 8500t, KM 9.0m, KG 8.3 , is listed 8 to stbd . The following cargo operation were carried out :
200 t discharge KG 4m from 5m stbd of CL .
300t discharged KG 5m from 2m port of CL .
100t loaded KG 2m, 4m to stbd of CL .
200t shifted up by 2m and port by 3m.
If the final KM is 9.3m , find the final list .
Solution:
Final W = 8100t Final VM = 68850tm FLM=289.23tm Final GM = (Final VM )/(Final W) Final KG = (68850 /8100) = 8.5m
Final GM = (KM – KG) = (9.3 – 8.5) = 0.8m tanθ = Final LM/(W x GM) = 289.23/(8100 x 0.8) = 2degree 33.3minute
14. A ship of 15000t displacement, KG 8.7m KM 9.5m is listed 10deg to port. The following cargo work was carried out:
500t loaded KG 8.0m, 5m stbd of CL .
300t discharged, KG 4.0m, 4 port of CL.
Find the quantity of SW ballast that must be transferred transversely to bring the vessel up right , the tank centres the being 12m apart.
NOTE : Since vessel is required upright , it is not necessary to calculate the final KG or GM unless specifically asked).
Solution:
Displacement (W) = 15000t, KG = 8.7m, KM = 9.5m & list = 10 degree (P) Final W = (15000 + 500) – 300 = 15200 t Initial LM = 15000 x 0.8 x tan 10 = 2115.9 tm (P) Listing moment caused LM(1) = (500 x 5) = 2500 tm (S)
LM (2) = (300 x 4) = 1200 tm (S)
Total listing moment = (2500 + 1200) tm = 3700 tm (S)
According to question and calculation done above , Initial listing moment = 1584.1 tm (S)
So, maintain vessel upright same LM must be created on (P) side. Again, LM = w x d 1584.1 tm = (w x 12) = (1584.1/12) = 132 t
Hence , 132t SW should transferred to (P) side to keel the vessel upright.
15. A bulk is carrier presently of 12250t, KM 9.8m , KG 9.0m has a list of 6dee to starboard .she then load 1250 t of ore (KG 8m , 2m to stbd of centre line ) and discharges 250t of ore (KG 2m 5m , from star board of centre line). 160t of SW ballast is then transferred from the stbd shoulder tank to the port DB tank vertically downwards by 9m and transversely by 10m ) Find the final list assuming that they are no slack tanks given the final KM is 9.6m .
Solution:
Final W = 13250t Final VM = 118310 tm FLM =766.57(S)
Final KG = (Final VM) /(Final W) = (118310/13250) = 8.929 m
Again,
Initial LM = ( W x GM tanθ) = (12250 x 0.8 x tan 6.50) = 1116.57 (S)
GM = ( KM – KG) = (9.6 – 8.929) = 0.671m
tanθ = ( Final LM)/( W x GM) = (766.57)/( 13250 x 0.671) = 4.93 degree(S)
16. From a ship of W 8000t KM 8.6m KG 8.0m some deck cargo was washed overboard KG 10m, 8m from the centre line .if the resultant list is 3 degree, find the quantity of cargo lost.
Solution:
Final W = (8000 – X) t Final VM =(64000 – 10X)tm Final KG = (Final VM)/(Final W) = (64000 – 10X)/(8000 –X)
Again, Final GM = (KM – Final KG) = 8.6 – (64000 – 10X )/(8000 – X) = 8.6 (8000 -X ) – (64000 – 10X) /(8000 –X) = (68800 – 8.6X – 64000 + 10X)/(8000 – X) = (4800 + 1.4X)/(8000 – X) Tan θ = (Final LM)/ (Final W x Final GM)
= 8X / ((8000 – X ) x (4800 + 1.4X)/(8000 –X)) = 8X /(4800 + 1.4X) 8X = tan 30 x (4800 + 1.4X) = (251.56 + 0.073X) (8X – 0 .073X) = 251.56
Hence, X = (251.56/7.927) = 31.73t
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 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
18. A ship displacing 12000t has KM 9.0m KG 7.25m. A 200t heavy lift is to be loaded by ship’s jumbo whose head is 24m above the Keel . Find
The list as soon as the derrick picks up the weight from the wharf on the stbd side with an outrech of 15m.
The list when the weight is placed on the upper deck KG 10m , 7m stbd of the centre line.
Solution:
Final W = 12200 Final VM = 91800 tm FLM = 3000 Final KG = (Final VM/Final W) Final KG = (91800/12200) = 7.524m
Again , GM = (KM – KG) = (9.0 – 7.524) = 1.476m tanθ = Final LM / (W x GM) = 3000/(12200 x 1.4476) = 9 degree 45min
Final W= 12200 t Final VM = 89200 FLM 1400 S
Final KG = (Final VM/Final W) Final KG = (89200/12200) = 7.311m
Again, Final GM = (KM –KG) = (9.0 – 7.311) = 1.689m
tanθ = 1400/(12200 x 1.689) = 3 degree 53.2min
19. A ship of W 10000t, KM 7.3m , KG 6.8m is listed 5degre to port . A heavy lift weighing 100t, lying 6m to port 0f the centre line and KG 10.0 m, is to be shifted to the lower hold KG 2.0m on the centre line of the ship , by the ship’s jumbo derrick whose head is 28m above the keel. Find
The list as soon as the derrick takes the load .
The list when the derrick swings to the load to the centre line.
The list after the shifting is over.
Solution:
Displacement (W) = 10000t, KM = 7.3m & KG = 6.8m , List = 5 degree(P) w = 100t, d = 6m to port of the CL KG = 10m & KG = 2.0m Derrick head = 28m . Initial LM = (W x GM x tanθ) = (10000 x 0.5 x tan50) = 437.44 P
Final W =10000t Final VM = 69800tm FLM=437.44 P Final KG = (Final VM/Final W) = (74800/10000) = 6.98m
Again , Final GM = ( KM – KG) = (7.3 – 6.98) = 0.32m
tanθ =LM/(W x GM) = 437.44/(10000 x 0.32) = 7.78 degree P
Case – 2 List = ( 100 x 6) = 600 tm (S)
Tan θ = (LM/ (W x GM) = 600/(10000 x 0.32) = 10.6 degree
Initial LM as we calculated = 437.44 tm(P) LM caused = 600 tm (S)
Now resultant listing moment = (600 – 437.44) = 162.56 tm S
Again,Tanθ = 162.56 /(10,000 x 0.32) = 2.9 degree S
Case – 3
Final W= 10000t Final VM= 67200tm FLM= 162.56 S Final KG = (Final VM/Final W) = (67200/10000) = 6.72m
Now, Final GM = (KM – KG) =(7.3 – 6.72)m = 0.58m
Again, tanθ = Final LM/ (W x GM) = 162.56 /(10000 x 0.58) = 1.6 degree (S)
20. A ship of W 13000t, KM 8.75m, KG 8.0m, has the list of 6degree to starboard . A heavy-lift weighing 150t lying on the upper deck 9m above the keel and 5m stbd of the centre line , is to be discharge using the ship’s jumbo derrick whose head is 22m above the keel . Calculate
The list as soon as the load is taken by the derrick.
When the load the hanging over the port side of the ship with an outreach of 10m from the line .
After discharging the heavy-lift.
Solution:
Given :
Displacement (W) = 13000t, KM = 8.75m & KG = 8.0m Initial GM =0.75m List = 6 degree stbd , w = 150t, KG = 9m lying = 5m to stbd ofC derrick = 22m
Case – 1
Final W= 13000t Final VM= 105950tm FLM =1024.76 Final KG = (Final VM/Final W) = (105950/13,000) Final KG = 8.15m
Again Final GM = (KM- KG) = (8.75 – 8.15) = 0.6m
Now, Tan =FLM /(W x GM) = 1024.76/(13000 x 0.6 ) = 7degree 48minute (S)
Case – 2
When the load is hanging over the portside of the ship with on outreach of 10m from CL. In this case the GM will be same but listing moment will change.
Listing moment caused =( W x d) = (150 x 15) = 2250 tm (P)
Hence, Final LM = (2250 – 1024.76) = 1225.24 P
Tanθ = 1225.24/(13000 x 0.6) = 8.9 degree P
Case – 3
After discharging the heavy lift.
Final W= 12850 Final VM = 102650 tm FLM = 274.76(S) Final KG = (Final VM/ final W) Final KG = (102650/12850) Final KG = 7.988m
Now,Final GM =( KM – KG) =(8.75 – 7.988) = 0.762m Tanθ = FLM/( W x GM) = 274.76/(12850 x 0.762) = 1.6 degree (S).
21. A ship of 10000t displacement is floating in SW and has KM of 10.8m and KG of 9.0m .she is listing 10degre to stbd. she has two rectangular deep tanks, one on either side , each 12m long ,12m wide and 9m deep . The stbd tank is full of FW while the port one is empty . If FW is to be transferred from the stbd tank to the port one , find
The quantity of FW to to bring the ship upright.
The list if one third of the original FW in the stbd tank is transferred to the port tank.
NOTE : Fluid GM should be used here.
Solution :
Displacement(W) = 10000t, KM = 10.8m, KG = 9.0M & GM = 1.8M List = 10 degree stbd
Volume of tank = (L x B x H) = (12 x 12 x 9 )m3
Initial LM = (W x GM x tan 100 ) = (10000 x 1.8 x tan 100) = 3173.88tm
Case -1
To bring the ship upright the stbd LM must be equal to port side LM.
LM = (Weight x Distance) 3173.88 tm = ( W x 12)
Hence W = (3173.88/ 12) = 264.49 t
Case -2
Volume of the tank = (L x B x D) = (12 x 12 x 9) = 1296m3
Now, Weight of FW on stbd tank = (Volume x Density) = (1296 x 1) = 1296 t.
Now, calculating 1/3 of mass of FW = (1296 x 1/3) = 432t.
Final W = 10000 Final VM = 87408 tm 2010.12 P
Final KG = (Final VM/Final W) Final KG = (87408/10000) = 8.741m
Solid GM = (KM – KG) = (10.8 – 8.741) = 2.059m
FSC = (LB3 x di) / (12 x W) = (12 x 123 x 1) /(12 x 10000) = 0.3456m
Fluid GM = (2.059 – 0.3456)
tanθ = (2010.12/(10,000 x 1.7134) =6.69 degree P
22. On a ship 8000t displacement , 50t is shifted transversely by 4m. Find the list if the total FSM is 1216tm, KM 7.0m , KG 6.4m.
Solution :
W = 8,000t , w = 50t , transverse list =4m FSM = 1216t , KM = 7.0m , KG = 6.4m
GM = 7 – 6.4 = 0.6m
LM = W x D = 50 x 4 = 200t
FSC = FSM / W = 1216 / 8000 = 0.152m
Fluid GM = 0.6 – 0.152 = 0.448m
tanθ = 200 / 8000 x 0.448 = 3.2deg11.6minut
23. A ship has W 10000 t, KM 7.8m , KG 7.075m, and is upright . NO3 port and stbd DB tanks are full of HFO RD 0.95. Each tank is rectangular , 15m long ,12 wide and 2m deep . calculate the list when HFO is consumed from NO3 stbd until the sounding is 1.2m.
Solution :
W = 10000t, KM = 7.8 , KG = 7.075, GM = 7.8 – 7.075 = 0.725m , LM = 0 RD 0f HFO = 0.95
, L x B x D = 15 x 12 x 2 Volume of stbd tank = 15 x 12 x 2 = 360m3
Wt of consumed HFO = 15 x 12 x 0.8 x 0.95 = 136.8t
LM = W x D = ? doubt why ‘d’ taken as (6m)
FSC = LB cub x di / 12W = 15 x 12 x 0.95 / 12 x 9863.2 = 0.208 m
Fluid GM = 0.649 – 0 .208 = 0.441m
24. A vessel displacing 9000t has KM 8.02m, KG 7.5m , and is up right . she loads 250t KG 12m, 3m to stbd of the centre line ; loads 1000t KG 3m, 1m to port of the centre line; discharges 250t KG 8m, 2m to stbd of the centreline.100t of cargo is then shifted transversly 3m to stbd . If the total FSM is 1200tm , Calculate the final list.
Solution :
FW = 10,000 FKG = FVM / FW 71.500 450p
FKG = 71500 / 10,000 FKG = 7.15m
Final solid GM = 8.02 – 7.15 = 0.87m
FSC = FSM / FW = 1200 / 10000 = 0.12m
Fluid GM = 0.87 – 0.12 = 0.75m
Tanthi = LM / W x GM = 450 / 10000 x 0.7 = 3.4p = 3d26.
25. A ship of 14000t displacement , KM 9.0m , KG 7.8m has a total FSM of 2100tm and is listed 8degre to port . How many tonnes must be shifted transversely by 10m to upright the ship?
Solution :
W = 14000t , KM = 9.0m KG = 7.8m , Initial GM = 9 – 7.8 = 1.2 FSM = 2100 tm thit = 8degp, d = 10m
FSC = 2100 / 1400 =0.15
Fluid GM = 1.2 – 0.15 = 1.05m
LM = W x GM tanthi = 1400 x 1.05 x tan thit = 2065.1tm
To upright the vessel we have to shift ballast from to which is to the same LM of on stbd
LM = W x D 2065.1 = W x 10 W = 2065.1 / 10 = 206.51t to ss