Stability – I : Chapter 11 |

# Stability – I : Chapter 11

###### Solution:

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

Listing moment = (20 x 5 )
= 100tm

We know that :
Tanθ  = ( Listing Moment) /(W x GM)
= 100/(5000 x 0.3)
= 3.8 degree.

1. ###### 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:-

We can calculate 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)

###### Solution :-

Given :
Cargo shifted (w)   = 200t,
Distance  = 10m,
d = 5m from CL to stbd
W = 10000 t ,
KG = 7.0m &
KM = 7.4m

 Ship’s wt KG VM 10000 t 7.0m 70000 tm 200t (shift) 10m (-) 2000 tm

Final W =10000t                                             Final VM= 68000 tm

We know that:
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

###### 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’.

###### 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 .

###### VM
4950 t 4.85m 24007.5 tm
(+)50t 1.25m (+ )62.5 tm

Final W = 5000 t                                                    Final VM = 24070 tm

We know that:
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

###### 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

We know that:
LM caused = (weight  x distance)
= (100 x 6) tm
= 600 tm (P)

###### VM
10000 t 7.45m 74500 tm
(-)100 t 2.45m (-) 245 tm

Final W = 9900 t                                                     Final VM = 74255 tm

We know that:
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.

###### Distance off centre line
200 D 10.0 5m port
800 D 2.3 4m stbd
500 D 5.2 3m port
250 L 8.0 Nil
250 L 12.0 Nil
###### LM
10000 t 8.3 83000
(-) 200(Dish) 10.0 (-) 2000 5P 1000 P
(- ) 800(Dish) 2.3 (-)  1840 4S 3200 S
(-) 500(Dish) 5.2 (-) 2600 3P 1500 P
(+) 250(load) 8.0 (+) 2000 NIL
(+) 250(load) 12.0 (+) 3000 NIL

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

We know that:
GM = (KM – KG)
= (9.6 – 9.062)
= 0.538m

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

###### LM
9000t 8.3 74700
(+) 600t 4.0 (+) 2400 3(P) 1800
(-) 400t 9.0 (-) 3600 5(P) 2000
200 t 5 (+) 1000 8 (S) 1600
300 t 1 (-) 300 4(P) 1200

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)

1. ###### 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:

###### LM
18000t 7.73m 139500
(-) 1500t 6.0m (-) 9000 3P 4500(S)
(+) 500t 10m (+) 5000 4P 2000(P)
500t 2m (+) 1000 4S 2000(S)
800t 2m (-) 1600 3P 2400(P)

Final W = 17000                      Final VM = 134900 tm                       FLM =2100(S)

We know that:
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

###### Solution :

List = 8 degree (P) ,
Displacement (W)  = 12000t,
KM = 7.54m &
KG = 6.8m

We know that:
GM = (KM – KG)
= (7.54 – 6.80)
= 0.74m

Again, listing moment (LM)= (W x GM x tanθ)
= (12000 x 0.74 x tan80)
= 1248 tm

We know that ,
Listing moment (LM) = (W x D)
1248 tm    = (W x 10m)
So,      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.

###### Solution :

Displacement (W) = 4000t,
GM = 1.0m (initial) ,
list  = 10 degree (P)
W = 16000t, G
M = 1.0m (final)

We know that:
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

Again we know  that :
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.

###### Solution :

Displacement (W) = 10000t,
GM = 1.5m,
list = 5 degree

Final GM = 0.5m

We know that :
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 .

We know that:
Tanθ  =  LM / (W x GM)
= 1312.33/(10000 x 0.5)
=    14.7dgree stbd

1. ###### 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:

###### LM
8500t 8.3m 70550tm 389.23tm
(-) 200t 4m (-) 800tm 5S 1000 P
(-)300t 5m ( -)1500tm 2P 600 S
(+)100t 2m (+) 200tm 4S 400 S
200t( Shift) 2m (+) 400tm 3P 600P

Final W = 8100t                                  Final VM = 68850tm                               FLM=289.23tm

We know that:
Final GM = (Final VM )/(Final W)
Final KG = (68850 /8100)
= 8.5m

Final GM = (KM – KG)
= (9.3 – 8.5)
=      0.8m

We know that:
tanθ = Final LM/(W x GM)
=   289.23/(8100 x 0.8)
= 2degree 33.3minute

1. ###### 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)

We know that:
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, we know that  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.

###### LM
12250t 9.0m 110250   1116.57(S)
(+) 1250t 8m (+) 10000 2S 2500(S)
(-) 250t 2m (-) 500 5S 1250(P)
160t (Shift) 9m (-) 1440 10P 1600(P)

Final W = 13250t                                     Final VM = 118310 tm                 FLM =766.57(S)

We know that :

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)

###### Solution:
 Ship’s wt KG VM LM 8000 t 8 64000 X t 10 (-) 10X

Final W = (8000 – X) t                                Final VM =(64000 – 10X)tm

We know that:
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)

We know that:
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

###### 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

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

1. ###### 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:

###### LM
12000 7.25 87000 0
(+)200 t 24 4800 (15 x200)s

Final W = 12200                                 Final VM = 91800 tm      FLM = 3000

We know that :
Final KG = (Final VM/Final W)
Final KG = (91800/12200)
= 7.524m

Again , we can calculate
GM = (KM – KG)
= (9.0 – 7.524)
= 1.476m

We know that :
tanθ     =  Final LM / (W x GM)
= 3000/(12200 x 1.4476)
=  9 degree 45min

###### LM
12000 7.25 87000 0
200 10 2000 1400 S

Final W= 12200 t                                  Final VM = 89200                 FLM 1400 S

We know that:

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

1. ###### 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

We can calculate :
Initial LM = (W x GM x tanθ)
= (10000 x 0.5 x tan50)
= 437.44 P

###### LM
10000 6.8 6800 437.44P
100t 18m (+) 1800 0

Final W =10000t                                Final VM = 69800tm         FLM=437.44 P

We know that:
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

We know that :
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

###### LM
10000 6.8 68000 437.44 P
100 8 – 800 600.00 S

Final W= 10000t                                     Final VM= 67200tm            FLM= 162.56 S

We know that:
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)

1. ###### 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 centre 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

###### LM
13000 8.0 104,000 1024.76
150 13.0 + 1950 0

Final W= 13000t                                       Final VM= 105950tm       FLM =1024.76

We know that:
Final KG = (Final VM/Final W)
= (105950/13,000)
Final KG = 8.15m

Again we can calculate
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.

###### LM
13000 8.0 104000 1024.76(S)
(-) 150t 9.0 (-) 1350 750 (P)

Final W= 12850                             Final VM = 102650 tm               FLM = 274.76(S)

We know that:
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

We know that :
Tanθ = FLM/( W x GM)
= 274.76/(12850 x 0.762)
= 1.6 degree (S).

1. ###### 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 transfer 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.

 Ship’s wt KG VM LM 10000 9.0 90000 3173.88 S 432 6t (-) 2592 5184.00 P

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

We know that :

FSC = (LB3 x di) / (12 x W)
= (12 x 12x 1) /(12 x 10000)
= 0.3456m

Fluid GM = (2.059 – 0.3456)

tanθ  = (2010.12/(10,000 x 1.7134)
=6.69 degree P

###### 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

###### Solution :

W = 10000t,
KM = 7.8 ,
KG = 7.075,
GM = 7.8 – 7.075 = 0.725m ,
LM = 0
RD 0f HFO = 0.95

We know that, 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

###### LM
9000t 7.5m 67500   0
+ 250t 12m + 3,000 3s 750s
+ 1000t 3m + 3,000 1p 1000p
-250t 8m – 2,000 2 (s) 500p
100        – 3s 300s

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.

###### 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

We know that

LM = W x D
2065.1  = W x 10
W = 2065.1 / 10 = 206.51t to ss 