Stability – I : Chapter 5 |

# Stability – I : Chapter 5

A ship of 16000t displacement and TPC is 20 floating in SW at a draft of 8.0m . Find her draft in FW .

###### Solution:

Displacement (W) = 16000 t
TPC = 20 & SW draft = 8.0m

We know that:

Displacement when in SW = ( L X B x draft) x  1.025
Displacement when in FW  =( L X B X D) x 1

It is understood that displacement of ship will remain constant , as displacement is independent of change in density ,(is referred as MASS).

So, (L x B x 8) x 1.025 =  (L x B x draft) x 1
Hence draft  = ( 8 x 1.025)
= 8.2 m

###### 2nd  Method :

As we know that

FWA = (W/40TPC)
= 16000/(40 x 20)
= 20cm

We can calculated:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1) x 20 / 0.025
= 20c
=0.2m

So new draft of ship in FW = (8.0 +0.2)
= 8.2 m

###### Solution :

FWA = 180mm = 18cm

We can calculate :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.008 – 1.025 ) x 18 / 0.025
=(0.017 x 18)/0.025
= 12.24c
= 0.12m

Here , Change in draft is  0.122m and it will be rise.

###### Solution :

FWA = 160mm = 16cm

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1)x 16 /0.025
= 0.01 x 16 /0.025
= 6.4 cm
= 0.064 m

Here change in draft would lead to sinkage.

###### Solution:

FWA = 175mm = 17.5cm

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.018 – 1.006) x 17.5 /(0 .025)
= (0.012 x 17.5) /(0 .025)
= 8.4cm

Here change in draft would cause rise of vessel.

###### Solution :

TPC =25.

We know that:
FWA = W/(40 TPC)
= 18000/( 40 x 25 )
=18cm.

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

=(1.025 – 1.018 ) x 18 /0.025
=5.04cm
= .05m

Since, Her load line should immersed to 0.05 m so that she will not be loaded.

###### Solution:

Volume of box shape vessel = (Lx B x H)
= (20 X 4 X 2)

Mean draft = 1.05m
Displacement  = (u/w volume )x density

Again, displacement can be calculated as (W)
=(L x B x D) x(density)
= (20 x 4 x 1.05) x( 1.025)
=86.1t

Let ‘X’ be the displacement at RD of 1.012
So, X = (L x B x D) x ( 1.012)

Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.
(20 x 4 x 1.05) x(1.025) = (20 x 4 x d) x(1.012)
d = (1.05 x 1.025) / 1.012
= 1.06m

###### Solution:

Volume of box shape vessel = (L x B x H )
= (18m x 5m x 2m)

RD of DW = 1.000 & Depth = 1.4m

Displacement at DW of RD 1.000 (W)
= (u/w volume)x (density)
=( Lx B x 1.4) x(1)

Displacement at SW (W1) = (u/w volume) x (density)
=(L x B x Draft) x (1.025)

Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.

W = W1
(L x B x 1.4) x (1) = (L x B x Draft) x (1.025)

Hence, Draft = (1.4/1.025)
= 1.365m

Total volume of the ship = (L x B x H)
= (18 x 5 x 2)
= 180m3

Again, U/w volume of SW = (L x B x draft)
=(18 x 5 x 1.365)
= 122.85m3

Above water volume = (180 -122.85)
= 57.15m3

Hence, RB % = (Above water volume/ Total volume) x 100
= (57.15/ 180) x 100
= 31.75%.

###### Solution:

Displacement(W) = 3000t,
Draft= 5m & RD = 1.018

W = (u/w volume) x (density)
= (L x B x draft) x(1.025)
Draft = W/ (L x B x 1.025)

Let W1 be the displacement at RD = 1.018

W1 = ( L x B x d) x (1.018 )
d1 = W1/( Lx B x 1.018)

But according to question ,

Draft = d1
W/ (L x B x1.025 )= W1 /(L x B x 1.018)
3000 / 1.025 = W1 / 1.018
W1 = 3000 x (1.018 / 1.025)
= 2979.51 t

###### Solution:

Displacement (W) = 4500t,

Displacement = (u/w volume ) x (density)
= (L x B x draft) x 1.025

Draft = W / (L x B x 1.025) m

Let W1 be the displacement at RD 1.020

So W1 =( L x B x d )x (1.020)
d = (W1 / (Lx B x 1.020)

according to question,  Draft  =  d

W /( L x B x 1.025)  =  W1 /( L x B x 1.020)
W1 = (4500 x 1.020) / 1.025
= 4478.04t.

###### Solution:

Area of Waterplane = (L X B)
= (100m x 20m)

Cb = 0.8, Depth = 8m

We know that :

Displacement = (u/w volume ) x (density)
= ( Lx B x draft ) x (0.8)
= (100 x 20 x 8) x(0.8)
= 12800m3

Displacement (W) = (u/w volume)x (density)
= (12800 x 1.025)
= 13120t ,

Draft = W / ( L x B x Cb x 1.025)

Let W1 be the displacement in FW
= (u/w volume) x (density)
=(L x B x d1 x Cb ) x ( density)

d1 = (W1 /( L x B x 1 )x (Cb)

According  to  question, Draft = d1

W /( L x B x Cb x 1.025) = ( W1 /( L x B x Cb)
W1 = ( 13120 x 1) /( 1.025)
= 12800t

So, Difference is displacement = (13120 – 12800)
= 320 t .

###### Solution:

Displacement (W )= 14500t
RD = 1.025

Displacement (W) =( u/w volume ) x (density)
= (L x B x draft)  x  (1.025)

Draft = W /( L x B x 1.025)

Let W1 be the displacement at RD of 1.010

So W1 = (u/w volume )x( density)
= (L x B x d1 ) x  (1.010)
Hence, d1 = W1 / (L x B x 1.010)

According to question, Draft  = d1

So, W / ( L x B x 1.025)  =  W1 / (L x B x 1.010)
14500 / (L x B x 1.025) =  W1 / (L x B x 1.010)
W1   = (14500 x 1.010) / 1.025
= 14287.8 t

Hence, Cargo to load = (14500 – 14287.8)
= 212.2t.

###### Solution:

Displacement (W)= 12000t ,
Depth = 10m & RD = 1.025

Let W1 be the displacement at RD 1.012

Now, W =( L x B x draft ) x( 1.025)
W1 = (L x B x d1) x (1.012)

According to question ,

Draft  =  d1
W/( L x B x 1.025)  = W1 / (L x B x 1.012)
W/ (L x B x 1.025)  = W1/ ( L x B x 1.025)

W1 = (12000 x 1.012) / 1.025
W1 = 11847.8 t

Hence, Cargo she has to discharge = (12000 – 11847.8) t
= 152.2 t

###### Solution:

RD = 1.0
Starboard side = 0cm above by below summer draft
Port side = 50mm (5cm above the water line to summer draft)

Mean draft   =    (0+5 ) / 2
= 5/ 2
= 2.5cm above the water line

FWA = 180mm
= 18cm,

TPC = 24

As we know :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.005 ) x 18 /0 .025
= (0.2 x 18) /0 .025
= 14.4cm

Total sinkage = (14.4 + 2.5) cm
=   16.9cm

TPC is SW = 24

TPC = (A / 100) x 1.025
= (24 x100)/ 1.025
A = 23.41 t/cm

Again ,

TPC in RD 1.005 = ( A / 100) x 1.005
= (23.41 x 1.005)
= 23.53t/cm

Cargo to load = sinkage x TPC
=   23.53 x 16.9
= 397.657 t

###### Solution:

Present draft = 7.8m,
RD of DW = 1.010 & TPC = 18 t/cm
FWA = 250mm = 25cm
Maximum permissible draft = 8.0m

As we know :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1.020 ) x FWA /0 .025
=(0.015 x 25) /0 .025
= 15cm

When vessel enters SW,  the draft will rise by 15 cm,

Hence ,actual draft in SW = (7.80 -0. 15)
=7.65m

Sinkage available = (8.00 – 7.65)
=0.35m

TPC = 18 ( given)

We know that :

TPC = (A/100)  x ( density)
18 = (A/100) x 1.025
So, A = (18 x 1.025)/ 100
= 17.56 m2

Now , TPC at RD 1.010 = A / 100 x 1.010
= 17.56 x 1.010
= 17.73 t/cm.

DWT available =( sinkage x TPC)
= (35 x 17.73)
= 620.5t.

###### Solution:

Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm

As we know :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm

Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m

Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm

TPC = 24 ( Given)

24  =  (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A   = (24 / 1.025)
= 23.414 m2

Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm

DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.

###### Solution:

RD of river = 1.010
Summer load line on starboard side = 20mm above
Port side = 50mm above

Mean draft = ( 20 + 50) / 2
= (70/2)
= 3.5cm above

W = 15000t,

FWA = (W/TPC)
= 15000/(40 x 25)
= 15cm

As we know:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.010 ) x 15 / 0.025
= 9cm

Total sinkage  = (9 + 3.5 )
= 12.5cm

TPC = 25( Given)

We know that:

TPC = (A/100) x 1.025
= (25 / 1.025)
= 24.39 m2

Now, TPC for RD 1.010 = (A /100) x( 1.010)
= (24.39 x 1.010)
= 24.63t/cm

Hence ,Cargo can be load = (sinkage x TPC)
= (12.5 x 24.63)
= 307.875 t

###### Solution:

Dockwater  RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line. Distance from WNA to water(W) = 50mm
= 5cm

So, sinkage available = (50 – 45)
= 5mm
= 0.5cm

We know that :
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm

Sinkage = (17.5 + 0.5 )
= 18cm

Total sinkage = (18 + 12.8)
= 30.8cm

As we know:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.005 )x 16 / 0.025
= 12.8cm

Again, TPC = (A/100) x density
30 = (A/100) x 1.025
A  =  30/ 1.025
A  = 29.268 m2

Now , TPC for density of 1.005
TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41  t/cm.

Hence ,cargo can be loaded = (30.8 x 29.41)
= 905.82 t

###### Solution:

Starboard winter load line  = 60mm above
Port winter load line = 20mm below
Mean draft = ( 60 – 20)/2
= (40/2)
= 20mm
= 2cm

Summer SW draft = 7.2m( given) We know that:

Waterline (W) = (1/48 of summer draft)
=(1/48 x 7.2)
= 0.15m
= 15cm

Total sinkage available to bring the vessel to her tropical load  line  = ( 2+ 15 + 15)
= 32cm

TPC =20( given)
So, Cargo can be load = (TPC x sinkage)
=(20 x 32)
= 640t.

###### Solution :

For, Port Side = (3 -2.8)  = 0.2m
= 20cm (above )

For , Starboard side =( 2.9 – 2.8) =0. 1m
= 10cm above

Hence ,Mean freeboard can be calculated as
= (20+ 10)/2
= 30 /2 cm
= 15cm available

As we know:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.020 ) x 20 /0 .025
= 4cm

Now, Total sinkage = (15 + 4)cm
= 19cm

TPC = 30( given)
TPC   =  (A/ 100) x density
30   = (A/100) x 1.025
A = 30/ 1.025
= 29.268m2

Again for calculating  ,TPC at RD 1.020
= (A/100) x ( 1.020)
=  29.85 t/cm

Now , Cargo can be load =( TPC x sinkage)
= (29.85 x 19)
= 567.15 t

###### Solution :

Present freeboard port = 1.68cm
Stbd side = 1.79m
Water of RD = 1.017,
Tropical SW freeboard = 1.63c
Tropical SW draft = 9.6m
FWA = 150mm
TPC = 20.4 • Shubham Sharma says: