Staility – I : Chapter 2

  1. Find the thrust exprienced by a flat keel plate 10m x 2m when the draft is 8m in SW.

Solution :  Area of the flat keel plate = (10 x 2)
                                                  = 20m2
               Pressure = (depth x density)
                                           =    (8 x 1.025)
                                          =  8.2t /m2
              Thrust = (pressure x area)
                                       = 8.2 x 20 =164 t .

  1. A box-shaped vessel 150m x20m x 12m is floating in a dock of RD 1.010 at an even keel draft of 10m. Find the total water pressure experienced by the hull.

 Solution: The total water pressure exerted on the hull to the water =  (Thrust of keel plate which acting   horizontally + Thrust of forward and aft  which act vertically +Thrust of  port and stbd side which act vertically)
Thrust of Keel plate = (pressure x area)
Pressure = (depth x density)
              =10 x1.010 t/m2

Area = 150m x 20m = 3000m2
Thrust =(10 x1.010 x 3000) = 30, 300 t

Thrust of forward and Aft =(pressure x area)

Pressure = depth x density
            =( 5 x1.010)t/m2

Area =L X B
           =(20 x10)
           = 200m2

Thrust = (P x A)
              = (5×1.010 x200)
              = 1010t

Thrust acting both forward and aft = (1010 x2)
                                                               = 2020 t

Thrust of port and stbd  side = (pressure x area)

Pressure  =( depth x density)
                   =( 5×1.010)

Area = (L X B)
          = 150m x 10 = 1500m2

Thrust = ( P X A )
              = 5 x 1.010 x1500
              = 7575 t

Thrust acting on Both sides = 7575 x 2
                                                =15150 t

Hence , Total pressure on the hull = (30300 +2020 + 15150)
                                                  =47,470 tonnes.

  1. A submarine has a surface area of 650m2 and can with stand a total water pressure of 1332500 t . Find at what approximate depth in SW she would collapse.

Solution:     Area = 650m2
                    Thrust = 1332500t
                     RD = 1.025
                    Depth =?

Thrust = ( P x A )
1332500 = (P x 650)
P = 2050 t/m2

Now,  P =(depth x density)
2050 = (depth x 1.025)

Hence ,  Depth = 2000m.

  1. A rectangular lock gate 40m wide and 20m high has water of RD 1.010 12m deep on one sideand water of RD 1.020 11m deep on the other. Find the resultant thrust experienced and direction in which it acts.

Solution: Water of RD =1.010
                Depth =12 m
                C = (12/2)
                  =6m

Pressure = (depth x density)
                  =(6 x1.010) t/m2

Thrust =(P x A)  =( 6 x 1.010 x40 x12)
                          = 2908.8t

Water of RD = 1.020
Depth = 11m
C = (11/2)
   =5.5m

Pressure =(depth x density)
              = (5.5 x 1.020 ) t/m2

Area =( L X B) =(40 x 11)
                      = 440m2

Thrust =( P X A) = (5.5 x 1.020 x 440)
                           = 2468.4 t

Thus we can Resultant thrust =(2908.8 – 2468.4)
                                                      = 440.4t

  1. A rectangular lock gate 36m wide and 20m high has FW on one side to a depth of 16m. Find what depth of SW on the other side will equalize the thrust.

Solution Inside the lock gate area =( 36 x 16)m2

Pressure = depth x density =(16/2 x 1)
                                                      = 8 t/m2

Thrust inside the lock gate = (pressure x area)
Thrust =(8 x 36 x 16)
       = 4608 t

Let ‘X’ depth of   salt water on other side will equalize the thrust .

Pressure     = depth x density =( X/2 x 1.025)
Area    =(36 x  X )
Thrust = (P x A)

4608 = (X/2 x 1.025 x 36X )
4608 = (1.025 x 36X  x  X/2)
4608 =   (1.025 x 36X2/2)
           =   ( 1.025 x 18X2)
X2   = (4608 /1.025 x 18)

X = 15.8 m

Hence, Depth of other side of lockgate =15.8m

  1. A collision bulkhead is triangular in shape. Its maximum breadth is 12m and its high 15m. Find the thrust experienced by it if the fore peak tank is pressed up to a head of 3m of SW.
Solution :

Breadth of collision bulkhead = 12m
Height       = 15m
Pressure   = (depth x density)
Thrust       = (pressure x area)
Breadth     = 12m
Height of the water inside the tank = 15mans 6
Water pressed up to ahead of  = 3m
‘C’ inside the tank = (1/3 x15)
                                 = 5 m
Outside the tank =    3m
so total =5 + 3 =8m

Now pressure =(depth x density)

                          =( 8 x 1.025)

Area = (1/2 x 12 x15)
         = 90m2

Thrust  experienced  = (pressure x area)
                                      = (90 x 8 x 1.025)
                                       =738t.

  1. A collision bulkhead is triangular shaped, having a breadth of 14m at the tank top and a height of 12m. As a result of a collision, the forepeak tank gets ruptured and SW enters the tank to a sounding of 9m. Calculate the thrust on the bulkhead.
Solution :

B = 14m, H =12mans 7
As we know that :
Pressure = (depth x density)
Thrust   = (pressure x area)

An right angle triangle PBC , PB =7m
Since, Height divide AB  equally into two parts.
Triangle PBC  and   FEC are similar
So by law of similar angle triangle

PB/ FE = PC / FC
7 / FE  =12 / 9

Hence, FE  =  (63 / 12)
              = 5.25m

In triangle DEC
DF = FE
DE = (2 X FE)
       = 2X 5.25
       =10.5m

Area of triangle DEC = (1/2 x10.5×9)
                                    =47.25m2

Pressure = (depth x density)
                  = (9 x 1/3 x 1.025)
                 = 3.075 t/m2

Hence Thrust experienced  = (P X A)
                                            = (3.075 x 47.25)
                                            =145. 29 t

  1. A tank has a triangular bulkhead, apex upwards. Its base is 14m and its sides, 15m each. It has a circular inspection hole of radius 0.5m the centre of a manhole 0.8m above the base and 1.6 from one corner. Find the thrust on the manhole cover when the tank contain oil of RD 0.95 to a sounding of 10m.(Assume π to be 3.1416).
Solution:

Base = 14m , Sides = 15m , Depth of the oil = 10mans 8

Pressure = (depth x density)
             =  (10 – 0.8 x 0.95)
            = 8.74 t/m2

Area of the circle = π r2
                 = (3.1416 x 0.5 x 0.5)
                 =0.7854m2

Thrust = (P X A)
          =( 8.74 x 0.7854)
          = 6.864 t.

 

 

  1. A rectangular deep tank is 22m x 20m x10m. Above the crown of the tank is a rectangular trunkway 0.2m high, 5m long and 4m wide. Find the thrust on the tank lid when the tank is pressed up with SW to a head of 2.64m above the crown of the tank .
Solution :

Rectangular deep tank  = 22m x 20m x10m
Trunk way = (5m x 4m x 0.2m)ans 9
Pressure = (depth x density)
              = (2.64m – 0.2) x1.025

Since , Depth is (2.64m – 0.2 )  and we have the calculate the thrust   act  on the trunkway and the water level above the trunkway will act as thrust )

If this trunkway would have been placed inside the tank then the depth will be   (10m + 2.64m – 0.2/2m ).

Area of the trunkway = (L X B)
                                = 4×5
                                 =20 m2

Thrust experienced = (P X  A)
                           = (2.501 x 20 )
                          = 50.02tonnes.

  1. A double bottom tank measures 25m x20m x 2m. Find the thrust on the tank top when pressed upto a head 16m of SW. Also find the resultant thrust on the tank bottom, and the direction that it acts, if the ship’s draft in SW is 10m.
Solution :

Volume of  tank  = (L X B X H)
                         = (25m x 20m x 2m)

Area    =     (25 x20)
          =   500m2ans 10

Pressure = (depth x density)
                = (16 x 1.025)
                =16.4  t/m2

Thrust on tank top  = (PX A)
                                      = (16.4 X 500)
                                     = 8200 t

Pressure on the tank bottom = (depth x density)
                                                            = (18 x1.025)              = 8.45t/m2

Thrust   act  on the tank bottom = (P X A)
                                                               =   (500 x 18 .45)
                                                               = 9225t

Pressure act on outside of the tank bottom = (depth x density)
                                                                            = (10 x 1.025)
                                                                           = 10.25t

Hence, Thrust acting = (P X A)
                                 = (10.25 X500)
                                 = 5125 tonnes.

Finally, Resultant thrust = (9225 -5125)
                                    = 4100 tonnes.

About the author

Amit Sharma

Graduated from M.E.R.I. Mumbai (Mumbai University), After a brief sailing founded this website with the idea to bring the maritime education online which must be free and available for all at all times and to find basic solutions that are of extreme importance to a seafarer by our innovative ideas.

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