Basic concept on Ship stability and free surface area

Stability Of Ships

Two Principal Forces Which Act On A Ship Floating Freely Are – Weight  &  Buoyancy.
For The Ship To Float, It Must Displace It’s Own Weight Of Water.
Then To Be In Equilibrium Condition, The Centre Of Weight & The Centre Of Buoyancy Must Be Vertically In Line.
External Or Internal Forces Can Move The Ship In Either Transverse Direction Or Longitudinal Direction. But It’s Ability To Return To It’s Original Stable Position I.E., Equilibrium Condition Is Related To The Stability.

Upright Position Of Ship :  A Vessel Is Said To Upright If It Is Rolling Slightly About The Upright Position.
Statical Stability : This Is The Measure Of The Tendency Of A Ship To Return To The Upright Position If Inclined By An External Force.
Upright Position Of Ship : In The Upright Position, Weight Of The Ship Acts Vertically Down Through The Centre Of Gravity ‘G’. While The Upthrust Acts Through The Centre Of Buoyancy ‘B’.
Equilibrium Condition Of Ship : When Weight Of The Ship Is Equal To The Upthrust And The Centre Of Gravity And The Centre Of Buoyancy Are In The Same Vertical Line, The Ship Is Said To Be In Equilibrium Condition.


Consider A Ship Inclined By External Force To An Angle ‘Ǿ’.
               In This Case Centre Of Gravity Remains In The Same Position, But The Centre Of Buoyancy Moves From ‘B’ To ‘B1’.
    Therefore Buoyancy Acts Through ‘B1’  And The Weight Still Acts Through ‘G’, This Creates
    The Righting Moment = Δg X Gz -------------------(1)
    This Moment Tends To Turn The Ship To It’s Upright Position.
    Righting Lever = Gz = Gm Sinǿ
    Now
The Vertical Through New Centre Of Buoyancy -‘B1’ Intersects The Centreline At ‘M’ Which Is Called The Metacentre. The Height  Of ‘M’ From The Centre Of Gravity Is Called The Metacentric Height- ‘GM’

Stable Ship : ‘GM’ Is Said To Be Positive When ‘G’ Lies Below ‘M’. In Such Case Ship Will Roll Back To Original Upright Position When Disturbed By External Or Internal Force.
Tender Ship : A Stable Ship With Small Metacentric Height Will Have Small Righting Lever At Any Angle And Roll Easily. In Such Case Ship Is Said To Be Tender Ship.
Stiff Ship : A Stable Ship With Large Metacentric Height Will Have A Large Righting Lever At Any Angle And Will Have Considerable Resistance To Rolling. In Such Case It Is Called The Stiff Ship.
Unstable Ship : When Metacentre ‘M’ Lies Below The Centre Of Gravity ‘G’, Then ‘GM’ Is Said To Negetive, Which Increases The Angle Of Heel. In Such Case The Vessel Is Said To Be Unstable And Will Not Return To Upright Positon.

Neutral Equilibrium : When Centre Of Gravity And The Transverdse Metacentre ‘M’ Coinside, The Righting Lever Is Zero And There Is No Righting Moment Acting On The Ship. In Such Case Ship Will Remain In Inclined Position With An Angle ‘Ǿ’, And The Ship Is Said To Be In Neutral Equilibrium Condition.
Stability At Small Angle Of Heel
Transverse Metacentre :
The Height Of Transverse Metacentre Above The Keel ‘KM’ May Be Found Out, By Considering The Small Inclination Of The Ship About It’s Centre Line.
For Small Angle Of Heel
Upright And Inclined Water Line Intersects.
Volumes Of The Emerged And The Immersed Wedges Are Equal For Constant Displacement.
    Now
    The Distance Of The Transverse Metacentre Above The Keel Is Given By
    Km = Kb + Bm -------------------------------------------(1)
    Note : ‘KB’ Is The Distance Of The Centre Of Buoyancy Above The Keel. This Can Be Found Out From The Hydrostatic Curve.


‘BM’ Can Be Found Out As Follows.
        Let Us Consider A Ship As Shown In Figure, Whose Volume Of Displacement Is ‘▼’, Lying Upright At Water Line ‘WL’. The Centre Of Buoyancy ‘B’ Being On The Centre Line Of The Ship.
        If The Ship Is Now Inclined By An Angle ‘Ǿ’, It Will Lie At The Water Line W1l1, Which Intersects The Original Water Line ‘WL’ At ‘S’. Since ‘Ǿ’ Is Small It May Be Assumed That ‘S’ Is Lying On The Centre Line.
        A Triangular Wedge Of Buoyancy ‘W1SW’ Has Been Moved Across The Ship To ‘L1SL’ Causing The Centre Of Buoyanmcy To Move From ‘B’ To ‘B1’
    Now
    Moment Of Shift Of Ship’s Buoyany From B To B1
    = Moment Due To Shift Of Buoyancy Wedge
Therefore
    ▼ X Bb1  =  V X Gg1
                                         V X Gg1                                        
                Bb1  =   ---------
                              ▼
                                   V X Gg1
             Bm Tanǿ = ---------- -----------------( As Bb1 = Bm Tanǿ )
                                      ▼
                     V X Gg1
        Bm = --------------  ----------------------------------(2)
                        ▼ X Tan∆

    Now To Determine The Value Of (v X Gg1) , Ship’s Length Is Devided In To Thin Strips Of Length
    Δx. The Half Width Of Original Water Line Is ‘Y’
    Therefore
    Immersed C/S Area Of Wedge = ½ Y X Y Tanǿ X Δx
                                                                   = ½ Y2tanǿ X Δx
    Volume Of Immersed Wedge  = Σ ½ Y2tanǿ X Δx

The Volume Of This Wedge Is Effectively Moved From One Side To The Other By A Distance Of (4Y/3)
    Therefore
    Total  Moment Of Shift Of Wedge
                        = Σ ½ Y2tanǿ X Δx  X (4Y/3)
                                                        = Tanǿ (2/3) Σ Y3 Δx
    We Know That,  (2/3) Σ Y3 Δx = Ixx
    (IXX = Second Moment Of Area Of Water Plane         About The Centre Line Of The Ship )
    Therefore
    Total  Moment Of Shift Of Wedge = Ixx X Tanǿ --(3)
    From Equation (2) And (3)
                                                                                     Ixx
    Height Of Metacentre From ‘B’  =  ------ ------------(4)
                                                                                      ▼
Metacentric Diagram
In This Diagram Kb (HEIGHT Of Centre Of Buoyancyfrom Keel ) & Bm ( Height Of Metacentre Above The Centre Of Buoyancy )  Are Plotted Against The Ship’s Draught As Shown In Figure.
As The Position Of Buoyancy-B & The Position Of Metacentre-M Depends Only Upon The Geometry Of The Ship & Draught Of The Ship At Which It Is Floating, Position Of ‘B’ & ‘M’ (HEIGHT Of Metacentre) Can Be Found Out Without The Knowledge Of Loading Of The Ship At Any Intermediate  Draught.


Stability At Large Angle Of Heel
Stability Discussed So Far Is For Small Angle Of Heel With Certain Assumptions ( Two Water Planes Intersect At The Centreline,Wedges Formed Are Right Angled Tringles) The Metacentric Height ‘GM’ Was Taken As The Measure Of Stabilty.
When Ship Heels To An Angle Greater Than 100, Above Assumptions Cannot Be Made & The Principle On Which Initial Stability Were Based Are No Longer True. Insteadthe Righting Lever ‘GZ’, Which Is The Peprendicular Distance Between Vertical Lines Through The Centre Of Gravity & The Inclined Centre Of Buoyancy, Is Used As Measure Of Stability.


Let Us Consider
A Ship Which Is Inclined To Some Angle ‘Ǿ’ From The Vertical.
‘WL’ Is The Initial Water Line And W1li New Water Line When Inclined.
The Volume Of Displacement In Each Case Is Same.
If The Side Of The Ship Were Vertical Along It’s Length, Then The Two Water Line Would Intersect At The Centreline At Point ‘P’
    Further,
Volume Wpw1 Which Has Emerged Will Be Equal To The Volume Which Has Been Immersed. Let This Volume Be ‘v’
The Centroid Of These Two Wedges ‘g’ And “g1’ Be Located At A Distance Of ‘d’

Now
    Moment Of Shift Of Ship’s Buoyancy
    = Moment Due To Shift Of Buoyancy Wedge
    Bc X ▼ = D X V
                       D X V
            Bc = -------  ------------------------------------------------(1)
                         ▼
    &   Gz = Bc – Bg Sinǿ
                       D X V
                 =  -------- - Bg Sinǿ          
                        ▼
                                                   D X V
     Righting Lever ‘GZ’ = -------- - Bg Sinǿ -------------(2)
                                                      ▼
    Note ;  This Is Called Atwood’s Formula. From This Formula If ‘v’ And ‘d’ Are Evaluated For A Range Of Angles Of Inclination ‘Ǿ’ , The Graph Of Righting Lever ‘GZ’ Verses The Angle Of Inclination ‘Ǿ’ Can Be Drawn. The Curve So Drawn Is Called ‘CURVE Of Statical Stability’

Curve Of Statical Stability


In This Curve Righting Lever Can Be Seen, Rising To A Maximum Value And Then Slowly Fall To Zero.
A Ship Inclined Beyond The Point Of Zero ‘GZ’ Will Be Unstable.
Angle Up To Zero ‘GZ’ Point Is The Range Of Ship Stability At That Particular Load Condition.
Ship Operator Should Know This For Safe Operation Of The Ship.

Free Surface Effect


If The Tank On A Ship Containing Liquid Is Not Fully Filled, The Liquid Moved Through The Tank In The Direction Same As The Heel. For This Reason, The Focus Of The Vessels Central Gravity Shifts Away From The Centre Reducing The Righting Lever "GZ" And Height Metacenter Which Results In Increase In Heel Angle. This Effect Is Known As The Effect Of The Free Surface.
We Consider A Full Tank Water In Ship Of Displacement "Δ" With Inclination To Any Angle Ǿ.
The Focus Of Vessel Moves From "G" To "G1", As Wedge Of Liquid Moves Through The Tank

Tank Divisions Use In Tankers:


Largest Ship With Free Surface Effect Must Be Left Space Of Oil Tank For Expansion.
Tanker Was Built Initially With Centeline Bulkhead And Expansion Tanks. Twin Length, Bulkhead Were Introduced Without Expansion Tanks And Successfully Proved Because It Deals With The Loss Of Metacentric Height Of Design Due To The Free Surface Effect.
It Is Not Possible To Design A Dry Cargo Ship In Same Way As C.G Position Varry With The Nature Of Deposition Of Cargo. Effect Of The Free Surface Is Dangerous For A Ship With Small Metacentric Height And Can Make The Ship Unstable.
In These Ship Tanks Required To Be Pressed Up. If The Ship Is Initially Unstable And Helling To Port, Then Any Attempt To Fill The Water Ballast Results In Reduction Of Stability.

# Various books, study material and other online sources has been refereed prior to writing this article but no part is copied or produced  from any of the source but explained same thing in better detailed way.

Author Arpit Singh and Amit                                                                 Article requested by: Deepak Kumar


Post a Comment