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Design of Fenders and Bollards

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    6/13/2013Prof.Dr. Tharwat Sarhan 1

    CHAPTER I

    FENDERS AND BOLLARDS

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    Mooring Forces

    The forces acting on a moored vessel are both

    environmental and operational.

    Environmental forces are caused by naturalphenomena such as wind, waves, currents and

    tides. Operational forces include those caused by

    passing ships, changes in the vessel trim,

    freeboard or draught and mooring line over-tension.

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    Shows an Optimized Mooring

    Arrangement

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    Wind and Current Forces

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    The depth of the water under the keel greatly

    affects current forces.

    As the clearance under the keel decreases, the

    forces due to currents increase.

    The magnitude of current force can be three

    times as great on vessels with very small

    underkeel clearances than for vessels indeepwater.

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    Vessel Motion

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    Design of Bollord Forces due to wind

    Fw = wind pressure * exposed area of ship (ton)

    Wind pressure = 0.00256 v2 ib/ft2,

    v = wind velocity (55 to 78 mile/hour)

    The current force = W / 2g 2

    W = weight / ft3 of water,

    g= gravitational acceleration = 32.2ft/sec2

    The current velocity ranges from 1 to 4 ft/sec.

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    Aex1

    Aex2

    B

    LBP

    dD

    F

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    EXAMPLE 1It is required to find the pull forces under the effect

    of both wind and current effect on a shipberthing on container quay ( closed structure).

    Ship characteristics are as given:

    - Dwt = 40,000 tons & LOA =237m , LBP 225 m,B = 32.2 m , D = 11.70m,

    F = 6.9 ms. Where Vs = 30 knots ,

    Vc = 1.0 m /sec., d = 13.0 m.

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    Wind direction

    Current direction

    Bollard

    Wind direction

    Current direction

    Wind direction

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    Solution Ship empty (Longitudinal direction)

    Aex1 = 225 x 11.70 = 2632.5 m2

    ship full (Longitudinal direction)

    Aex1 = 225 x 6.9 = 1552.5 m2

    ship empty (Cross direction)

    Aex2 = 32.2 x 11.7 = 376.75 m2

    ship full (Cross direction)

    Aex2 = 32.2 x 6.9 = 222.18 m

    2

    Cf1 = Aex1/ L2 empty < 0.50

    Cf1 = 1.2

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    Cf2 = Aex2/ (B)2 =376.75 / (32.2)2 = 0.363

    Cf2 = 0.70 from tables

    Wind force calculations Pw = 0.066 Vs2

    Vs = 30 x 0.5144 = 15.432 m/sec.

    Vs = Vs x S1 x S2 x S3 S1 = coefficient due to location & exposed

    for wind ~ 1.0 for normal condition

    S2 = coefficient depends on the shape ofvessel , take S2 = 0.96 from tables

    S3 = 1.0

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    Vs = 15.432 x 1.0 x 0.96 x 1.0 =

    14.815 m/sec. Pw = 0.066 x (14.815)2 = 14.490 Kg / m2

    Fw1(maximum) = Cf1 x Pw x Aex1 (empty)

    = 45.8 tons.

    Fw1(minimum) = Cf1 x Pw x Aex1 (full)

    = 27 tons.

    Fw2 (max) = Cf2 x Pw x Aex2 (empty)

    = 3.82 tons

    Fw2 (min) = Cf2 x Pw x Aex2 (full)

    = 2.25 tons

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    Current calculation

    Current force calculations

    Pc = current intensity = 52 Vc2 Kg/m2

    Vc = 1.0 m /sec. Pc = 52 Kg/m2

    Current parallel to the quay.

    Fc1 = 0.6 x B x D x [1 + D/d]3 x Pc

    Fc1 (min) = 12.4 tons

    Fc1 (max) = 80.6 tons.

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    Forces on Berth Case (I) Current is normal to the berth, closed structure

    Force on the berth F final1 = 45.8 tons

    Case (II) Current is parallel to the berth

    Ffinal2 = Fc1 max + Fw2 (min) =

    80.6 + 2.25 =82.85 tons

    No. of Required Bollards as min. are 4 units

    Bollard capacity = 82.85 /4 = 20.71 x 2 (S.F.) = 41.42 ton

    Take Required Bollard with 50 ton tension force

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    Design of Fender System

    References

    * Code of practice for design of Fendering and Mooring systems

    BS 6349 : Part 4 : 1994 (ISBN 0-580.22653-0)

    * PIANC WG33 Guidelines for the Design of Fenders: 2002 (ISBN

    2-87223-125-0)

    * Recommendations of the Committee for Waterfront Structures

    EAU 2004 8th Edition (ISBN 3-433-01790-5)

    * Technical Notes of the Port and Harbor Research Instuite,

    Ministry of Transport, Japan. No. 911, Sept 1998 (ISSN 0454-

    4668)

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    Structures The jetty structure will have a large influence on the choice of

    fendering system, and sometimes vice versa. Structure design will

    depend to a large degree on local practice, geology and materials.The right choice of the fender, when considered at an early stage,

    can often have a significant effect on the overall cost of the berth.

    The structures can be classified to

    i) Open Pile Jetties Ii) Dolphin

    Iii) Monopile

    Iv) Mass Structure

    V) Sheet Pile

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    Location

    Berthing Structures are sited in a variety of location,

    from sheltered basins to unprotected open waters.

    Local conditions will play a large part in deciding the

    berthing speeds and approach angles, in turnaffecting the type and size of suitable fenders.

    1) Non-Tidal Basins

    2) Tidal Basins

    3) River Berths 4) Coastal Berths

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    TidesTides vary greatly with location and may have extremes

    of just a few centimeters (Mediterranean, Baltic, etc.)up to of 15 meters (part of UK and Canada).

    Tidal variations will influence the structure design and

    selection of fenders.

    HRT Highest Recorded Tide

    HAT Highest Astronomical TideMHWS Mean High Water Spring

    MHWN Mean High Water Neap

    MSL Mean Sea Level

    MLWN Mean Low Water Neap

    MLWS Mean Low Water Spring

    LAT Lowest Astronomical Tide

    LRT Lowest Recorded Tide

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    Berthing Energy Calculations

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    Dolphin Berthing

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    End Berthing

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    Lock Entrance

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    Ship to Ship Berthing

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    Berthing Velocity

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    Berthing Velocities

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    Calculation of Berthing Velocity

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    Added Mass Coefficient CM

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    Block Coefficient CB

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    Eccentricity Coefficient CE

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    CE = K2 / { K2 + R2}

    K = [(0.19 CB

    ) + 0.11 LBP

    R = [ LBP/2 x]2 + [B/2] 2

    Where x = LBP/4 for Quarter Point berthingX = LBP/3 for third- point berthingX = LBP/2 for mid-ships berthing

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    Berth Configuration Coefficient

    Cc

    Closed Structures

    Kc/D 0.50 Cc ~ 0.80

    Kc/D > 0.50 Cc ~ 0.90

    For > 5o Cc = 1.00

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    Semi- Closed Structures

    Kc/D 0.50 Cc ~ 0.90

    Kc/D > 0.50 Cc ~ 1.00

    For > 5o Cc = 1.00

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    Open Structures

    Cc = 1.00

    Where

    Kc = Under keel Clearance

    D = Draft (m)

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    )sSoftness Coefficient (C

    The softness coefficient (Cs) allows for the energy absorbed by

    elastic deformation of the ship hull or by its rubber belting. When

    a soft fender is used (defined as having a deflection, F of more

    than 150 mm) then Cs is ignored.

    ForF 150 mm Cs ~ 0.90

    ForF > 150 mm Cs = 1.00

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    Normal Berthing Energy (EN)

    SIDE BERTHING

    EN = 0.5 MD. (VB)2.CM.CE.CS.CC

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    Normal Berthing Energy (EN)

    DOLPHIN BERTHING

    EN = 0.5 MD. (VB)2.CM.CE.CS.CC

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    Normal Berthing Energy (EN)

    END BERTHING

    EN = 0.5 MD. (VB)2

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    LOAD FACTORS In Limit State Design, The load factors applied to fender reactions

    under normal berthing are higher than those applied underabnormal berthing.

    Fenders are generally designed to absorb the full abnormal

    energy, and the reaction will be similar for both normal and

    abnormal impacts. In this instance, factored normal reactions

    often yield the worst design case.

    It is important to check both the normal and abnormal cases to

    determine which results in the highest structural loads, moments

    and stresses.

    It is sometimes possible, depending on the fender and structuretype, to balance the normal and abnormal reactions of the fender.

    This will optimize the design of both fender and structure.

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    ABNORMAL BERTHING ENERGY(EA)

    Abnormal impacts may occur for many reasons engine failure,

    breakage of towing lines, sudden weather changes or human error.The following table summaries the safety factors according to

    PIANC11

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    Safety FactorVesselType of Berth

    1.251.75LargestSmallestTankers and Bulk Cargo1.50

    2.00LargestSmallestContainer1.75General Cargo

    2.0 or higherRo/Ro and Ferries

    2.00Tugs, Workboats etc.

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    Design of Fenders Example

    It is required to design a fender system for agrain ship with the following specifications;

    DWT = 100,000 tons & Mass displacement

    = 125,000 tons. LBP = 280 ms, B ship = 41ms, draft = 15.0

    ms, approach velocity = 0.10 m/sec.

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    Solution :-

    Cm = 1 + (2x15/ 41) = 1.73

    Ce = K2 / (K2 + R2) = (56)2 / ( 562 + 702 ) = 0.39

    Cs = 1.0 for soft fenders

    Cc = 0.8 for solid quays.

    The absorbed energy EN EN= 0.5 x M x V

    2 x Cm x Ce x Cs x Cc

    EN = 0.5 x 125,000 x (0.10)2 x 1.73 x 0.39 x 1.0

    x 0.8 = 337.35 K J = 34.39 ton . m.


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