Given deck loading, can we find deadweight?

Discussion in 'Stability' started by ejim, May 7, 2010.

  1. ejim
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    ejim Not Really Junior Member

    Hi guys...need little help here, as per above subject, could we find deadweight by deck loading (20MT/m2) given?

    if can, what is the formula?

    thanks in advance
     
  2. Perm Stress
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    Perm Stress Senior Member

    20MT/m2 mean 20ton per square meter?
    that's hell of a lot for deck loading!
    5t/m2 is already real pain to design supporting structure for...

    If there you talk about inner bottom load of large cargo ship, 20t/m2 is perfectly possible, but this will mean "heavy cargo" situation, i.e., only part of hold area will be taken by some heavy items, like steel rolls, locomotives, ore, etc..
    So the answer is: "most probably, no".

    Dead weight could be found by subtracting lightship weight from displacement at full summer draught.
     
  3. ejim
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    ejim Not Really Junior Member

    Dear Perm Stress,

    Thanks for your quick reply and answer. Forget to mention that im talking about barge...simple form...which is 76m length.

    the answer is no, so why on earth it state Deck Loading: 20MT/m2 in GA?what its purpose then?

    "Dead weight could be found by subtracting lightship weight from displacement at full summer draught."
    next question: how to find lightship eh? :?:

    Again, thank you very much
     
  4. Perm Stress
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    Perm Stress Senior Member

    probably, if it is not typing error, 20t/m2 (2.0t/m2 is quite in the range for ro-ro carrier) mean very strong structure, suited for transprt of heavy items and cargoes to dense to fill all the available cargo space without sinking the boat.

    lightship weight= metal hull+outfitting+paint and corrosion protection+machinery+accommodations+anchors and other equipment...

    no one said it is easy.

    on the other hand, you can use statistics, especially for inland barges, as their weight is pretty stable figure for each size in question.
     
  5. jehardiman
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    jehardiman Senior Member

    Question here

    If the deck loading is 20MT/m^2, what is the total load? Total will determine the barge size which scales displacement.

    Or is it that the barge deck has to be designed to support 20MT/m^2 since you already have a length? That will determine the structure and therefore lightship weight.
     
  6. conceptia
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    conceptia Naval Architect

    jehardiman,

    Barge you may be looking for is a deck loading barge. It will be used to carry equipments overboard. In some cases, for example in a reactor carrier barge, the reactor(lets say 500 tons) wont be directly placed on the deck. They will be provided with deck stools, on which the reactor structure is going to sit. So, the total area on which the load is acting is less, thus increasing the pressure load(stress). -[remember the formula stress = total load acting/total area on which the load is acting]. Hope you got the point.
    Now regarding the deadweight(I assume this is a dumb deck-loading barge)- in the GA you will find the equipments, please find the weight of them, total them and it is your deadweight.
     
  7. conceptia
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    conceptia Naval Architect

    I meant the deck stools will be inducing the load onto the deck. This will be produce a high stress as the area of contact will be less, and concentration of the load will be higher.
     
  8. jehardiman
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    jehardiman Senior Member

    Oh yes, I understand heavy deck barges, I just wonder if the OP does.

    FWIW, a distributed load of 20MT/m^2 is large but manageable, less than 2ST/ft^2 (a common heavy barge deck) or ~30 psi: equal to the loading of a 7m pile of taconite pellets, or twice normal wave slap. But for a 75m x 25(?)m barge that is a total deck load of 37,500 MT which is unreasonable for a 75m barge. More reasonable is deck load 20MT/m^2 over a 100 m^2 area giving a total deck load of 2,000MT for a 75mx25mx~4m barge which is about right.
     
  9. conceptia
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    conceptia Naval Architect

    the first thing you must understand is the deck strength is the strength of your deck on accounts of the max load it can carry. That is how its significance has to be understood. It wont help you to find the total dead weight.
    To find the lightship mass, it better you calculate the steel mass.
     
  10. glitgirl
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    glitgirl Junior Member

    hi,I think 20T/M2 just a local strength of the main deck,the deadweight is decided the freeboard and the lightship.my company have maked many barge.I want to learn the FASTSHIP ,anyone can provide me some manual,tank u.
     
  11. ejim
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    ejim Not Really Junior Member

    Hi to all senior member and all naval architect

    thank you for your info and knowledge sharing, but i have another question here...
    let say, the deadweight of this barge is approximately 3000 tonnes, how many BHP of tug boat required to tow this barge eh?
    how to calculate required BHP to pull this barge?
     
  12. conceptia
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    conceptia Naval Architect

    pal for that you need to do the resistance calculation. That does not depend upon da deadweight. It depends on da hull form(under water surface area, LCB, VCG, mass displacement of the vessel). So better download freeship and try doin resistance calculation.
     

  13. petlily
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    petlily Junior Member

    There is a good reference on how to calculate the total resistance of a barge that takes into consideration the bow shape and hull surface;

    Oilfield Publication, Volume 4, "Towing" by Michael Hancox, page 365-366.

    There are formulas for Frictional resistance, wave resistance and wind resistance (hull & deck cargo) that sum up the total resistance in those pages.
     
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