Solving a Naval Architecture problem

Discussion in 'Boat Design' started by Grimorum, Apr 20, 2016.

  1. Grimorum
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    Grimorum Junior Member

    Hi all,

    I encounter this problem while studying Basic Ship Theory by KJ Rawson and unable to solve it to reach the stated answer.

    info: Displacement of ship = 5588 tonnef,
    Forward draughts: 4.57m,
    Aft draught: 5.03m,
    Forward marks from midship = 70.1m
    Aft marks from midship: 51.8m
    LBP: 146m
    TPC: .204MN/cm
    CF: 4.57 AFT OF MIDSHIP
    MCT BP: 1560 MN per metre

    The vessel grounds on rock at 22.9m forward of midship, what is the force at the keel and new draughts when the tide falls by 0.3m?
     
  2. jehardiman
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    jehardiman Senior Member

    So ask yourself...

    What was the draft at the rock when she grounded?

    What will be the draft at the rock when the tide falls?
     
  3. Grimorum
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    Grimorum Junior Member

    Hi jehardiman, I have done the calculations that the change of draught at the rock when tide falls is 0.3m.

    Step 1: Assume the force at Keel is P, Parallel rise = P/20.4MN

    Step 2: Moment induced at CF is 27.47P, hence change of trim at rock= 27.47/146 * 27.47P/1560

    Step 3: Total change of draught at rock = P/20.4 + 27.47/146 * 27.47P/1560 = 0.3

    Answer I got P = 5.73MN, while the answer is 3.72MN.
    Hence I am not sure where did i went wrong?
     
  4. jehardiman
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    jehardiman Senior Member

    You never calculated the change in mean draft and corresponding sinkage. Draw a free body diagram remembering that the sum of the forces and moments have to equal zero.
     
  5. Grimorum
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    Grimorum Junior Member

    Hi there, pardon me but I dont quite understand what you meant there?
    If I can't get the force at the keel, how do I know what is the change in draught?
     
  6. jehardiman
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    jehardiman Senior Member

    Per your other thread:

    Yes this is problem 18 from Chapter 3 of BST and yes the answer given in the text is correct. Before you plug and chug read the problem and draw a diagram showing all the points.

    Notice that the MCT is between PERPENDICULARS, not between draft marks.

    Soooooo….

    First calculate state before grounding
    draft at midships:
    16.5 –(16.5-15)*(170)/(230+170)= 15.8625 = 15’ 10.35”
    drafts at the perpendiculars:
    FWD: 15.8625-(16.5-15)*240/(230+170) = 14.9625 = 14’ 11.55”
    AFT: 15.8625+(16.5-15)*240/(230+170) = 16.7625 = 16’ 9.15”
    Trim between Perps = 1.8’ = 21.6”
    Draft at center of flotation = 15.8625+(16.5-15)*15/(230+170)= 15.91875 = 15’ 11.025”

    Now ground the ship
    Draft at grounding point: 15.8625-(16.5-15)*75/(230+170) = 15.58125 = 15’ 6.975”
    Total change of draft at the grounding point
    (P/52)+((75+15)/480)*((75+15)*P/1300) = 12”
    P(0.0192+0.0129)=12

    P=373.8 tons

    Parallel Rise in draft @ center of flotation = 373.8/52 = 7.188” = 0.599’
    New draft at center of flotation = 15.91875-0.599 =15.31975 = 15’ 3.875”
    Change in trim between perps = 373.8*(75+15)/1300 = 25.87”
    New trim between perps = 21.6 +25.87 = 47.47” = 3.95’

    New draft at
    Midships = 15.31975 - 15*3.95/480 = 15.196 = 15’ 2.35”
    FWD = 15.31975 – (15+230)*3.95/480 = 13.303 = 13’ 3.6”
    AFT = 15.31975 + (170-15)*3.95/480 = 16.595 = 16’ 7.14”

    And as a check
    Draft at grounding point: 15.31975 – (15+75)*3.95/480 = 14.579 = 14’ 6.94”
    So with rounding the difference is 0.035”

    Note, however the problem with this method. The change in draft at midships is 7.998" (15.8625-15.196 = 0.6665') and therefore implies a grounding force or decrease of displacement of 415.9 tons, not 373.8 tons. This is the issue with a normal set of D&O's when the ship is way off trim. Personally, as a salvor's agent, I'd do this with the Bonjeans or the digital lines that most ships now carry.
     
  7. Rabah
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    Rabah Senior Member

  8. Grimorum
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    Grimorum Junior Member

    Gentlemen, I felt so honoured to have receive so much help from all the experienced NA here. It is really encouraging.

    So after all, the MCT value in the book is overvalued when made comparison with the US edition. This really helps me to boost my confidence in this topic as I was thinking day and night about this problem, and thanks to Rabah, The error you pointed out made all the confusion disappear.

    Thanks for all who posted here and I look forward to learn from you guys again.

    Regards,
    Grimorum
     
  9. Rabah
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    Rabah Senior Member

    Hello everybody,
    For those from you which are interested in computer calculation of the grounding I want to show the capabilities of the program Maxsurf Stability.
    See the file from the Maxsurf Stability Manual.
    ______________________
    NA Razmik Baharyan
     

    Attached Files:

  10. TANSL
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    TANSL Senior Member

    A very nice figure but that does not explain how to solve the problem that has arisen here. The issue is what happens when the boat "floats" partially in mud. What does MaxSurf about it ?. That would be interesting to know.
     
  11. jehardiman
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    jehardiman Senior Member

    That is covered in the Handbook for Marine Geotechnical Engineering, Chapter 4 Shallow Foundations. It is also touched on in other salvage manuals.
     

  12. Rabah
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    Rabah Senior Member

    Hi Grimorum,
    I am glad that together with the colleagues have had time to help you to find out the puzzle. That I published for Maxsurf Stability it is only an additional information how to make calculation in an event if you have a computer model. Do not pay attention on Tansl. As well as in other events he did not read all publications on your problems and for this purpose does not know or has not understood that the problem is already clarified.
    _____________________
    NA Razmik Baharyan
     
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