Initial stability calculation

Discussion in 'Stability' started by Optymista, Sep 8, 2021.

  1. Optymista
    Joined: Sep 2021
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    Optymista New Member

    Hi,

    Just started a course on ship design. I'm having really hard time calculating initial stability of the homogenous barge. Attaching my calculations and sketch.

    "We consider a flat bottom barge with dimensions l x b x h of 60.0m x 20.0m x 4.0m.
    Max draught 3.2m, (DWT 3375t) The mass is 1200t.
    The barge shall carry 2 pieces of equipment/ (modules):
    Module 1: mass 1200 t with height 8 m, length 20m, width 10m
    Module 2: mass 600 t with height 5 m, length 10m, width 10m

    Assuming the centers of gravity for the modules are in the centers of the modules, how much can the weight of the heaviest module increase without need for ballast, requiring a GM of 0.3m? Check that the freeboard is 0.8m"


    I let 'x' be the weight of module 1. (See attached calcs)
    Even when I try to calculate GM, using f=0.8m, I get GM = 6,13m. What am I missing?
     

    Attached Files:

  2. jehardiman
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    jehardiman Senior Member

    Check your calculation of BM. Also the think about trim and or list and the effect on freeboard.
     
  3. bajansailor
    Joined: Oct 2007
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    bajansailor Marine Surveyor

    Welcome to the Forum Optymysta.

    Here is the full size version of your calculations, for easier reference by other readers.

    Following on from JEHardiman's suggestions above, here are a few more thoughts.
    The questions does not say if the barge is in fresh or salt water, hence I presume it must be in fresh water (?).
    Re the two modules that are added - JEH has mentioned trim and list, but the question does not state where the modules are positioned on the barge.
    Are you supposed to assume that the two modules will be positioned on the deck such that the barge has level trim and is not heeling (ie no trim or list) once they are in place?

    I think that you need to start from the very beginning, and initially calculate the GM of the barge in the light ship condition, and then move on from there re adding the weights (?)

    Optymysta stability question.jpg
     
  4. Optymista
    Joined: Sep 2021
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    Optymista New Member

    Thank You for the suggestions.

    Sorry, forgot to mention that all the calculations are based on barge in fresh water.
    To clarify it better, let me state some facts about this exercise.

    I already calculated initial GM without two modules. Initial GM1 = 31.83m and freeboard, f1 =3m.

    After loading two modules, the draft is 2.5m, so new freeboard is fnew=1.5m
    Check for initial stability gives values for BM, KB nad KG wich are:
    BM = 13.33m
    KB = 1.25m
    KG = 5.3m
    So that new GM, GM2 = BM+KB-KG = 9.28m

    I was also asked to calculate position of second module, assuming that the heavier module (nr1) is placed with its CoG 10m from the mid of the barge (measured along the barge).
    I let 'x' be the distance from mid of the barge to CoG of module(arm). Here we must ensure equilibrium around the center of floatation.
    Got that x=20m. That copes well with the model, as the weight of Module 2 is half of Module 1, so the arm of module 2 has to be twice of arm for module 1.

    I even plotted the GM as a fuction of CoG of the modules. (same % of height for both)
    upload_2021-9-8_23-2-55.png


    Still can't figure out what am I missing on the one where I have to calculate the max increase in weight for heaviest module, requiring sufficient GM of 0.3m, while freeboard is 0.8m. I get GM = 6.13, which is approx. 20 times higher than the value I was supposed to get. 20 is the width of the ship. Maybe my equation for calculation of BM or GM is wrong?
     

    Attached Files:

  5. jehardiman
    Joined: Aug 2004
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    jehardiman Senior Member

    Your calculation of BM is incorrect. This is the simple stuff. Do it like the professional mariner. Don't calculate GM. Calculate KM from 1200t to 3840t then work out limiting KG with trim and list.
     
    Ad Hoc and bajansailor like this.
  6. Optymista
    Joined: Sep 2021
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    Optymista New Member

    Thank You for the advice about calculating KM, @jehardiman . I know this is simple stuff and I want to understand it better, hence I asked. As I mentioned, I only started the course and there are many terms I'm not familiar with yet.

    I varied the weight of heavier module from 1200t to the total weight of both modules of 3840t (DWT) and calculated corresponding KM, KG.

    upload_2021-9-9_10-27-7.png

    Since KM = KB + BM and my calculation of BM is incorrect, I will still get a wrong answer.

    BM is metacenter radius, depends on width and draft of the barge. For small inclination angles, we have:
    upload_2021-9-9_10-33-20.png

    I can also use moment of inertia to calculate BM. So that I = (lb^3)/12 = 40000 m^4
    Volume of submerged part: ∇ = lbd = 60*20*((1,8+x)/1,2)) = 1200((1,8+x)/1,2)
    BM = I/∇ = 40000/(1800+1000x), so basically the same result, as I've got previously. :(:confused:




     

    Attached Files:

  7. jehardiman
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    jehardiman Senior Member

    Your simplification of BM only works for zero trim and list. Keep to the basics and don't get "smart" unless you understand the basics first.
     

  8. Optymista
    Joined: Sep 2021
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    Location: Ballabubaland

    Optymista New Member

    Task states:
    Assuming the centers of gravity for the modules are in the centers of the modules, how much can the weight of the heaviest module increase without need for ballast, requiring a GM of 0.3m? Check that the freeboard is 0.8m and discuss the placement of the modules (to avoid trim in this case).

    Here, we consider a homogeneous barge on even keel, so there won't be any list.
    Since it's a homogeneous barge, the volume of the submerged part of
    the vessel is b · d when the barge is at rest horizontally, and bT when the barge is inclining. Since the submerged volume is always the same, it follows that d = T.
     
    Last edited: Sep 10, 2021
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