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Stability of Floating Bodies

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    The question is regarding the stability of a large barge.
    Overall width=4m
    Uniform weight per meter length of M= 6327.19kg/m (This empty weight is uniformly distributed along the length of the barge.
    Center of gravity of the empty barge is at position G, 2m above the flat bottom of the vessel.
    Barge has a rectangular water-line area and the density of seawater is 1025 kg/m^3
    Neglect the thickness of the hull.

    Determine that the empty barge is not stable.


    2. Relevant Equations

    GM(bar)=Io/V(submerged)- BG(bar)

    GM(bar) is bigger than zero if stable, otherwise not stable.




    3. The attempt at a solution

    Volume of displaced seawater= hx2xl= 2hl m^3

    Thus the weight of the displaced seawater= 6327.19g=ρ(seawater)g(2hl)

    From rearranging h= 3.0864/l .... the problem is i don't know how to use the weight per meter length so I am stuck with this unknown l (length of the barge)

    I am aware of the fact that i need to find the draft h of the buoy but I am unfamiliar with the term 'weight per meter length' so cannot proceed further unfortunately..

    Any help would be greatly appreciated
     
    Last edited: Oct 8, 2011
  2. jcsd
  3. Oct 8, 2011 #2

    SteamKing

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    I don't follow your buoyancy calculation. The problem stated the barge was 4 m in width.
     
  4. Oct 8, 2011 #3
    I followed some worked examples and in most cases they seem to use the half the distance of the width which is why I used the 2 instead of the 4. The reasoning was that of because of axis of symmetry. On the document I uploaded they have put dotted lines and put 2m and 2m markings on the diagram. I used 'l' to represent the length of the barge as it wasn't given and 'h' is the depth it is in water.

    So in the above calculation I equaled buoyancy force of 6327.19g (mass x gravity) to the density x gravity x volume which is ρ(seawater)g(2hl)..

    My only problem was that because I was confused with the 6327.19 representing weight per meter length I had two unknowns, one being l (length of barge) and h (depth of of barge in water). I am trying to solve to h so I can proceed further in finding the stability but I am really stuck because i have 2 unknowns!! PLEASE HELP!!!
     
  5. Oct 8, 2011 #4
    This is the document I am referring to
     

    Attached Files:

  6. Oct 9, 2011 #5

    SteamKing

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    As another esteemed member of this forum has stated, "the factor of 2 will get you every time."

    For now, forget the other examples you looked at. They haven't helped you but they have added to your confusion.

    Since the weight of the barge is 6327.19 kg/m, this means the total weight of the empty barge will be its length times 6327.19 kg /m = 6327.19 L kg

    Due to the shape of the immersed hull, the buoyancy likewise will be 6327.19 L kg.
    What is the draft h?
     
  7. Oct 9, 2011 #6
    I tried using the width of 4m and got the draft value h=1.5432m . I used the same logic as above and equated the buoyancy force to the weight of the displaced seawater.
     
  8. Oct 9, 2011 #7

    SteamKing

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    The draft looks good. Now try to evaluate the stability of the empty barge.
     
  9. Oct 9, 2011 #8
    I Found the GM (Prime)= -0.978.. So I concluded that GM(prime) being smaller than zero, this implies that the metacenter is below the center of gravity and thus the barge is unstable.

    Now my main struggle is the next section. Ballast of mass Mb kg per meter length is added to the barge in order to effect the stability. The weight of the ballast acts at point A (on the barge centerline at the bottom of the hull) and is uniformly distributed along the length of the barge. The addition of the ballast will shift the center of gravity from G to addition of the ballast will shift the center of gravity from G to G(prime). Write out an expression for the height of the new center of gravity above the bottom of the hull in terms of Mb(distance AG(prime))

    I am aware that adding the Mb will shift G to G(prime). G(prime) is lower that G. Because the weight was added below and not above. Therefore G(prime) is always below G.In order to find G(prime) we need to use moments. I just don't know where to start it from.. Thoughts?

    Thanks heaps for the help :)
     
  10. Oct 9, 2011 #9

    SteamKing

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    Write your moment equation for vertical moments with the origin at A. The vertical moment is simply the mass * distance above A
     
  11. Oct 9, 2011 #10

    nvn

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    db725: I think this currently looks incorrect. (And I am unsure what you mean by "prime" here, because I thought this was without ballast yet.) Please show us how you computed this GM value, so we can check your math.
     
  12. Oct 10, 2011 #11
    Sorry for the confusion the GM(prime) is suppose to be GM(bar)..

    So what I actually found is the GM(bar) value to assess the stability of the barge. G(prime) comes in later for the next question.

    In order to find this I did the following calculations.(Please refer to attached diagram above called to follow through my calculations easier)

    h=1.5432m
    Assume that:
    - the distance Lg of the center of gravity G from the base of the flat bottom of the vessel is 2m. So Lg=2m
    - the distance Lb of the center of buoyancy B from the base of buoy is the midpoint of the displaced volume of fluid. So Lb=1.5432/2=0.7716m

    Thus the distance BG(bar)=2-0.7716=1.2284m
    Second moment of area Io=bh^3/12 since its a rectangular shape.

    So Io=(4x(1.5432)^3)/12=1.2250
    Vsubmerged= area x h =4x1x1.5432=4.900 m^3

    So substituting it into

    GM(bar)= Io/(Vsubmerged) - BG(bar) gives the following:

    GM(bar)= (1.2250/4.900) -1.2284 =-0.978 which is approx -1
    so therefore since GM(bar) is smaller than zero the barge is unstable.. Hope my calculations makes sense..

    I am really struggling with moments. Tried all day and got really weird answers. What's the best way to start of the next part ?

    Cannot thank enough for the help guys :)
     
  13. Oct 10, 2011 #12

    SteamKing

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    In your calculation of GM, the inertia used in I/V must be for the waterplane rather than the cross section of the submerged portion of the barge. In other words, the dimensions of the waterplane are the length of the barge by the width.
     
  14. Oct 10, 2011 #13
    I was very confused with the Io calculations actually. So for the h value that we use for Io, i used h=1.5432 as an assumption but wasn't entirely sure with that.

    Are we suppose to use h as 2m or 2+0.7=2.7m since that the actual height of the entire barge?
     
  15. Oct 10, 2011 #14

    SteamKing

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    The Io is the moment of inertia for the waterplane about its longitudinal axis. In other words, Io is calculated for a rectangle with dimensions L = length of barge x B = breadth of barge, or 4 m. Since the barge will roll to one side if it is unstable, the axis about which Io is calculated is the longitudinal centerline of the barge.

    Once Io is calculated, BM = Io / V, where V is the volume of the displaced water,
    or L x B x h
     
  16. Oct 10, 2011 #15

    SteamKing

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  17. Oct 10, 2011 #16
    Thanks for the clarification those moment of intertia calcs are always tricky..Makes a lot of sense. I am working on another section now which is ' Calculate the minimum ballast per unit length (Mb) required to stabilize the barge when floating in sea water.'

    My guess would be that this is similar to first part of this where we looked at proving the instability of the barge and the use of GM(bar)=Io/Vsub -BG(bar). So would working backwards to find the Mb value work? Or should I take another approach? not sure if this is right.
     
  18. Oct 10, 2011 #17
    I thought it will help me and the other person in understanding the questions better, considering they are doing the same questions.
     
  19. Oct 10, 2011 #18

    nvn

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    db725: With your and SteamKing's permission, I would like to change the barge breadth (width) to lowercase b, so it will not be confused with point B, center of buoyancy. Also, point B' is currently missing from the diagram. For clarity, add point B', which is the new center of buoyancy after adding ballast mass Mb. To determine Mb, basically work the problem backwards.
    This is currently incorrect. Change b to L, and change h to b, then recompute Io. Show your work, so we can see if you do it correctly. It is explained in post 14. Hint 1: L = barge length = 1 m.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 0.7716 m, not 0.7716m. See the international standard for writing units (ISO 31-0).
     
  20. Oct 10, 2011 #19

    SteamKing

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    In order to calculate the amount of ballast required to stabilize the barge in seawater, you will need to repeat the calculations you used to determine that the GM of the empty barge was negative. You will need to add ballast and recalculate the draft of the barge and its ballasted center of gravity and check the G'M value. When G'M becomes positive, the barge will be stable.
     
  21. Oct 10, 2011 #20
    Nvn: With the changes I need to make to Io, if i understood correctly by changing b to L, and changing h to b I get he following:

    b=4 m
    L=1 m

    So Io = (1 x 4^3)/12= 5.33

    I did a few questions out of the textbook and found that they do it the opposite way.

    In that method they used it this way.. Io=(4 X 1^3)/12=0.333

    which one is correct?
     
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