Static stability(Academic problem)

Discussion in 'Stability' started by diegoperrera, Oct 1, 2014.

  1. TANSL
    Joined: Sep 2011
    Posts: 5,864
    Likes: 220, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Perhaps the following example will help the OP in his quest for truth. It is controlling the stability during the construction of a concrete block on a floating pontoon. As you increase the height of the block, the set goes deeply. Finally you have to enter ballast water in the pontoon tanks for it to sink and the concrete block floats freely .
    At no time is "stability lost"
    (Block dimensions in mm)
     

    Attached Files:

  2. Rastapop
    Joined: Mar 2014
    Posts: 278
    Likes: 5, Points: 18, Legacy Rep: 46
    Location: Australia

    Rastapop Naval Architect

    I graduate as an NA in four weeks, I'm not in this thread to learn something.

    If you were kidding around when you said there were NAs here saying there was no solution that's cool, my bad, it just wasn't apparent in what you'd written - so I looked and found no NAs saying that.

    It won't help him, the answer he needed was Ad Hoc's in post #2.
     
  3. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Well, strictly speaking, those (like TANSL) who say that there is no solution to the problem as it is described are right. No solution can be found without making some arbitrary assumptions, which might be correct or they might not be. For example, the assumption that the ship's mass is uniformly distributed on the cross-section, hence making it's CoG coincide with the geometric center of the cross-section area. In reality, ballast is used to shift the CoG into a position necessary to ensure a desired stability margin. Another assumption is the uniform longitudinal distribution of the ship and cargo volumes, thus reducing the problem to a 2-D case.
    All things which could have been cleared by the author of the exercise with just few more words in the description of the problem. The way it is written, no unambiguous solution can be given. Place the CoG of the ship closer to the keel and you have a different yet perfectly valid result. The professor will have a hard time in explaining why did he accept some of the results and did not accept some others.
    Cheers
     
  4. TANSL
    Joined: Sep 2011
    Posts: 5,864
    Likes: 220, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Amazing. Congratulations.
     
  5. TANSL
    Joined: Sep 2011
    Posts: 5,864
    Likes: 220, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Of course. But again, with information on the statement, the problem has no solution.
    I know well, as I explain in my previous post, when the data is complete, with a study as I've shown, that at last, after all is to be putting weights higher and higher on a pontoon, stability studies are possible. My experience tells me that if the weights are controlled properly, stability is not lost. (I repeat that we should define what is meant by "losing stability")
     
  6. gonzo
    Joined: Aug 2002
    Posts: 13,786
    Likes: 436, Points: 93, Legacy Rep: 2031
    Location: Milwaukee, WI

    gonzo Senior Member

    As long as the Center of Gravity and the Center of Flotation are aligned vertically, there is no loss of stability.
     
  7. TANSL
    Joined: Sep 2011
    Posts: 5,864
    Likes: 220, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    No, Gonzo, not exactly. See post #19. In addition to what you say, which is correct, on a ship that is not a submarine, the center of buoyancy must be below the center of gravity. But also, if the distance between them is small, the ship can be very unstable, ie, flip with a very small heeling moment.
    And it is not the center of Flotation, but the center of buoyancy.
    In addition, the alignment of two points is required to achieve equilibrium, but that does not mean that equilibrium is stable. Again, a light breeze can unbalance.
    Correction: any floating device the position of the center of buoyancy and center of gravity must be such that a righting moment is formed.
     
    Last edited: Oct 5, 2014
  8. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

  9. Nick_Sinev
    Joined: Aug 2014
    Posts: 63
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Sydney

    Nick_Sinev Junior Member

    I posted my variant of the problem ( for density = 1/3 ) in my LiveJournal and got a really interesting comment from "hyperpov": We'll get much more fun when the density is 1/4, i.e. anything from the interval from (3-sqrt(3))/6 to 9/32. http://biglebowsky.livejournal.com/94265.html?thread=1366329#t1366329

    Update.
    "hyperpov" added some kind of animation for density 1 ... 999 kg/m^3 (999 frames). To watch you will need either move scroll bar or hold PgUp / PgDn
    https://www.dropbox.com/s/q5b5z2i2tyl82do/brus1.pdf
     
  10. sharpii2
    Joined: May 2004
    Posts: 1,884
    Likes: 93, Points: 48, Legacy Rep: 611
    Location: Michigan, USA

    sharpii2 Senior Member

    You are absolutely correct. I have made a mistake.

    I transposed each box being worth 0.25 m of ship depth to being worth 0.25 m of ship draft.

    My bad.

    Actually, each added blue box would increase the ship's draft by only 0.0833 m. So, it would take three boxes to increase the ship's draft, as much as I said one would.
     
  11. diegoperrera
    Joined: Sep 2014
    Posts: 4
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: South Korea

    diegoperrera New Member

    HelloCmckesson, how did you find this answer? Did you make any arbitrary assumptions? I couldn't calculate it because I don't have the weight. Do you agree with me?

    Thank you again :)

    Diego
     
  12. daiquiri
    Joined: May 2004
    Posts: 5,372
    Likes: 239, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    You have weights per unit length (through densities), but you don't have the length of the ship and containers, and the KG of the ship. So, yes, some arbitrary assumptions are necessary to get the result.

    Cheers
     
  13. Nick_Sinev
    Joined: Aug 2014
    Posts: 63
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Sydney

    Nick_Sinev Junior Member

    I'm not cmckesson, but I can answer your question.

    Assumptions:

    1) The density of barge is uniform.
    2) The density of each box is uniform.
    3) The length of the box = the length of the barge.
    4) Cross section is constant on length.
    5) The length of the barge > the width.
     
  14. sharpii2
    Joined: May 2004
    Posts: 1,884
    Likes: 93, Points: 48, Legacy Rep: 611
    Location: Michigan, USA

    sharpii2 Senior Member

    I actually broke down an did the problem myself.

    The first thing I did was assign a length to the ‘barge ship‘.

    Next, I assumed the ‘blue boxes’ were as long as the ship.

    Using this assignment and this assumption, along with the ’densities’ given in the problem, I came up with a displacement for the barge ship and a weight for the blue boxes.

    Here’s what I came up with:

    Barge Ship

    Length = 6.0 m
    Beam = 3.0 m
    Draft = 0.333 m
    Displ. = 6.0 m^3

    Blue Box

    Length = 6.0 m
    Width = 0.5 m
    Depth = 0.25 m
    Weight = 1.5 mt

    Now that I have the actors in this play, it’s time for the drama to begin.

    The first step is to find the inertia of the barge ship.

    I hadn’t done this in a long while, so I had to look up the procedure.
    1.)
    I divided the barge ship in half, lengthwise and drew six stations (seven actual lines).
    2.)
    Next, I measured the length of each, from the center line to its end, at the side of the ship.

    Since the ship is a rectangle in plan form, all of these were the same length, 1.5 m.
    3.)
    Next I cubed them (1.5 * 1.5 * 1.5) and ended up with 3.375 m^3.
    4.)
    I then multiplied each by 1/3 the station spacing (0.333m), except for the first and last stations, which I used 1/6 station spacing for (0.166 m)
    5.)
    I then added these products up to get 6.725 m^4.
    Because the barge ship has two halves, I doubled this number to get 13.5 m^4. This is the Inertia of this barge ship, which doesn’t change, because it has vertical sides and a vertical bow and stern.
    6.)
    Next, I found the Meta Center Radius (MCR) for the barge ship, by dividing its Inertia by its Displacement (13.5 m^4/ 6.0 m^3) to get a MCR of 2.25 m. Now this number does change, as more weight is added to the ship.

    Now I need to find the Meta Center Height (MCH). To do this, I need to find the distance between the Center of Buoyancy (CB) and the Center of Gravity (CG).
    7.)
    Since the ship has rectangular sections that are all alike, the CB is going to be at half the draft of the Barge ship (0.333 m/2) which will be 0.167 m from the bottom. Since the Barge ship is assumed to have uniform density, its CG has to be halfway between the bottom and the deck, or 0.5 m, from the bottom.
    8.)
    By subtracting The CB from the CG (0.50 - 0.167), I get the distance between the two, which is 0.333 m.
    9.)
    This, I subtract from the MCR to get the Meta Center Height (MCH) (2.25 m - 0.333 m) to get a MCH of 1.92 m.

    Now, each time I add weight to the Barge Ship, four things are going to happen:

    a.) the displacement is going to increase,
    b.) the draft is going to increase,
    c.) the CG is going to change, and
    d.) the CB is going to change.

    All of these affect the MCH of the Barge Ship, so each time a blue box, or number of blue boxes are added, a new MCR and a new MCH are going to have to be calculated. So steps 6.) through 9.) are going to have to be repeated.

    I tried one blue box, two blue boxes, then six blue boxes.

    I found that, with six blue boxes, the CG ended up about 5.0 cm above the MCR of the Barge ship. This means the ship is unstable, at this point, and will capsize.

    The MCH can never be a negative number, unless some form of dynamic stabilizing is added. Such could be vertical thrusters or lifting fins, if the barge ship is moving through the water.

    So. My number of blue boxes is five.
     

  15. sharpii2
    Joined: May 2004
    Posts: 1,884
    Likes: 93, Points: 48, Legacy Rep: 611
    Location: Michigan, USA

    sharpii2 Senior Member

    For me, "loss of stability" is loss of righting moment.

    It is like standing a ruler on its short edge.

    It may stand there a moment or two, but will inevitably fall over.
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.