Help on calculation - BST I, chapter 3, problem 5

Discussion in 'Stability' started by roydoky, May 8, 2016.

  1. roydoky
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    roydoky Junior Member

    Hi,

    I am a new naval architecture student and I have a huge difficulty grasping the concept of adding/removing weight to making ship stable in transverse direction. I spent so many hours and I just can't seem to get the correct answer. I am trapped on the question from BST I, chapter 3, problem 5 shown below. I appreciate your help in advance.

    A model of a ship is built to float in water of 9823 N/m3 weight density. It has the following geometric properties:

    Length BP = 1.83 m
    Area of waterplane = 4461 cm2
    Draught = 14 cm
    Trans MI of waterplane = 5.182 x 10^-3 m4
    KB = 7.62cm
    Volume of displacement = 68 x 10^-3 m3
    KG = 18.29 cm

    Where must a weight of 111 N be placed to bring G below M, presuming the model to be wallsided at the waterline?

    - End of question -

    I can't figure out why LBP is given in this question...

    Thank you,

    roydoky
     
  2. NavalSArtichoke
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    NavalSArtichoke Senior Member

    Sometimes, authors slip extraneous information into questions to trip up unwary students into wasting their time wondering why that extra piece of data is there.

    If you don't need the LBP in your calculations, ignore it and proceed with solving the problem.
     
  3. roydoky
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    roydoky Junior Member

    Here is how I did it and not getting the correct answer

    Thank you for your reply,

    I didn't use LBP in my calculation but nonetheless, i am not getting the correct answer which is stated in the back of the book. It says that the correct answer for this question is "Approx 3cm below the keel".

    The below is how I calculated. Could you or anyone verify my calculation?

    1) convert all properties in 'm'

    LBP = 1.83 m
    Awp (Area of waterplane) = 0.4461 m2
    Draught = 0.14 m
    IT (Trans. MI of waterplane) = 5.182 x 10^-3 m4
    KB = 0.0762 m
    Vdispl = 68 x 10^-3 m3
    KG = 0.1829 m
    W (Weight added) = 111 N
    w (density of water) = 9823 N/m3
    Wdispl = 9823 * 68 x 10^3 = 668 N

    2) G'M' = KB' + B'M' - KG' and set G'M' = 0

    Let G', M', B', G' to represent the new points after adding the weight of 111 N.
    Let b, g represent center of buoyancy and gravity for the additional weight
    Set G'M' = 0.

    2-1) KB' = KB + BB'
    = 0.0762 + W(Bb)/(Wdispl+W)

    Find Bb:
    the added volume of layer = 111 N / 9823 N/m3 = 0.0113 m3
    (Assuming wallsided) the thickness of the added layer = 0.0113 m3 / 0.4461 m2 = 0.0253 m.
    Then, Kb = 0.14 + 0.0253/2 = 0.15265 m.
    Bb = 0.15265 - 0.0762 = 0.076465 m

    Then, KB' = 0.0762 + 111 (0.076465) / (668+111)
    = 0.0871 m

    2-2) B'M' = IT for new waterplane / (Vdispl +Vadded)

    Assuming wallsided, IT for new water plane remains the same.
    B'M' = 5.182 x 10^-3 / (68 x 10^-3 + 0.0113) = 0.06535 m

    2-3) KG' = KG + GG'

    GG' = W(Gg)/(Wdispl +W)
    = 111/(111+668) x Gg
    = 0.1425 Gg
    KG' = 0.1829 m + 0.1425 Gg


    3) Back to 2) to find Gg:
    0 = 0.0871 + 0.06535 - 0.1829 - 0.1425 Gg
    Gg = -0.21368 m

    4) Substitute Gg back into 2-3) to find KG'
    KG' = 0.1829 + 0.1425 (-0.21368) = 0.1829 - 0.0304494 = 0.152 m

    Thus, the answer is 0.152 m above the keel.

    But the correct answer is 0.03 m below the keel.
    I am not sure what I am doing wrong here.

    I appreciate your help.

    Thank you,
     
  4. NavalSArtichoke
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    NavalSArtichoke Senior Member

    This is what the new KG of the model must be after adding the 111 N weight
    so that GM > 0.

    However, the question asked where the 111 N weight must be placed in the
    model so that GM > 0.
    See comment above. You only need to do a short calculation to finish
    this problem.
     
  5. roydoky
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    roydoky Junior Member

    Thank you! NavalSArtichoke

    Hi NavalSArtichoke,

    Thank you for pointing it out.
    KG + Gg = -0.03 m from keel.
     
  6. roydoky
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    roydoky Junior Member

    BST, Chapter 5, Problem 1

    Now, I am stuck on chapter 5 problem 1 in BST.
    I would appreciate your input once again. The question is as below:

    1. A water lighter is in the form of a rectangular barge, 42.7 m long by 5.5 m beam, floating at an even draught of 2.44 m in sea water. Its centre of gravity is 1.83 m above the keel. What will be the new draughts at the four corners if 20 m3 of fresh water are pumped out from an amidships tank whose center of volume is 0.3 m from the port side and 0.91 m above the keel? Ignore free surface effects.
    ----end of question ----

    I convert the 20 m3 of fresh water to weight ---> w = 20 tonne
    Since this weight is removed from amidship, there would be no trim.
    I can get the parallel rise due to the removal of 20 tonnes and resulting B' (center of buoyancy). The resulting B' is vertically (along z-axis) down from B and there is no movement horizontally (along y-axis) since the center of flotation remains in the ship's xz-plane since it's wall-sided.
    I can get the resulting G' (center of gravity) from the removal of 20 tonnes.

    Now, I just can't figure out how to calculate the angle of heel to obtain the new drafts at STBD and PORT.

    I appreciate your help in advance!
     
  7. NavalSArtichoke
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    NavalSArtichoke Senior Member

    Think about why the transverse GM of a floating vessel is important.

    The estimation of angle of heel is covered in earlier chapters of BST; you just need to look for it. ;)
     
  8. roydoky
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    roydoky Junior Member

    Thank you again for your quick reply.

    I can find only the graphical trial and error method to get the angle of heel (pg 151, Chapter 5, Fifth edition) or angle of trim (pg 70, Chapter 3, Fifth edition).
    Is this the only way to solve this question? I feel like there is a computation method to solve this problem. If time permits, could you elaborate on how you would solve this problem?

    Thank you!
     
  9. NavalSArtichoke
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    NavalSArtichoke Senior Member

    You haven't looked far enough back in the earlier chapters of R&T. Go back to where the concept of GM is first developed in Chapter 4. You'll find the necessary formulas there.

    Remember, the basic idea of a stability test for a vessel comes from moving weights back and forth across the beam. Your problem is similar.
     
  10. roydoky
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    roydoky Junior Member

    Hi, NavalArtichoke

    I have followed your advice and I was able to obtain answer that is close to the correct answer in the back of the book. I am off by 0.28 degrees for angle of heel corresponding to 20 cm off on STBD side and 10 cm off on port side compared to the correct answers. I am afraid that I may have missed something in my calculation as I think this amount of error can’t be coming from round up/off, or can it???

    Here is how I have obtained my answers:

    First I found the new G’ and B’ due to parallel rise due to the removal of fresh water and obtain G’M’.
    Second, using formula, G’M’ = (w x d) / ((Wdispl – w) x tan (Theta)), obtain the angle of heel, theta.
    Then finally using the angle to get the new port draft and STBD draft.

    Wdispl. = 5.5 m x 42.7 m x 2.44 m x 1.025 tonne/m3 = 587.4 tonne  before removing the fresh water
    w = 20 m3 x 1 tonne/m3 = 20 tonne <- weight of fresh water removed.

    1) Find the vertical movement of G to G’ (new G vertical) = GG’
    GG’ = w*(KG-Kg) / (Wdispl – w) = 20 tonne x (1.83 m – 0.91 m) / (587.4 tonne – 20 tonne)
    = 0.0324 m UP (because the weight is removed)

    2) Find the vertical movement of B to B’ (new B vertical) = BB’

    Parallel rise = 20 / (1.025 x 5.5 x 42.7) = 0.0831 m DOWN

    3) Find B’M’

    B’M’ = I / Vdispl = B^2/(12 x T) = 5.5^2 / (12 x (2.44 – 0.0831)) = 1.0696 m

    4) Find G’M’

    G’M’ = B’M’ + KB’ – KG’
    = 1.0696 + (2.44/2 – 0.04154) – (1.83 + 0.0324) = 0.38566 m

    5) Find Angle of heel

    Theta = tan-1 [(20*2.45)/ (567.4*0.38566)] = 12.62 degrees
    6) Find the new draft at port and draft at STBD.

    Trim = B x tan 12.62 = 1.2314 m
    Since the center of flotation remains in the centerline,
    Tport = 2.44 – 0.0831 – 1.2314/2 = 2.3569 – 0.6157 = 1.74 m
    Tstbd = 2.3569 + 0.6157 = 2.97 m

    The correct answer is 1.73 m Port side and 2.99 m STBD side.
    The correct angle of heel is then, tan-1[(2.99-1.73) / 5.5] = 12.9 degrees.
    The error is then 10 cm on port side and 20 cm on STBD side. The difference in angle of heel is 0.28 degrees.

    Is my method getting the answers correct and the error is due to round up/off in calculation?
    Or, Have I missed to consider something additional in my calculation?

    Thank you so much for your feedback once again!
    You have been a great help on my studies.
     
  11. Heimfried
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    Heimfried Senior Member


    You are setting the difference of VCB eaqual to the difference of draft?
     
  12. roydoky
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    roydoky Junior Member

    I forgot to include the calculation of BB'. I apologize.
    BB' = 20*(2.44 - 1.22 - 0.0831/2) / (587.4 - 20) = 0.04154 m DOWN

    KB' = 2.44/2 - 0.04154 = 1.17846 m.

    This was used for obtaining G'M' = 0.38566 m.

    Thank you,
     
  13. Rabah
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    Rabah Senior Member

    Hi roydoky,
    About calculation of Problem 1, Chapter 5, BST volume 1- Rawson and Tupper, 2001, Fifth edition
    First we update the coordinates of the CG of the unloaded cargo p = - 20 t:
    Xp = 0, Yp = -2,45m /PS/ and Zp = 0,91m
    At the following positive directions of the axises of the coordinate system:
    X –from the midship to the bow, Y – from the CL to the SB, Z – up from the keel
    The beginning of the coordinate system - in the midship, at X=0, Y=0 and Z=0
    The solution:
    1. Displacement before discharging the cargo - old
    D = γ*Lbp*B*T = 1,025 x 42,7 x 5,5 x 2,44 = 587,36 t
    p/D % = 20 x 100 / 587,36 = 3,4% < 10% - therefore the cargo is "small"
    2. Change the draft after discharging the cargo
    ΔT = p / (γS) = - 20 / (1,025 x 234,85) = - 0,083 m
    S = Lbp * B = 42,7 x 5,5 = 234,85 m2
    3. The new mean draft in CL
    T1 = T + ΔT = 2,44 – 0,083 = 2,357 m
    4. A transversal metacentric height before discharging the cargo - old
    h= Zc + r – Zg = 1,22 +1,033 – 1 83 = 0,423 m
    r = Jx / V ; Jx = Lbp* B^3 / 12; V = Lbp*B*T
    or r = B^2 / (12 x T) = 5,5^2 / (12 x 2,44) = 1,033 m
    Zc = T / 2 = 2,44 / 2 = 1,22m
    5. Change of the metacentric height h at discharging a small cargo for a vessel with vertical sides:
    Δh =[ p / (D + p) ] * [ T + (ΔT/2) – h – Zp ] =
    = [ - 20 / ( 587,36 -20 ) ] x [ 2,44 + ( - 0,083 / 2 ) – 0,423 – 0,91 ] = - 0,03756 m
    6. Definition of the heel angle after discharging the cargo
    θ= p * Yp / [ (D + p) * (h + Δh) ] = (-20) x (- 2.45) / [ ( 587,36 – 20 ) x ( 0,423 – 0,03756 ) =
    = + 0,224 rad = + 12,838 deg /heel to SB/
    h1 = h + Δh = 0,423 - 0,03756 = 0,3854 m – new metacentric height
    D1 = D + p = 587,36 – 20 = 567,36 t – new displacement
    7. Definition of the drafts
    a = (B /2) * tg θ = ( 5,5/2) x tg 12,84° = 0,2279 x 5,5/2 = 0,6267 m
    T SB = T1 + a = 2,357 + 0,6267 = 2,984 m
    T PS =T1 – a = 2,357 - 0,6267 = 1,73 m
    As the cargo is unloaded in the midship Xf = 0 and there is no trim!

    ______________________
    NA Razmik Baharyan
     
    Last edited: Sep 10, 2016

  14. Rabah
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    Rabah Senior Member

    Hi roydoky,
    About calculation of Problem 5, Chapter 3 from BST volume 1- Rawson and Tupper, 2001, Fifth edition
    Introduction:
    Old displacement D = γ*V = 9823 x 68 x 10^-3 = 668 N
    Added Weight p = 111 N
    p/D % = 111 / 668 % = 16,6% > 10% - Follow that the adopted cargo is "major"
    If it is necessary to be precise in calculations, it was necessary to use the Bonjean diagram if the adopted cargo generates trim or Hydrostatic curves if to accept that the cargo is posed in the centroid of the square of WL Xf = Xp, that is misses trim.
    But in the statement of the problem it is accepted that within the limits of new and old WL the sides are vertical. At this position and if it is allowable that no trim, follows that it is possible to apply formulas to the reception of a small cargo on a vessel with vertical sides without trim.
    The solution:
    1. A metacentric height before to the cargo reception
    h = Zc + r – Zg
    Zc = 7,62 cm = 0,0762 m
    r = Jx / V = 5,182 x 10^-3 / 68 x 10^-3 = 0,0762 m
    Zg = 18,29 cm = 0,1829 m
    h = 0,0762 + 0,0762 – 0,1829 = - 0,0305 m
    As the old metacentric height is negative, follows that the model floats with heel before to the cargo reception. The cargo p = 111 N should be put with Yp from CL to the side being opposite heeled.
    2. Increase of mean draft
    ΔT = p / γS
    S = 4461 cm^2 = 4461 x 10^-4 m^2 = 0,4461 m^2
    ΔT = 111 / ( 9823 x 0,4461 ) = 0,02533 m
    3. Change of a metacentric height
    Δh =[ p / (D + p) ] * [ T + (ΔT/2) – h – Zp ] =
    = [ 111 / ( 668 + 111 ) ] x [ 0,14 + ( 0,02533 / 2 ) – ( - 0,0305) – Zp ]
    Δh = 0,14249 x ( 0,18317 – Zp )
    T = 14 cm = 0,14 m
    4. On condition:
    h1 = h + Δh > 0
    or - 0,0305 + 0,14249 x ( 0,18317 – Zp ) > 0
    - 0,14249 x Zp > + 0,0044
    Zp > + 0,0044 / - 0,14249

    Zp > - 0,03088 m

    Examples:

    При Zp = - 0,03088m Δh = 0,0305m h1 = - 0,0305 + 0,0305 = 0

    При Zp = - 4 cm = - 0,04 m Δh = 0,0318m h1 = - 0,0305 + 0,0318 = 0,0013 m

    При Zp = - 5 cm = - 0,05 m Δh = 0,0332m h1 = - 0,0305 + 0,0332 = 0,0027 m


    ______________________
    NA Razmik Baharyan
     
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