Some questions about ISO 12215-5: 2019

Discussion in 'Class Societies' started by TANSL, May 7, 2021.

  1. Alik
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    Alik Senior Member

    In 2008 the approach is correct. Shear strength is increasing with length.
    In 2019 the approach is opposite...
     
  2. Ad Hoc
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    Ad Hoc Naval Architect

    I understand that, and your comment.
    But I don't have the 2019 rules to check the definitions and if there are any caveats, as previously noted.

    Since the formula you wrote simply means if L is less than 5.8m, it defaults to the minimum of 0.3MPa.
    But if the max L is 24m, there is then a max up to 0.58MPa.

    Thus there must be a means for interpolation between the 5.8m to 24m? for the min of 0.30 to max of 0.58MPa.

    The previous one was
    below 10m, = 0.25 - so this is an increase.
    Then 10 - 15m a sliding scale from 0.25 - 0.40, by interpolation,
    Then 15 - 24m = 0.4MPa.

    So, other than a means for interpolation between 5.8 to 24m, it appears consistent., with the lower and upper values greater than the 2008 version.
     
  3. Alik
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    Alik Senior Member

    Understood, thnaks, but interpolation should have physical sense...
     
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  4. rxcomposite
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    rxcomposite Senior Member

    It is table 13 in my 2008 version. Minimum core shear according to length. The values are;
    Less than 10m= 0.25
    10 to 15m = 0.25 to 0.40 Formula 0.25+ .03*(LH-10). max at 0.4 for 15m.
    15 to 24m =0.4

    Finding the extrapolation formula for 15 to 24m. with an upper value of 0.58, I used problem solver in Excel.

    1. Readjusting, it should read 16 to 24 m and have 9 division each for every meter.
    2. Looking at the first formula, it is a linear progression. 0.03*(LH-10)
    3. The first term is an offset.
    Thus it is;

    0.28 + 0.02*(LH-9)

    for values of 16 to 24 m max at 0.58. Works for a length and a fraction.
     
  5. Alik
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    Alik Senior Member

    Yes this is the key: in 2008 option LH is with plus, in 2019 is with minus. I believe formula from 2019 is faulty.
     
  6. Alik
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    Alik Senior Member

    To add, I also believe that linking compressive strength of core with base pressure is nonsense.
    What is the relation of bottom pressure to the core used for deck and cabin?

    Say, for high speed boats, one won't be able to use DIAB H80 core in the deck, becasue it does not pass this criteria. Most of classification societies just specify min properties of the core not linking it to pressure.
     
  7. rxcomposite
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    rxcomposite Senior Member

    How was it worded?
     
  8. Alik
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    Alik Senior Member

    This is in Table A.7, right column.
     
  9. rxcomposite
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    rxcomposite Senior Member

    Mine goes only up to table A.2. Can you post a snapshot? Maybe I have a different tag for the rule.
     
    Last edited: Mar 23, 2022
  10. Ad Hoc
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    Ad Hoc Naval Architect

    RX
    Alik is referring to 2019 version, not the 2008 version which we have.
    So unless someone can quote the rules exactly as written from 2019 edition, or just copy it... we have no visibility of table A.7 and the words/formula used.
     

  11. rxcomposite
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    rxcomposite Senior Member

    Alik,

    I ran the calculations several times and all came out to be the same. The formula we discussed seems to be applicable only to bottom pressure. In ISO 2008, there is a formula for motorcraft deck pressure PDM eq 15 of which you need to define first PDM Base eq 17. PDM Base is dependent on Lwl, which is a fraction of LH. For a 24 meter displacement boat, with my given parameters, I get 6 kN/m2 of pressure. Just slightly above the 5 kN/m2 PD MIN eq. 16.

    You are right, compression property of core is not needed but pressure is necessary.

    ISO does not have a tabulated table for cored laminate. I used Table H2 ply calculation as a guide. Given only pressure and b short dimension and the formula Design shear force = pb^10^-4, I get 0.15 N/mm2. With only p (6kN/m2) and b(500mm) given, I validated this formula using the standard mechanical engineering formula for finding shear stress Max shear = (V*Ib) *b(tc/2)*y where:

    V= Vertical shear (750 N)

    I= second moment of Inertia (41667 mm4)

    b= width of core (500mm)

    tc= thickness of core (10mm)

    y= distance to N.A. (Zcrit) from centroid of upper core. (2.5 mm)

    This gave me 0.23 N/mm2 max shear. Note that this is max shear. The formula can be simplified to Max shear = (3/2) V/bt. Average shear is V/bt which is 0.15 N/mm2, the same as ISO result.

    Validating further, using the same inputs, LR gave me a core shear of 0.15 N/mm2 with a slightly different formula, Shear = (pb /2tc) x10^-3. Any of the formula is valid.

    Knowing the thickness for the required shear, I can now choose among the core candidates from balsa, PVC, SAN, ect. With 0.15 N/mm2 of design shear and 10 mm core thickness, you have plenty of choices.

    If I don’t like the 10mm thickness, or if the size is not available, I choose a different thickness and recalculate.
     
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