Scantling Moment of Inertia

Discussion in 'Boat Design' started by Mitch1990, Jun 3, 2020.

  1. Mitch1990
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    Mitch1990 Junior Member

    Hi,

    I am developing a spreadsheet for midship frame scantlings. I am struggling with the use of angle for plate stiffening, specifically when it comes to moment of inertia. My notes state that when finding Icl (moment of inertia of the vertical neutral axis), that moment of inertia of vertical components such as side shell plate have a negligible impact on Icl and are ignored when calculating moment of inertia of the entire cross-section. If a strength deck was longitudinally stiffened with vertical flat bar, would the moment of inertia of these stiffeners be neglected from finding Icl?

    If the same strength deck was longitudinally stiffened with angle, would the moment of inertia of these stiffeners still be neglected when finding Icl?

    My thoughts are that the portion of the angle that is parallel to the strength deck increases the plates stiffness when resisting a horizontal bending moment (caused by heeling, rolling etc).

    Mitch
     
  2. gonzo
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    gonzo Senior Member

    Is this to comply with class regulations? I expect that any software for ship design would address class requirements instead of engineering first principles. Even if part of a structural member increases the stiffness of the deck, What matters most is if it is in compliance. You can always show that a different method is equivalent. Is that what you are trying to do?
     
  3. Mitch1990
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    Mitch1990 Junior Member

    Hi Gonzo, it is to comply with ABS but it is an assignment not a real design, hence having to use a spreadsheet. The question is more of a basic understanding issue I am having.
     
  4. Barry
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    Barry Senior Member

    You made the comment that "your notes state that the shell plate would have negligible impact on Icl". This is at odds with most strength analysis of beams.

    In the current thread on this forum, "Attached Plate Effective Width" Post #12, Adhoc mentions that as a rule of thumb that you can use 50t the web thickness to determine how much of the material that you can use
    in your calculation. I have not heard this approximation but say the distance between stringers was an excessive 4 feet. You would not expect that the skin flange to not deform and reduce the ability of the beam to carry the
    theoretical design load. A generally accepted concept is that you can use half the distance to the next stringer and if the skin plate is quite thick and the half the distance is not enough to cause premature deflection, you could run with that

    But the 50t sounds reasonable and perhaps increases the factor of safety somewhat

    What is important that you utilize some of the skin material as one of the flanges to come up with the moment of inertia IF you intend to cut access holes in the web of the angle. Normally, you would want to cut
    access holes at the neutral axis to not compromise the beam and without including some of the skin as one of the flange, if you use just the angles NA to cut access holes, the hole will not be on the true NA of the
    skin and angle made up beam.

    I forgot to add that I am making the assumption that the stiffener/skin combination
    would be designed for equal positive or negative bending. If the outer skin is the flange in compression and it is thin, then deflection is more likely to occur at a lower bending moment loading than if the thin flange is in tension.
     
  5. Ad Hoc
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    Ad Hoc Naval Architect

    Something is not correct here.
    Thus to avoid any confusion, you'll have to provide a sketch of what you're referring to, as it may be something simple that has been misunderstood.
     
  6. Mitch1990
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    Mitch1990 Junior Member

    Agreed, I found it difficult to explain. The moment of inertia of the vertical components in the below midship frame are not included when calculating Icl which corresponds to what is said in my notes. If the vertical element circled was changed to an angle section, would the moment of inertia of the new section need to be included when calculating Icl?
    upload_2020-6-3_19-5-3.png
     
  7. Ad Hoc
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    Ad Hoc Naval Architect

    Ok.. now it is clear.

    If you calculate the modulus of just the items noted below:

    upload_2020-6-4_8-10-21.png

    Without that area circled, see what the value is... and then do the same but WITH it... you will see that the contribution of the that one small item is very minor.

    If you have a whole sides bottom, side and deck with many of them.. then it does provide a good contribution. But one single FB/angle bar... not really.
     
  8. Mitch1990
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    Mitch1990 Junior Member

    Understood, in a more realistic situation where you might have multiple longitudinal stiffeners that are angle bar, does the moment of inertia of the vertical components contribute to finding the Icl. This is from my course notes in the section for "Tabular Calculation for Moment of Inertia about the Vertical Neutral Axis (Icl)"
    upload_2020-6-3_19-17-12.png
     
  9. Ad Hoc
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    Ad Hoc Naval Architect

    This is something - now you understand how to do this calculations - you can do yourself.

    Draw a simple box of dimensions 10m wide and say 3m deep.
    Make the bottom 10mm the sides 6mm and the deck 4mm.
    Caluate the I.

    Then add one FB of say 10x100mm on the CL on the main deck.. then reclaute the I... what is the difference?

    Then repeat, but have a FB spaced every 250mm on the deck.. what is the difference?
    Then repeat and do the same now using the sides.... calculate the I
    And finally.. repeat but now adding on the bottom...calculate the I

    Once you have set up your spread sheet.. this should take you no more than about 10mins.

    Compare each I you calculate and see for yourself.. how much difference 1 FB makes, or a deck full of them.. or a deck + sides or deck + sides + bottom makes.

    And there is your answer...
     
  10. Mitch1990
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    Mitch1990 Junior Member

    Thanks for your help
     
  11. TANSL
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    TANSL Senior Member

    I suppose you have taken into account that all those moments must be taken with respect to the neutral axis of the transversal section, so the first thing you should calculate, before launching into calculating moment of inertia (second moments of areas), is the position of the neutral axis of the section.
    Once you are clear on the procedure, if you have AutoCAD and you are interested, I can teach you a procedure to perform all these calculations with total precision and in a few minutes.
     
  12. Mitch1990
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    Mitch1990 Junior Member

    I do indeed have autocad and spent a good part of the day trying to figure it out. I mainly use solidworks. If you could that wouod be great
     
  13. TANSL
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    TANSL Senior Member

    You have to draw each piece, sheet metal panel or reinforcement, exactly as it is, with its true dimensions and thickness. Each piece must form a closed polyline. You convert all the pieces into regions and give the "union" command to generate a single region. Done this the command "massprop" will give you all the properties of that region, area, center of gravity, ... This is an example:

    ---------------- REGIONS ----------------

    Area: 8189656.0000
    Perimeter: 17314.0000
    Bounding box: X: 71386.0000 - 78962.0000
    Y: 31207.0000 - 32288.0000
    Centroid: X: 75174.0000
    Y: 31747.5000
    Moments of inertia: X: 8255182554179100
    Y: 4.6320E + 16
    Product of inertia: XY: -1.9545E + 16
    Radii of gyration: X: 31749.0336
    Y: 75205.8060
    Principal moments and X-Y directions about centroid:
    I: 7.9751E + 11 along [1.0000 0.0000]
    J: 3.9171E + 13 along [0.0000 1.0000]

    Now what you should do is translate the region so that its centroid is at the coordinate origin, and repeat the command "massprop". Now yes, the values will be what you were looking for.
     

  14. Mitch1990
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    Mitch1990 Junior Member

    Thanks for your help
     
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