moment of inertia for inclined plates

Discussion in 'Boat Design' started by abhishek, Apr 24, 2012.

  1. abhishek
    Joined: Sep 2009
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    abhishek Junior Member

    What is the moment of inertia of a rectangular plate making an angle except right angle or multiple of right angle with it's axis which is passing through the centroid of the plate?


    SIR,.

    Please help me in this regard, I have a doubt in finding out the moment of inertia (Iown) for bottom plate and side shell plate as the bottom shell is making 18 degree with the base line and the side shell is making 5 degree with the vertical reference line, the sizes are (bottom is 15300mm x 6 mm ) and (side shell is 14250 mm x 6 mm ), suppose if the bottom shell is parallel to base line it is straight to horizontal and not making any angle, it is {(15300) x (6)3 }/ 12 = 275400 mm4 .

    If it is side shell {(14250) 3 x (6) }/ 12 = 1.44 x 1012 mm4

    But here both the bottom and side shell are not horizontal and vertical.
    So I know the formulae should be in sin and cos as they are making some angles.

    For bottom shell moment of inertia is :
    [{(15300) x (6)3 }/ 12] x cos2 18° + [{(6) x (15300)3 }/ 12] x sin2 18° = 249101.64 mm4

    For side shell moment of inertia is:
    [{(14250) x (6)3 }/ 12] x cos2 (90-5° )+ [{(6) x (14250)3 }/ 12] x sin2(90- 5°) = 1.435 x 1012
    mm4

    IS the calculation correct?


    Or else could any one solve for me please.
    Please find the attached copy of the section .
     

    Attached Files:

  2. abhishek
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    abhishek Junior Member

    hope i could get an answer from adhoc sir.........
     
  3. TANSL
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    TANSL Senior Member

    For each plate you have to consider the following coordinate system: X axis in the direction of the plate, Y-axis in the direction perpendicular to the plate. With this you can calculate Ix and Iy, according to the traditional formula, and the situation of the neutral axis, shear area and other properties necessary for the calculation of scantlings.
    shell side Ix : (1425 x 6^3) / 12 = 25650 mm4

    CAUTION: Check the dimensions on your midship section
     
  4. abhishek
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    abhishek Junior Member

    thank you fro your early reply sir....i shall have a check
     

  5. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    You’re over thinking this.

    Take the lower bottom shell plate of 15.3m at 18 degrees.

    The inertia of this is the inertia if itself and then the inertia about a given axis.

    Itself, is simply I=A.k^2 where k^2 = n^2/12, where n = the vertical span (distance) of the member.

    So this is A = 15300x6mm = 91800mm^2, or in proper numbers = 918cm^2

    And k^2 = 430^2/12 = 15408 cm^2.

    Therefore I = AK^2 = 918 x 15408 = 14144544 cm^4 or 0.1414 m^4

    The I about an axis, is the then area x lever^2, (in the usual manner from an assumed axis to its centroid) = 918 x (1/2 x 4.3m)^2 = 0.4243m^4.

    So, for this one member the total is the sum of the 2 inertias, and then you need to readjust to the Neutral axis, found in the usual way.
     
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