rudder scantling

Discussion in 'Boat Design' started by magwas, Feb 1, 2010.

  1. magwas
    Joined: Oct 2009
    Posts: 287
    Likes: 8, Points: 18, Legacy Rep: 47
    Location: Hungary

    magwas Senior Member

    I am trying to size my steering arrangement, using GL scantling rules here:
    http://www.gl-group.com/infoServices/rules/pdfs/english/schiffst/teil-1/kap-1/englisch/abschn14.pdf

    (I will omit choosing the various factors because it would make the post waay too long. You can look them up in the rulebook above.)

    rudder area:
    1, 75 L T
    A = c1 c2 c3 c4 -----------
    100

    A = 1.0 * 1.0 * 1.0 * 1.5 * 1.75 * 3.0 * 0.22/100 = 0.0173 m^2 = 173 cm^2

    area from freeship is 0.06, but it is needed to be halved.
    highest point is 0.38
    above waterline: (0.38-0.22)*0.1=0.016
    below waterline=0.03-0.016=0.014 < 0.0173

    So I have increased the rudder with 2cm to astern.
    now the area is 0.073
    (0.073/2)-(0.38-0.22)*0.12=0.0173, perfect.

    This calculation has been made without taking trim into account, but I expect the trim to be positive, because the hull is symmetrical longitudinally and the crew will be aft to midship. Because trim being positive the rudder area shall be greater than computed.

    rudder force is
    CR = 132 A v^2 k1 k2 k3 kt
    where k1=4/3, k2=1.1, k3=1.0, kt=1.0

    I take the design speed as 18 kn, a very conservative overestimate I guess.

    CR=132*(0.073/2)*(18**2)*4/3*1.1=2290N

    Rudder torque is
    QR = CR * r
    where
    r = c ( alfa - kb ) , alfa=0.33, kb=0.08

    Now c is the mean breadth of rudder area.
    At first approximation it would be 12cm, if I choose a rudder arrangement which cannot rotate through a transversal axis. And I do just that: one design goal is to make it simple and stupid. Rotation would make the rudder more complex.
    To attack the problem of rudder catching something I would make an arrangement which give up for moderate fore-to-aft or lifting forces.
    However I would compute the rotating case also: when my rudder first breaks while hauling the vessel, or I loose steering in a storm I might want to change it without a major redesign.
    I take the height of the full rudder as c for now as a conservative estimate.

    QR(hard case) = 2290*0.12*0.25=68.7Nm,
    QR(rotating case) = 2290*0.35*0.25=200.4Nm

    I will use the rotating case from now on.

    The rudder stock diameter:

    Dt = 4.2 (QR kr)^(1/3) [mm]

    kr is 1 for yield strength of 235 MPa, and 235/Reh for lower, while (235/Reh)^0.75 for higher tensile strengths
    I have looked up normal values and found that yield strength of "Low Alloy Steel" is between 110 and 2400 MPa

    I have found that the rudder stock diameter should be
    52.6 mm for Reh=110MPa
    24.6 mm for Reh=235MPa, and
    13.8 mm for Reh=2400MPa
    according to these scantling rules.

    To be honest I find these values way too big.
    To see whether the rotating case was too much I computed it for the hard case also, getting 22.2, 17.2 and 9.6 for it, which still seems way too high.

    I can think of three reasons:

    a) I cannot do such simple calculations correctly
    b) The GL scantling rules are not applicable to such small vessel
    c) these are indeed correct numbers

    Which one?

    (I have calculated the same values for the hull speed which is 8.2 knots, and got
    5.8-13.1 for the hard case and 8.1-18.7 for the rotating one. )
     
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