Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Keel design, high tensile steel elongation

  1. Jul 3, 2011 #1
    To simplify things, I have assumed the keel section is a rectangle and not a foil.

    (Also, these specs are taken from an existing design.)

    dimensions:

    keel 38cm (width/chord) X 3.8cm (height/thick), length 1.64m,

    bulb 445Kg


    Using beamboy, (cantilever with weight at end) the deflection is:-

    Using a mass of 445Kg, Moment of Inertia of 380mmX38mm = 1740000, Gpa 210, distance to mid 19mm

    Gives a deflection of 17mm.


    What I would like to know, is this considered a failure. ie the steel was stressed more the 0.2% allowed.



    About this allowed 0.2% elongation:-

    From a rod rigging spec, 1mm^2 can support 140Kg. = 0.7% elongation. So how does HT steel develop its strength. All steel has approx. the same Modulus of Elasticity, but different ultimate strength, then does it mean that HT steel is allowed to elongate more, ie to 0.7%?

    But the yield of point of HT is not well defined, so 0.2% is used????

    I'm a little muddled.

    thanks,
     
  2. jcsd
  3. Jul 3, 2011 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    For different steels, the key parameter used in stress analysis is the yield stress. This stress value represents the upper limit of elastic behavior in the material. Elastic behavior in this context means that when the material is loaded, a certain strain is created. Once the load is removed, the previous strain disappears, with no permanent set being created.

    Typical mild steel (ASTM A-36) has a min. yield of 36000 psi (248 MPa). AISI 4340 steel has a min. yield of about 78000 psi (540 MPa). Both steels have the same Young's modulus (about 209 GPa).

    Different HT steels have different yield points, due to alloy composition, heat treatment, etc. If you are going to design using HT steel, you should know which kind you are dealing with.

    I can't comment on your keel design calculation because it is not clear what you are trying to analyze.
     
  4. Jul 3, 2011 #3
    Thanks SteamKing,

    Have a cantilevered beam 1.64m long thats deflected 17mm, has it failed? ie. more than 0.2% elongation?

    HT steel has a 0.7% allowed elongation?
     
  5. Jul 3, 2011 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I calculate a bending stress of about 78 MPa (11300 psi) in the keel, assuming the 445 kg bulb is located 1.64 m from the fixed point of the keel. Based on these assumptions, it does not appear that the keel has been overstressed. I am also assuming that the keel is solid steel 38 mm thick.

    The .2% elongation is usually referred to when a tensile test is performed on a specially machined bar-shaped test piece. Again, to emphasize my earlier comments, steel design is based on a calculation of stress rather than a measurement of strain. Due to the large number of different steel alloys in use, a proper structural evaluation requires that the particular alloy from which the member is fabricated should be known.
     
  6. Jul 4, 2011 #5

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    idfka909: You can assume you have mild steel (tensile yield strength, Sty = 248 MPa). I think the modulus of elasticity might be as low as E = 200 000 MPa. As mentioned by SteamKing, the bending stress is sigma = 78 MPa. Therefore, if we pretend you have a uniaxial stress state, then your strain (elongation) is epsilon = sigma/E = (78 MPa)/(200 000 MPa) = 0.04 %, which is much less than the yield strain (Sty/E). Therefore, these values indicate the keel is not overstressed nor overstrained. This is not considered a failure. And the cantilever deflection you computed is correct.
     
    Last edited: Jul 4, 2011
  7. Sep 16, 2011 #6
    Many thanks nvn,
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook