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Revision on Past Exam Paper

  1. May 10, 2007 #1
    Revision on Past Exam Paper !!

    Hey there, i am stuck on this question in my past exam paper, and i have tried many ways of doing it but just simply not getting any where !!
    So please would u mind helping me out a little here !!??

    A rectangular cross section hollow steel tube (outer diameter 50 mm and 65 mm) is to be used as a strut. In initial designs the ends of the strut are held in position but not restrained against rotation. The distance between the restraints is 2m, and the load it has to carry is 180kN.

    (a) The wallthickness is 4mm. Is the design safe? (you may assume Euler's buckling theory is valid) (v = 0.3, E=205GPa for steel; sigmaY = 250MPa)

    (b) The tube is now restrained against rotation at one end while being pinned at the other. Find the maxiumum allowable load using both Euler and AISC formula (given below). Fully explain design process.

    NOTE: The AISC rules state for L/K (effective length / radius of gyration) values less than (L/K)c, where (L/K)c = squareroot {(2*pi^2*E)/(SigmaY)} plot the parabolic curve: sigma = sigmaY*{1-((L/K)^2)/(2*(L/K)c)}
    Smoothly join this parabola into the Euler curve for L/k values greater than (L/K)c.

    Here is the question, sorry it is a bit long, but i just dont know where to start lol !!

    Thanks !!
     
  2. jcsd
  3. May 10, 2007 #2

    FredGarvin

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    Buckling at it's finest. What is the basic equation for buckling of a column? That's the place to start.

    [tex]P_{cr} = \frac{\pi^2 E I}{L_e^2}[/tex]

    The end connections dictate what the effective length of the column is.
     
  4. May 10, 2007 #3
    Re: Revision on Past Exam Paper !!

    So just get the numbers for this formula and only to get the LOAD !!

    What does the question mean by is the design safe !?? i don't know what to caluclate for that or wrtie about??


    how do i caulate the maximum allowable load when the strut is restrainted is there a different formula ???

    Thanks lot !!
     
  5. May 11, 2007 #4

    FredGarvin

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    You can decide the "safeness" by calculating the safety factor. The 4mm wall thickness dictates your cross sectional moment. Use that to calculate your allowable critical load and then compare that to the 180 kN applied load. Safe is a relative term. At the bare minimum, you want a safety factor of at least 1.0 which means your applied load is exactly the same as the critical load.

    Usually on questions like this, you need to justify with your own judgment what is considered safe.

    BTW, your text should have a table of some kind showing different end conditions and the associated effective lengths that is used in the calculation like this:

    http://en.wikipedia.org/wiki/Image:ColumnEffectiveLength.png
     
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