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Static engineering v -- stress/buckling

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A column has the dimensions shown below...

    untitled.GIF


    Column material properties:

    young's Modulus E = 200 GN m^-2

    Yield stress = 140 MN m^-2




    2. Relevant equations


    a) what is the minimum length of the column at which buckling is likely to occur?



    3. The attempt at a solution


    i got an answer of: 2*((118.74 * Sqrt((2.33x10^3)/(2800x10^-3))) = 6850.56 M
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 19, 2012 #2

    nvn

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    oxon88: Your current answer is wrong. Try again. First, list relevant equations. And then show how you computed each numerical value used in your final equation.

    By the way, MN*m^-2 is called MPa. And GN*m^-2 is called GPa. Always use the correct, special name for a unit. E.g., 140 MPa, not 140 MN*m^-2.
     
  4. Jan 20, 2012 #3
    1. The problem statement, all variables and given/known data

    a) what is the minimum length of the column at which buckling is likely to occur?


    3. The attempt at a solution


    least second moment of area. I = (B^4 - b^4)/12 = (80^4 - 60^4)/12 = 2.33x10^6

    area. A = B^2 - b^2 = 80^2 - 60^2 = 2800

    σ = 140x10^6

    E = 200x10^9

    effective slenderness ratio. E.S.R = sqrt(pi^2.E/σ) = sqrt(pi^2.(200x10^9)/(140x10^6)) = 118.74

    effective length. L = (E.S.R). Sqrt(I/A) = 118.74(sqrt(2.33x10^3)/(2800)) = 3427.7


    effective length = 1/2 column length. = 3427 x 2 = 6854 mm


    --------------------------------------------------------------------------------------


    I also tried another method and got the same answer......


    Least value of radius of gyration. K = (1/2).sqrt((B^2+b^2)/3) = 1/2 sqrt((80^2+60^2)/3) = 28.868

    effective length. L = (E.S.R x k) = 28.868 x 118.74 = 3427.73

    effective length = 1/2 column length. = 3427 x 2 = 6854 mm
     
  5. Jan 20, 2012 #4
    ok. so i worked it out.

    the answer is 5.94m
     
  6. Jan 20, 2012 #5

    nvn

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    Your answer in post 4 is correct.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 5.94 m, not 5.94m. See the international standard for writing units (ISO 31-0).
     
    Last edited: Jan 20, 2012
  7. Jan 20, 2012 #6
    ok. So if the column was to be 5.94 m. What would be the mode of failure? and at what load would you expect the failure to occur?

    I guess it would be buckling?
     
  8. Jan 23, 2012 #7
    any ideas?
     
  9. Jan 24, 2012 #8
    useless forum....
     
  10. Jan 24, 2012 #9

    nvn

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    We are not allowed to answer your homework questions for you.
     
  11. Mar 28, 2013 #10
    Hi all, I have the same question and was getting the same answers originally. I assume that your new answer is based on changing the formula to a hollow tube. I was wondering without being told in the question how we are supposed to determine what shape it is... Or am I being a bit thick...
     
  12. Jul 2, 2013 #11
    Wrong formula

    The least second moment of area I about the xx axis is not what you stated so your value is incorrect. I get L = 5.236389088 m
     
  13. Jul 4, 2015 #12

    David J

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    I am just wondering if this thread is still active as I am studying this question and facing similar problems to those on here
     
  14. Jul 4, 2015 #13

    SteamKing

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    The last post was two years ago, so, no, it's not still active. :frown:

    However, you can and should create your own, fresh thread. It's not like you can create only so many threads at PF before you run out. :wink:
     
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