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Homework Help: Question involving Slenderness Ratio and E.S.R.

  1. Sep 15, 2018 at 12:42 PM #1
    1. The problem statement, all variables and given/known data

    A column has the dimensions shown in the diagram below. (a) What is the minimum length of the column at which buckling is likely to occur? (b) If the column is the length determined in (a), (i) What will be the mode of failure? (ii) At what load would you expect failure to occur?

    upload_2015-10-26_15-10-26-png.png

    The specific problem I am having with this question is for b)i) how to determine what the mode of failure will be from the E.S.R. and the S.R. Admittedly this question has been asked many times and people seem to say it will buckle at the length determined in a). Now reading my course material, it states that if your E.S.R is below your S.R. then it will be a crushing mode of failure and above will be buckling. Now I have an E.S.R of 118.741 and a S.R. of 237.52. Am I correct in thinking that this will be a crushing mode of failure with the length of 5.938 m as we have just worked out that the minimum length at which buckling is likely to occur is 5.938 m, so how can this length be the minimum for bucking when the slenderness ratio is 237.52. I would like to make sense of this and I feel like I am chasing my own tail at the moment. Thank you very much for any assistance.
    2. Relevant equations

    Euler's equation. Fc= ((pi^2)*E*I)/Le^2

    E.S.R. = Le/k

    S.R. = L/K


    3. The attempt at a solution

    Effective Length = 2.969 m
    Length = 5.938 m
    k = 0.025
    Fc = 306 kN
    I = 13.7*10^-7
    E.S.R. = 118.741
    S.R. = 237.52
     
  2. jcsd
  3. Sep 16, 2018 at 5:47 AM #2

    PhanthomJay

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    The ESR is
    kL/r, where k is a function of the boundary conditions and equal to 0.5 for the fixed-fixed ends, and r is the radius of gyration which is equal to sq rt of (I/A). I don’t know where you are getting your formulas. Also, the SR L/r has no place here. The column will crush instead of buckle when it’s length is shorter than the minimum buckling length determined at the yield stress of the steel.
     
  4. Sep 16, 2018 at 8:09 AM #3
    Thank you for the reply and clarification. I have taken a screen shot of a couple of pages from the learning materials which can show you where I get some formulas from and the part about slenderness. In particular the paragraph below the E.S.R. formula on the euler's equation page. This is the one talking about slenderness ratio of a material, i.e. steel, so if I don't know the material being used, does that mean the way it fails will change with different material choices?
    upload_2018-9-16_14-9-28.png
    upload_2018-9-16_14-9-44.png
     

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  5. Sep 17, 2018 at 7:29 AM #4

    PhanthomJay

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    Yes. As long as the geometric properties of the column are the same, the material used does not change the ESR. However , since bucking depends on E, and type of failures depends on the yield stress of the material, then it makes a big difference when using other materials. Steel for example has an E modulus about 20 times greater than E for wood, so for a given long length its critical buckling load is 20 times greater than wood. And in terms of the minimum length for buckling at yield stress, the wood , which has a breaking stress about 10 percent that of steel, will buckle at about a half the min length of the steel. Except for short columns, the wood will always buckle before it crushes.
     
    Last edited: Sep 17, 2018 at 11:04 AM
  6. Sep 18, 2018 at 3:23 AM #5
    Thank you very much PhanthomJay. I feel far more confident with my understanding of this now.

    Best regards.
     
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