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Math Help - Can anyone help with this problem on Force??

  1. #1
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    Can anyone help with this problem on Force??

    A circulur column has a diameter of 80mm outside edge and 60mm inside edge.
    Both ends are fixed with equal force being applied top and bottom.
    Young's Modulus E = 200GN m-2
    Yield Stress deltay = 140 MN m-2

    What is the minimum length of column at which buckling is likely to happen?
    If the column was this length what load would you expect failure to occur?
    If the column is half this length what load would you expect failure to occur?

    Really struggling with this one if anyone can help id really appreciate it along with the working of how they came to their final answers......
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    Are you familiar with Euler's buckling equation?

    F_c = \frac {\pi^2 EI}{(KL)^2}

    where the factor 'K' is defined by the type of end supports - for both ends fixed K = 0.5.

    If you divide through by the cross-sectional area of the column you get stress:

     \sigma_c = \frac {F_c} A = \frac {\pi^2 EI}{A(KL)^2 } = \frac {\pi^2 E}{(KL/r)^2} where r is the radius of gyration (ie I = Ar^2)

    Given the value for yield stress this lets you find the value L that would result in buckling. Then use the first equation to find the resulting critical load. As for the third question - from dimensional analysis you can see what happens to F if L is divided by 2.
    Last edited by ebaines; September 4th 2013 at 01:46 PM.
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    Re: Can anyone help with this problem on Force??

    I seen this in an earlier post but didn't really follow it that great I am afraid and was looking for more detail with the info I had given plug into to follow it better if possible?....
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    From this equation:

     \sigma_c = \frac {F_c} A = \frac {\pi^2 EI}{A(KL)^2 }

    Rearrange to make 'L' the subject:

     L = \sqrt{\frac {\pi^2 EI}{\sigma_c AK^2}}

    From the dimensions you were given you will need to calculate values for I and A, and you have been given E and \sigma_c. What do you get for L?
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    Re: Can anyone help with this problem on Force??

    Area= pi(R^2-r^2) Ixx=pi(R^4-r^4)/4
    = pi(40^2-30^2) =pi(40^4-30^4)/4
    =700pi which equals 2119.114858 =1374446.786 which equals 1.37x10^4

    Is this correct so far?? As you can see I am really struggling with this!!!!
    Last edited by Jock; September 4th 2013 at 02:07 PM.
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    Re: Can anyone help with this problem on Force??

    After further working I got
    Area=pi(0.04^2-0.03^2)
    =2.20x10^-3


    I=pi(0.04^4-0.03^4)/4
    =1.37x10^-6


    ESR=[sqrt]pi^200x10^9/140x10^6
    =118.74

    Then I did 118.74x[sqrt] 1.37x10^-6/2.20x10^-3
    =2.96

    Is this any closer or along the right lines now??
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  7. #7
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    Re: Can anyone help with this problem on Force??

    After lots of number crunching I get the minimum length of buckling in the column to be 5.93 m is that correct?
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  8. #8
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    Re: Can anyone help with this problem on Force??

    Quote Originally Posted by Jock View Post
    After further working I got
    Area=pi(0.04^2-0.03^2)
    =2.20x10^-3

    I=pi(0.04^4-0.03^4)/4
    =1.37x10^-6
    So far so good.

    Quote Originally Posted by Jock View Post
    ESR=[sqrt]pi^200x10^9/140x10^6
    =118.74
    You forgot about I and the factor K.

    Quote Originally Posted by Jock View Post
    Then I did 118.74x[sqrt] 1.37x10^-6/2.20x10^-3
    =2.96
    Is this any closer or along the right lines now??
    I don't understand what you are attempting here.
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  9. #9
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    Quote Originally Posted by Jock View Post
    After lots of number crunching I get the minimum length of buckling in the column to be 5.93 m is that correct?
    Yes, I believe it is.
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    Re: Can anyone help with this problem on Force??

    so for mode of failure for a the beam stated I found the beam will buckle I got a load of failure of 3.143x10^-19. And if the column is half the length of 5.93m the load I would expect failure to occur would be 1.257x10^-20 therefore failure by yielding. Does that all seem correct?
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  11. #11
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    I get a different result. If you get a result for buckling load that is on the order of 10^-19 newtons you should recognize immediately that it can't possibly be correct.
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    Re: Can anyone help with this problem on Force??

    I thought that and the same with the second. I cant seem to see where I am going wrong though and hence struggling to work out the right answer.....
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  13. #13
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    Show us the math. You're using

     F_c = \frac {\pi^2 E I}{(KL)^2}

    right? You found the length L, so plug and chug. Alternatively you can simpl use  F_c = \sigma_c A. Also, whenever you do egineering calculations you should always include units, because messing up units is a sure-fire way to go wrong.
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    Re: Can anyone help with this problem on Force??

    sure I have pi^2x200x10^9x140^6/(0.5x5.93)^2 for the first one and for the one half the length I have pi^2x200x10^9x140x10^6/(0.5x2.965)^2 hope you find the errors of my ways.....
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  15. #15
    MHF Contributor ebaines's Avatar
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    Re: Can anyone help with this problem on Force??

    Quote Originally Posted by Jock View Post
    sure I have pi^2x200x10^9x140^6/(0.5x5.93)^2 for the first one
    That would give you 3.14 x 10^19, which is cearly not right either. Why did you include the factor 140x10^6? You should be using the value for I there, not stress. I'll say it again - include units in your calculation and you won't make this type of error.

    Quote Originally Posted by Jock View Post
    and for the one half the length I have pi^2x200x10^9x140x10^6/(0.5x2.965)^2 hope you find the errors of my ways.....
    Same problem as above. You can actually do a bit of a shortcut on this, recognizing that there is an L^2 term in the denominator. If you cut L in half, what does 1/ L^2 become?
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