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Stress concentrator

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A plate with a hole in the centre is loaded in tension It has a height of 86mm (W), a thickness of 12 mm, a hole diameter of 33mm and a yield stress of 346MPa. The tensile load is 117kN.
    A) Determine the stress on the top edge of the hole
    B)Determine the stress on the top edge of the plate above the hole
    C)Determine the maximum moment that the plate can withstand based on the stress at the top edge of the hole
    D)Determine the maximum moment that the plate can withstand based on the stress at the top edge of the plate above the hole

    2. Relevant equations
    [itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks
    σref = [itex]\frac{P}{Wt}[/itex]

    3. The attempt at a solution
    So for part A and B i have a chart i can use to find the stress concentrator factor, i get [itex]\frac{d}{W}[/itex]=0.384 which leads to to values for K1 the inside of the hole as 3.65 and K2 the top of the plate 0.85( From My chart) I then used σref = [itex]\frac{P}{Wt}[/itex] to find my reference stress.
    σref = [itex]\frac{117x10^3}{0.086*0.012}[/itex] = 113.37x10^6 Pascals
    Then using
    [itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks >> σs = K*σref
    For the top of the hole, σs = 3.65 * 113.37*10[itex]^{6}[/itex] = 413.8*10[itex]^{6}[/itex] Pa
    For the Top of the plate above the hole, 0.85*113.37*10[itex]^{6}[/itex]=96.36*10[itex]^{6}[/itex]Pa

    I believe i am correct up to this point. For part C and D i don't really know exactly what its asking for, i have another chart which has stress concentrator values for a plate in bending or i have the bending equation σ=-[itex]\frac{My}{I}[/itex]
    I'm using the chart, but i'm not quite sure if its correct.

    Using the other chart i find that the stress concentrators are:
    K2 = 1 ( top of the plate)
    K1 = 0.75 ( Bottom of the plate)
    This chart has 2 equations attached with it :
    [itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks & [itex]\frac{6M}{W^{2}t}[/itex] = [itex]\sigma[/itex]ref
    So the Reference stress is biggest when K is its smallest value, therefore K1 is used. So for part C, based off the stress at the top of the hole σref = [itex]\frac{413.8*10^6}{0.75}[/itex]
    [itex]\sigma[/itex]ref = 551.73*10^6 Pa
    [itex]\frac{6M}{W^2 * t}[/itex] = [itex]\sigma[/itex]ref
    M = [itex]\frac{551.73*10^6}{0.086^2 * 0.012}[/itex] = 8161Nm
    This answer isn't correct according to my sheet. I'm not sure what I'm doing wrong, for part D i would just do the same just with different stress value. Maybe i'm missing something as i haven't used the yield stress presented to me in the question.

    Any help would be much appreciated.
     
    Last edited: Nov 2, 2013
  2. jcsd
  3. Nov 3, 2013 #2

    PhanthomJay

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    Check stress concentration factors in bending, and correct your equation for ref stress you used 6M/W^2t instead of 6M/Wt^2. I once at the University had to solve for stress concentration factor at a hole using finite elements and partial differential equations that would challenge even the most skilled of the calculus gurus. And it was done by hand without any fancy computers. And I all the better for it I guess, having survived that awesome task. All bragging rights reserved.
     
  4. Nov 10, 2013 #3
    I think it is better to use the yield stress instead of the stress that you had found in the first part.
    Try fs = 346MPa
    at the top of the hole σref = 346/0.75= 461.3 MPa
    σ_ref=6M/(w^2 t)
    from this equation M = 6.8235496 KN.m
    same for the top plate
    K2 = 1
    σref = 346 MPa
    so M = 5.118 KN.m
    I think it is better to use the yield stress because you have been asked to determine the maximum moment and that occurs when the stress equal to the yield stress before failure could happen
    I hope you got the answer now
    cheers
     
  5. Nov 11, 2013 #4

    nvn

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    steve2510: You failed to attach the diagrams that go with the question, and you failed to attach your stress concentration charts. Therefore, we have no way of knowing whether you are referring to in-plane or out-of-plane bending. And you did not state whether you are referring to in-plane or out-of-plane bending. Besides (even if we assume in-plane bending), we have no way of knowing exactly how your charts are defined. Therefore, we do not really know what you are talking about, nor whether or not you used the charts correctly. Furthermore, for a plate with a hole subjected to axial tension, sigma_nom (sigma_ref) is usually defined as P/[t*(W - d)], not P/(W*t). Or, for in-plane bending, sigma_nom (sigma_ref) is usually defined as perhaps something like 6*M*W/[t*(W^3 - d^3)], not 6*M/(t*W^2). That is how most charts are defined; but it completely depends on your particular charts, which we would need to see, in case you have nonstandard charts.

    Also, I am not too familiar with K values less than 1.00. Therefore, some of your numbers do not seem to make sense yet. But like I said, we would need to see how your charts are defined, and what you are talking about, in order to check your work.
     
    Last edited: Nov 11, 2013
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