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Magnitude fo the compressive stress

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data
    two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude fo the compressive stress in rod BC.

    [​IMG]
    edit: p is supposed to be pointing outward (left)

    2. Relevant equations



    3. The attempt at a solution

    im not sure how to do this because i thought stress was calculated at a certain point/area. im not sure how to calculate stress for the whole rod.
    stress = load/area

    how would i do this?
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2009 #2
    Re: stress

    does anybody know how to do this?
     
  4. Jan 25, 2009 #3

    tiny-tim

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    Hi jaredmt! :smile:

    (best to use links, not attachments, because attachments take too long to get approved, and by then your thread is too far down the list :wink:)

    tensile stress = tension force per area,

    compressive stress = compression force per area,

    so if the tensile stress in rod AB is twice the magnitude fo the compressive stress in rod BC, what is the proportion of tension force over compression force? :smile:
     
  5. Jan 25, 2009 #4
    Re: stress

    Tbc = 2Tab
    is that what u mean?
    im pretty sure im calculating at least one of the tensions incorrectly. here's what i have:
    Tbc = load/area = (60,000 - P)/(pi*1.5^2)
    Tab = load/area = (60,000 + P)/(pi*1^2)

    do they look right at all? then i set Tbc = 2Tab to find P but i get a negative answer and its wrong. the answer should be 28.2kips
     
  6. Jan 26, 2009 #5
    Re: stress

    I have a hunch that at the step, that is just 30 kips total, not 60 kips acting there.

    sigma_ab=P/(pi*2^2)/4
    sigma_bc=(B-P)/(pi*3^2)/4
    where B is the load acting at the step, either 30 or 60 kips.

    Then
    abs(sigma_ab)/abs(sigma_bc) = 2
    so that
    abs(P/(pi*2^2)/4) = 2*abs((B-P)/(pi*3^2)/4)
    P/pi = 2*abs(B-P)/(9*pi/4)
    and so forth ...
     
  7. Jan 26, 2009 #6

    tiny-tim

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    Hi jaredmt! :smile:

    (have a pi: π :wink:)
    Nooo …

    Hint: if P was 0, how much would the tension force be in ab? :wink:
     
  8. Jan 26, 2009 #7
    Re: stress

    ah ok got it, thanks :)
     
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