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Solid Mechanics(Stress Question)

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    In the picture below


    2. Relevant equations
    Stress=P/A


    3. The attempt at a solution
    I equated stress AB=-stress BC
    p/pi=-(p-60 kips)/2.25pi

    P=18.4615kips
    This is not the right answer according to the book
     

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    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 23, 2012 #2
    Does anybody know whether my answer is right or wrong?

    Thank you
     
  4. Feb 23, 2012 #3

    PhanthomJay

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    Well you have the correct equations....so let's check the math;;;
    P/pi = (60-P)/2.25(pi)
    2.25P = 60 -P
    3.25P = 60
    P = 18.5 K

    Looks Good!
     
  5. Feb 23, 2012 #4
    But for my homework, I got this question wrong and in the book the answer is 28.2kips
     
  6. Feb 23, 2012 #5

    PhanthomJay

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    I don't know where that answer comes from...If P =28.2 K, then the force in the fat shaft would be woul be 31.8 K compression; and the tensile stress in the thin shaft would be 9 ksi, while the compressive stress in the fat one would be 4.5 ksi...far from equal in magnitude.... off by a factor of 2..
     
  7. Feb 24, 2012 #6

    nvn

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    zack7: Nice work. I currently agree with your answer in post 1, and with the answer by PhanthomJay. The answer in the book currently appears wrong.
     
    Last edited: Feb 24, 2012
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