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Finding elongation of bar and maximum tensile stress

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    wk1xkz.jpg
    L=52 in
    A=2.76 in^2
    E=10.4*10^6 psi

    2. Relevant equations

    σ=F/A
    ε=σ/E
    δ=εL


    3. The attempt at a solution

    4) σAB = (3P)/A
    ε=(3P)/(AE)
    δAB=(3PL)/(6AE) → δAB=(PL)/(2AE)
    solving for P
    P=[0.17*2*2.76*(10.4*106)]/52 → P=187680 lb → P=187.7 kip

    5) Because AB and CD are in tension i did this...
    σmaxABCD
    σmax=[(-2P)/A]+[P/A]
    solving for P and using 5000psi for σmax i get
    P=-5000*2.76 → P=13800 lb → P=13.8kip

    I tried looking for an example in the book to follow, but they were completely different. I hope i didn't mess up too bad.
     
    Last edited: Jan 30, 2013
  2. jcsd
  3. Jan 30, 2013 #2
    Don't add the stresses
     
  4. Jan 30, 2013 #3
    So i should only do one of them? σmaxAB
    or do the entire bar?
    Can I get a better hint than that?
     
  5. Jan 31, 2013 #4
    I confirmed that I did part 4 right. I still need help with part 5. Anyone?
     
  6. Jan 31, 2013 #5
    I made a mistake on part 5. I plugged in the value of σBC (-2P/A) in place for σAB (3P/A) in σmaxABCD

    Now i get σmax=3450 lb or 3.45 kip. But still don't know if it's right.
     
  7. Jan 31, 2013 #6
    Part 5:

    You correctly determined the stresses in the 3 sections of the bar.

    AB = 3P/A
    BC = -2P/A
    CD = P/A

    So, you already know that the maximum tensile stress in the bar is 3P/A. This cannot exceed 5000 psi = 5 ksi

    chet
     
  8. Jan 31, 2013 #7
    So what I'm getting from this is that the maximum tension occurs at AB so I only set σmax=σAB and don't add them with the other member in tension.

    Thanks :)
     
  9. Jan 31, 2013 #8
    Yes. That's right. The other sections will be less prone to failure.
     
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