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The rigid bar AC is supported by two axial bars

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data
    The rigid bar AC is supported by two axial bars (1) and (2). Both axial bars are made of bronze [E = 100 GPa; α = 18 × 10−6 mm/mm/°C]. The cross-sectional area of bar (1) is A1= 211 mm2 and the cross-sectional area of bar (2) is A2 = 303 mm2. After load P has been applied and the temperature of the entire assembly has increased by 21°C, the total strain in bar (2) is measured as 1210 με (elongation). Determine:
    (a) the magnitude of load P.
    (b) the vertical displacement of pin A.


    2. Relevant equations

    εT = αΔT
    εσ = εtotal - εT
    σ = Eε
    σ = F/A

    3. The attempt at a solution

    I started by using the first equation and getting the strain caused by temp change in bar (2) → (18x10-6)(21) = 0.000378
    Then, I plugged that result into the second equation to get strain caused by normal stress in bar (2) → 0.00121 - 0.000378 = 0.00032
    I plugged this result into the third equation to get the stress in bar (2) → (0.00032)(100) = 0.032 GPa = 32 MPa
    Then I plugged this into the fourth equation to solve for the force F2 → (32 N/mm2)(303 mm2) = 6752 N

    To find P, I used the moment about point A → -480P - (9696)(1210) = 0 ⇒ P = 24442 N = 24.4 kN

    My answer was incorrect, and I really have no idea what to do.

    As for part b), I'm not really sure where to begin.
    Any help would be appreciated, I really struggle at FBD's and correctly labelling moments and forces so if I could see one for this problem it would be a big help :smile:
  2. jcsd
  3. Feb 24, 2015 #2
    When you took moments about point A, what was your sign convention?
  4. Feb 24, 2015 #3
    I took ccw as negative and cw as positive. Moments are not my forte so I'm pretty sure I'm doing something wrong here.
  5. Feb 24, 2015 #4
    OK, can you explain then how you determined the signs in your last equation? A free body diagram might help as well.

    I think you made a simple arithmetic mistake in step 4 but seem to have corrected for it in the last equation.

    Check your arithmetic in the 2nd step.
    Last edited: Feb 24, 2015
  6. Feb 25, 2015 #5
    Yeah I re did my calculations and it turns out I wrote down the wrong number o0).
    I solved it, thanks for your help!
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