1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    upload_2015-2-23_19-13-16.gif

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: The rigid bar AC is supported by two axial bars
Loading...