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Homework Help: Is it well analyzed? Doubt about a bar: tension/compresion

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data

    Calcular the force P when a bar starts yield, when 2 bars start yield, and when the 3 bars start yield
    Diameter = 25 mm (for every bar), Yield Stress of Steel = 250 MPa.

    2. Relevant equations

    3. The attempt at a solution

    [F][/AD] + [F][/BE] + [F][/CF] = P

    2[Δ[/AD] - 3[Δ][/BE] + [Δ][/AC]= 0

    I calculate first what happend when the 3 bars were in elastic zone, with the Hooke law I obtain:

    [F][/AD]= 0.14P
    [F][/BE]= 0.29 P
    [F][/CF] = 0.57P

    So the firs bar in reach yield stress is CF.

    0.57P = [σ][/y] * A => P=215, 295.54 N

    When the 2 bars (BE and CF) are with the yield stress, [F][/BE]=[F][/CF]

    From equations of equilibrium, sumatory of moments in C:

    0.4P = 1.2 [F][/AD] + 0.8 [F][/BE]

    and I know from the eq (1) [F][/AB] + [F][/BE] + [F][/CF] = P, so 2[F][/BE] + [F][/AD]= P. Solving this 2 eq. I obtain P=243, 605.30 N and [F][/AD]= -1831.62 N ....So I dont know if this answer is correct, because it says that the bar AD now is in compresion, when before it was in tension
    Last edited by a moderator: Apr 23, 2015
  2. jcsd
  3. Apr 23, 2015 #2
    Can you explain how you got this equation?
    2[Δ[/AD] - 3[Δ][/BE] + [Δ][/AC]= 0
  4. Apr 23, 2015 #3
    From semejant triangles, ( delta AD - delta BE) / 0.4 = (delta AD- delta CF) /1.2. Is the C.G.D compatibility geometry of deformation
  5. Apr 23, 2015 #4
    What is Δ? Maybe a picture would help.
    Last edited by a moderator: Apr 23, 2015
  6. Apr 23, 2015 #5
  7. Apr 23, 2015 #6
    OK, I see now. The ΔAC was a typo then.

    Can you show how you used Hooke's Law to get the forces?
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