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Homework Help: Need help with a Statically Indeterminate System!

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Three steel bars are pin-connected to a rigid member K. Determine the force developed in each bar. Determine the load carried be each of the tension members and the elongation of each member

    http://img3.imageshack.us/img3/6850/problemdiagram.jpg [Broken]

    [tex]A_A_B[/tex] = 0.10 in^2
    [tex]E_A_B[/tex] = 30E6 psi
    [tex]A_C_D[/tex] = 0.20 in^2
    [tex]E_C_D[/tex] = 15E6 psi
    [tex]A_F_H[/tex] = 0.30 in^2
    [tex]E_F_H[/tex] = 10E6 psi

    2. Relevant equations

    [tex]\delta[/tex] = (PL) / (AE)

    3. The attempt at a solution

    Finding the moment about B:
    10[tex]P_C_D[/tex] + 20[tex]P_F_H[/tex] = 15(15000)
    [tex]P_F_H[/tex] = 7500 - (1/3)[tex]P_C_D[/tex]

    The equation [tex]\delta[/tex] = (PL) / (AE) yields:
    [tex]P_A_B[/tex] = 150,000[tex]\delta_A_B[/tex]
    [tex]P_C_D[/tex] = 200,000[tex]\delta_C_D[/tex]
    [tex]P_F_H[/tex] = 300,000[tex]\delta_F_H[/tex]

    And the sum of forces in the Y direction gives:
    [tex]P_A_B[/tex] + [tex]P_C_D[/tex] + [tex]P_F_H[/tex] = 15000

    This is where I'm stuck. If any point along K was fixed it would be easy; K is rigid, so then the distance from the fixed point can be turned into a ratio to find the other [tex]\delta[/tex] values. I think all 3 points (B, D, and H) are pulled downward, but I'm not sure what there relation is to each other. Any clues?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 8, 2009 #2
    Yikes, I think my TEX tags are all messed up. Not sure how to fix it, let me know if you have any questions!
  4. Jul 8, 2009 #3
    Now if it was something like this:
    http://img194.imageshack.us/img194/8167/fixedpoint.jpg [Broken]
    I would say sigma_CD = 10(theta) and sigma_FH = 30(theta) and all would be good.

    But since it's like this:
    http://img199.imageshack.us/img199/9023/nonfixedpoint.jpg [Broken]
    I have another variable in there with x. Now sigma_AB = (x)(theta), sigma_CD = (10+x)(theta), and sigma_FH = (30+x)(theta)

    Last edited by a moderator: May 4, 2017
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