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FBD for tri-pod jack stand?

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A jack-stand with three equally spaced legs (120 degrees) are connected to a center tube is loaded axially with 36,000 pounds. Each leg has a horizontal cross member 11 inches from the ground. All connections are welded. Set up FBD and find reaction forces.

    2. Relevant equations
    ∑Fy = 0
    ∑Fx = 0
    ∑M_x = 0

    3. The attempt at a solution
    I guess I'm confused about the connections being rigid and welded. Most structures I have seen in examples are assumed to be pinned and therefore do not transmit moment/shear loads. Solving globally for

    ∑Fy = 0
    36,000 = 3*Na = 0
    Na = 12,000 (Vertical reaction at each foot contacting ground)
    The horizontal reaction at each foot is the opposition of friction:
    Fa = μsNa
    Fa = (0.5*12000) = 6000 (I use a μs = 0.5 for concrete/steel)

    Summing moments about A:
    ∑Ma = (15*R_dy) + (15*R_cy) - (28*R_cx) = 0
    ∑Fy = Na - R_dy - R_cy = 0
    ∑Fx = R_cx - Fa = 0

    R_cx = 6000

    Plugging everything in I get R_dy = 0?? So DB is a zero force member then if CB is in compression? I would think the 36,000 would transform some load to the member DB. The other thing I'm confused about is there a reaction at point B between the angled leg and the cross-member? I don't know if each of these points would have a bending moment associated with it either since they are rigid and not pinned.
     

    Attached Files:

    Last edited by a moderator: May 16, 2016
  2. jcsd
  3. May 18, 2016 #2

    CWatters

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    See previous thread.
     
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