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Tension in rods of different materials and diameters

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A rigid weightless horizontal beam is hinged at a wall and is supported by two rods. The beam has a 8000 pound load applied at the end. Calculate the tension in each rod. The left rod is made of brass and has diameter of 1.25" and the right rod is made of aluminum and has a diameter of 1"

    2. Relevant equations
    σ = εE
    ε = ΔL/L
    F = σA = εEA
    A = (∏D^2)/4 = ∏R^2
    E(Brass)=15,000,000 psi

    3. The attempt at a solution
    A(Brass rod)= ∏/4 * (1.25)^2 = 25∏/64 in^2
    A(Aluminum rod)= ∏/4 * (1)^2= ∏/4 in^2

    I have been trying to figure out this problem for close to two hours and I am no further than where I began. I have done a few problems before where the load is applied to a beam, but never where the beam is hinged in the wall as such. I was told that I need to use similar triangles to find the lengths the rods change by but do not know how to implement this. I would appreciate any help, I am more concerned about how to do the problem than the answer itself.

    Attached Files:

  2. jcsd
  3. Nov 8, 2013 #2
    If the bar is totally rigid and the end where the 8000 lb load is applied moves downward δ, from geometry, how much does the base of the left rod move down, and how much does the base of the right rod move down?
  4. Nov 8, 2013 #3
    So the right rod would just move down 31δ/39 and the left 15δ/39? I still am unsure of where to go from here. The sum of the forces for each rod (F1 and F2) I presume would equal the 8000 lbs. When F1+F2=8000 is expanded you have the unknowns of the initial length L (from ε=ΔL/L) which is the same for both rods and then the value of δ to find the individual ΔL values. I feel like I'm missing an equation somewhere because the epsilons can't be set equal to each other nor can the forces, which was how previous examples I've done were solved.
  5. Nov 9, 2013 #4


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    The sum of the forces in the rods would only equal 8000 lbs if there were no reaction at the pin, which is unlikely.
    You still should write equations of static equilibrium for the hinged bar. Perhaps that is what you're missing.
  6. Nov 9, 2013 #5
    Express F1 and F2 in terms of δ, and then take moments about the hinge pin. This will give you the equation you need to solve for δ.
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