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How to force forces of the floors

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data
    The steel column is used to support the symmetric loads from two floors of an officer building. Determine the loads P1 and P2 if A moves downward 0.12 in and B moves downward 0.09 in under the applied loading. The steel column has a cross-sectional area of 23.4 in2.
    Length of A from the ground = 22ft
    Length of B from the ground = 12 ft


    2. Relevant equations
    δ=(Force*Length)/(Area*Modulus of Elasticity)
    Force = (δ*Area*Modulus of Elasticity)/(Length)


    3. The attempt at a solution
    Force of A = (-0.12in*23.4in2*30000Kpsi)/(264in) = -319Kip
    Force of B = (-0.09in*23.4in2*30000KPsi)/(144in) = -438.75 Kip
    The correct answer is Force of A = -90 kip and Force of B = -130 Kip
    What did I do wrong?
     
  2. jcsd
  3. Jun 6, 2012 #2
    Well that is not written correctly.
    It should be P1 = -90 kip and P2 = -130 Kip ( or vice versa )

    You should home in on two things.
    Force of B = (-0.09in*23.4in2*30000KPsi)/(144in) = -438.75 Kip
    which surprisingly is twice of P1+P2 so you have a factor of 2 that you are not taking into account. You should read every word of the problem. ( symmetrical )

    and the displacement.
    If I move 10 inches to the right and you move 4 inches to the right we are now seperated by 6 inches. Take seperation and/or movement of the floors into account, And maybe the distance between the floors would be helpful in one of your calculations.
     
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