How to force forces of the floors

[texasfight]

Homework Statement

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

Homework Equations

δ=(Force*Length)/(Area*Modulus of Elasticity)
Force = (δ*Area*Modulus of Elasticity)/(Length)

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?

 

[256bits]

The correct answer is Force of A = -90 kip and Force of B = -130 Kip

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 separated by 6 inches. Take separation and/or movement of the floors into account, And maybe the distance between the floors would be helpful in one of your calculations.