A new tapered foundation is designed to support a distributed area load of 6 kip/in². The top 10 inches has a constant width of 2 inches, but the bottom 20 inches tapers from 2 inches to 4 inches, as shown in the diagram. The foundation length is large when compared to the height or width. What is the vertical deflection of the foundation top? The foundation is constructed from concrete, E = 4×10⁶ psi.

Compression Load on
Tapered Foundation

The foundation can be analyzed using a 1 inch unit thickness which will have a load of

     P = (6 kip/in²) (1 in) (2 in) = 12 kip

This load will push down on the foundation from the top to the bottom.

Next, it will be easier to analyze the deflection if the foundation is split into two sections, 1 and 2. The deflection of the top section can be determined by

     δ₁ = PL₁/A₁E₁
         = (-12,000) (10) / [(2)(4×10⁶)] = -0.0150 in

However, since the cross sectional area of the bottom section varies, the deflection needs to be integrated over its length using

The load P and stiffness E are constant, but the area, A changes over its length. The rate of change is linear, and can be modeled as

     A(x) = 1 (4 - x/10) = (40 - x)/10

The variable of integration, x, is zero at the base where the area 4 in² and goes to 20 in at the top of the bottom section where the area is 2 in². This gives,

   

   = 0.030 (ln 20 - ln 40) = -0.02079 in

The total deflection is the sum of both sections,

δ_total = δ₁ + δ₂
        = -0.01500 - 0.02079 = -0.03577 in

Loads in Steel Rods

From statics, the forces acting on members AB and BC can be determined as

     ΣF_y = 0
     F_AB sin36.87 - 10,000 = 0
     F_AB = 10,000 / sin36.87
     F_AB = 16,670 lb (Tension)

     ΣF_x = 0
     F_AB cos36.87 + F_BC = 0
     F_BC = -16,670 cos36.87
     F_BC = -13,330 lb (Compression)