Load Diagram

In order to integrate the distributed wind load, determine a function f(y) that describes the wind load.

     f(y) = 40 + 80/500 y   lb/ft

The force resultant is given by

     
          = 40(500) + 0.08(500)² = 40,000 lb

Its location is given by

     
        = 20(500)²/40,000 + 0.05333(500)³/40,000

        = 125.0 + 166.7

        = 291.7 ft

Wind Distributed Load Split
into Two Parts

Composite Parts Method

The force resultant of the uniform load and its location is

     F₁ = 40 (500) = 20,000 lb

     y₁ = 250 ft

The force resultant of the triangular load is

     F₂ = 0.5 (80) (500) = 20,000 lb

Its location is two-thirds of the distance from the vertex to the peak value

     y₂ = 2/3 (500) = 333.3 ft

Nest, combine these two forces into a single force F_R located at the position y'.

     F_R = ΣF = F₁ + F₂ = 40,000 lb

Its location is given by

Substituting F₁, y₁, F₂, and y₂ and simplifying the equation gives

     y' = 1.166 x 10⁷ / 40,000

        = 291.7 ft