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STATICS - CASE STUDY SOLUTION

    Integration Method


Integration Method


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

     
    Composite Parts Method


Wind Distributed Load Split
into Two Parts


Composite Parts Method

  For the composite parts method, first break the wind load into two parts: a uniform load of 40 lb/ft and a triangular load of 0 lb/ft at the base and 80 lb/ft at the top of the building.

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 FR located at the position y'.

     FR = 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

     
   
 
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