Ch 4. Moments/Equivalent Systems Multimedia Engineering Statics Moment2-D Scalar Moment3-D Scalar Moment3-D Vector Couples and Equiv. System Distributed Loads, Intro
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

 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

Practice Homework and Test problems now available in the 'Eng Statics' mobile app
Includes over 500 problems with complete detailed solutions.
Available now at the Google Play Store and Apple App Store.