 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 - THEORY General Distributed Load with Load Intensity of f(x) (units force/distance) In many static problems, applied loads are given as distributed force loads. This is similar to stacking sand bags on a beam so that the load is distributed across the beam instead of at one location (point load). To help make the problem easier to solve, it is convenient to convert the distributed load into equivalent point loads. Discrete Distributed Force Forces Parallel to the Y axis Forces Parallel to the X-Y Plane If the loading on the object is a set of parallel discrete forces, the resultant force is simply the sum of all the forces, or      FR = ΣFi In the previous section, three scalar equations were derived that determine the position r of a force FR that represents the force resultant to a system of discrete forces. If all forces are parallel to the y direction then the three scalar equations simplify to      Σ(zF) = z' ΣF      Σ(xF) = x' ΣF Here x and z are the Cartesian coordinates of the individual forces, and x' and z' are the coordinates of the force resultant FR. Rearranging these two equations gives If the forces are further restricted so that they all lie in the x-y plane, then z' = 0 and only the second equation applies. Distributed Forces Forces Distributed over a Line Continuous Distributed Force If instead of a system of point loads, consider a continuous distributed force f(x) that acts in the x-y plane and is parallel to the y axis, then through calculus the second equation (x') above becomes The force resultant is simply the force magnitude FR given by The force magnitude FR is located a distance x' from the origin. Uniform and Triangular Line Loads Resultant Force and Location for Two Distributed Loading Types In the case of a uniform line load as shown, it is unnecessary to perform the integrations because the force resultant is always the value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is always the center point (centroid) of the distributed load. For a triangular line load, it can be shown that the force resultant is one half of the peak value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is two-thirds of the distance from the vertex to the peak value of the load. Composite Line Load Distributed Load Simplification Composite Loading A load of the type shown is said to be a composite load since it can be generated with a combination of simpler loads such as a uniform and a triangular line load. Loads of this nature can be converted to force resultants by splitting the load into its composite parts, solving for the force resultant of each part, and then combining the forces into a force resultant for the entire load. As an example, the diagram at the left can be splitting to a triangular and rectangular distributed load. The two separate loadings can be combined as,      FR = F1 + F2      xR FR = x1 F1 + x2 F2

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