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Unformatted text preview: Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 111 Chapter 11 Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids To be truly ignorant, be content with your own knowledge. Chuang Tzu (c. 360 BC  c. 275 BC) 11.1 Overview Thus far we have covered a wide range of topics, but they all have one common theme in the sense that you are learning to apply physics and mathematics to more and more realistic problems in mechanics. We began this course with some preliminary vector concepts and moved quickly into the particle equilibrium. After solving some interesting, but relatively simple problems, we extended our capabilities by studying moments and distributed loads. That allowed us to consider the equilibrium of rigid bodies and solve a much wider range of problems. We applied the concepts of particle and rigid body equilibrium to analyze structures including trusses, frames, and machines. Then friction, where we were able to incorporate slipping/tipping, wedges, and belts. Now we will consider equivalent systems, distributed loads, centers of mass and centroids. A proper understanding of these concepts will make it possible for the student to apply their knowledge of statics and structures to complicated shapes that are subjected to complicated loadings. After completing this part of the course, the student should be able to, 1. Replace one set of forces and moments with an equivalent single force and single mo 112 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids ment. 2. Use integral formulations to transform distributed loads into a single equivalent force. 3. Use composite parts to transform simple distributed loads into a single equivalent force. 4. Calculate centers of mass and centroids using integral formulations. 5. Calculate the centers of mass and centroids using the method of composite parts. 11.2 Important Points 1. Equivalent systems form the basis for replacing multiple forces and moments or dis tributed loads with a simplified set of reactions. 2. For a distributed load described by the function w ( x ), the force, F , is, F = Z L w ( x ) dx, and the location of that force, X , F = R L xw ( x ) dx F = R L xw ( x ) dx R L w ( x ) dx . Please note that you must integrate the numerator and denominator separately. 3. The center of mass is a physical property, the centroid is a geometric property. 4. The center of mass is based on the concept of equivalent systems. What we would like to do is replace the complicated distribution of material with a single force so that the single force generates the same total force and same total moment as the original material distribution, x = R xdm R dm y = R ydm R dm . Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 113 5. The centroid effectively defines the geometric center of an object, x = R xdA R dA y = R ydA R dA ....
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This note was uploaded on 12/21/2011 for the course ME 270 taught by Professor Murphy during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 MURPHY

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