Lecture 5 part II

Lecture 5 part II - LECTURE 5 DISTROBUTED FORCES: CENTROIDS...

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24 LECTURE 5 DISTROBUTED FORCES: CENTROIDS AND CENTER OF GRAVITY PART II CONTENTS DISTRIBUTED LOADS ON BEAMS FORCES ON SUBMERGED SURFACES ‡ : Reading using the same effort as doing your homework.
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25 DISTRIBUTED LOADS ON BEAMS When a load is applied to a rigid body, it is often distributed along a line or over an area A . In many instances, such as solving the unknown reaction forces on a body via the equilibrium equations , it is convenient to replace the distributed load with a resultant force that is equivalent to the distributed load. The quantities that need to be solved are the magnitude of the resultant force and the location of its line of action. Fig. 5-8 Replacing a distributed load on a beam with a resultant force R Consider the beam , defined as the component to resist lateral load and to be placed horizontally in most cases as shown in Fig. 5-8, subject to distributed load intensity across the length, the magnitude of the differential force dR exerted on the beam by the distributed load in an increment of length dx is dR wdx = (24) The magnitude of the resultant force R is a ) b ) c ) w=f ( x )
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26 LL R dR wdx == (25)
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Lecture 5 part II - LECTURE 5 DISTROBUTED FORCES: CENTROIDS...

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