HES1125 Unit 3 L1[1]

HES1125 Unit 3 L1[1] - UNIT3INTERNAL LOADINGS Lecture1...

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UNIT 3 INTERNAL  LOADINGS Lecture 1  Section 6.4  Resultant of a Distributed Force System
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2 Objective To determine  an equivalent  force for a  distributed load
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3 DISTRIBUTED LOADING In many situations a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body’s surface. We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body. In such cases, w is a function of x and has units of force per length.
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4 MAGNITUDE OF RESULTANT FORCE Consider an element of length dx. The force magnitude dF acting on it is given as dF = w(x) dx The net force on the beam is given by + F R = L dF = L w(x) dx = A Here A is the area under the loading curve w(x) .
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5 LOCATION OF THE RESULTANT FORCE The force dF will produce a moment of (x)(dF) about point O. The total moment about point O is given as + M RO = L x dF = L x w(x) dx Assuming that F R acts at , it will produce the moment about point O as + M RO = ( ) (F R ) = L w(x) dx x x x
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6 Comparing the last two equations, we get LOCATION OF THE RESULTANT FORCE F R acts through a point “C,” which is called the geometric center or centroid of the area under the loading curve w(x).
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7 This equation is the  distance x to the  centroid of an area Unit 4 Section  Properties of  Structural Members  will be covering  centroids and other  section properties in  more detail
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HES1125 Unit 3 L1[1] - UNIT3INTERNAL LOADINGS Lecture1...

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