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ME 201 Engineering Mechanics: Statics Chapter 4 – Part F 4.9 Reduction of a Simple Distributed Loading Distributed Loads Thus far we’ve been working with loads that are concentrated at a point: Many times in engineering we need to be concerned with another type of loading referred to as distributed loading: 6 kN/m 10 N/m Distributed Loads Instead of being concentrated at a point, a distributed load is spread out over a distance It can be thought of as a collection of smaller loads Distributed Loads To compute the resultant, FR, of a distributed load, consider the following: To find FR, we need to sum an infinite number of small forces Distributed Loads Consider a small differential element, dF with a width of dx and a height of w(x) w dF w(x) x dx x Distributed Loads w dF w(x) x dx x The area of the element is dF w( x)dx Since infinite number of forces, need to integrate to find FR FR w( x)dx dA A l A Distributed Loads Where does FR act? Can be determined by equating the moments of the force resultant and the force distribution FR w( x)dx l xFR xw( x)dx l xw( x)dx xw( x)dx xdA x l l FR w( x)dx l This is also the centroid of the area A dA A Centroids For simple shapes, centroid can be found in a table (see back cover of textbook) Where is the centroid for these common shapes? h h b b b xc 2 h yc 2 b xc 3 h yc 3 Class Exercise Given: w=100x N/m Find: FR, x 600 N/m A B 6m Example Problem Solution Given: w=100x N/m Find: FR, x Solution: FBD FR x 600 N/m A B 6m FR A x B 6m 1 FR 600 N / m 6 m 2 1800 Nm or 1.8 kNm 2 x 6m 3 4m Example Problem Given: trapezoid Find: FR, x w=60x2 N/m 2m Example Problem Solution Given: trapezoid Find: FR, x Solution: FR integral w=60x2 N/m 2m FR w( x)dx L 2 60 x dx 2 0 x3 2 60 | 3 0 160 N Example Problem Solution Given: trapezoid Find: FR, x Solution: FR integral x integral w=60x2 N/m 2m xw( x) dx x w( x)dx x60 x dx 60 x dx x 60 | 4 L L 2 2 0 2 2 0 4 2 0 3 x 60 3 2 | 0 240 160 1.5 m