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: 10 N/m 6 kN/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, F R, of a distributed load, consider the following:  To find F R, 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(x) w dF x dx x

Distributed Loads  The area of the element is  Since infinite number of forces, need to integrate to find F R w(x) w dF x dx x

Distributed Loads  Where does F R act?  Can be determined by equating the moments of the force resultant and the force distribution This is also the centroid of the area

Centroids  For simple shapes, centroid can be found in a table (see back cover of textbook)  Where is the centroid for these common shapes? b h b h

Class Exercise Given: w=100x N/m Find: F R, 600 N/m 6 m A B

Example Problem Solution Given: w=100x N/m Find: FR,FR, Solution: FBD FRFR 600 N/m 6 m A B A B FRFR

Example Problem Given: trapezoid Find: F R, w=60x 2 N/m 2 m

Example Problem Solution Given: trapezoid Find: FR,FR, Solution: F R integral w=60x 2 N/m 2 m

Example Problem Solution Given: trapezoid Find: FR,FR, Solution: F R integral integral w=60x 2 N/m 2 m