Determining Equivalent Force and Location for Distributed Loads on Beams, Exams of Acting

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REDUCTION OF A SIMPLE DISTRIBUTED LOADING

Today’s Objectives: Students will be able to determine an equivalent force for a distributed load.

APPLICATIONS

A distributed load on the beam exists due to the weight of the lumber. Is it possible to reduce this force system to a single force that will have the same external effect? If yes, how?

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.

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).

Comparing the last two equations, we get LOCATION OF THE RESULTANT FORCE (continued) You will learn later that F R acts through a point “C,” which is called the geometric center or centroid of the area under the loading curve w(x).

EXAMPLES

Until you learn more about centroids, we will consider only rectangular and triangular loading diagrams whose centroids are well defined and shown on the inside back cover of your textbook. In a rectangular loading, F R = 400  10 = 4,000 lb and 𝑥 = 5 ft. In a triangular loading , F R = (0.5) (6000) (6) = 1,800 N and 𝑥 = 6 – (1/3) 6 = 4 m. Please note that the centroid in a right triangle is at a distance one third the width of the triangle as measured from its base.

GROUP PROBLEM SOLVING

Given : The loading on the beam as shown. Find : The equivalent force and its location from point A. Plan:

1. Consider the trapezoidal loading as two separate loads (one rectangular and one triangular).
2. Find F R and 𝑥 for each of the two distributed loads.
3. Determine the overall F R and for the three point loadings.

GROUP PROBLEM SOLVING (continued) For the combined loading of the three forces, F R = 1.5 kN + 3 kN + 1.5 kN = 6 kN

• M RA = (1.5) (1.5) + 3 (1) + (1.5) 4 = 11.25 kN • m Now, FR 𝑥 = 11.25 kN • m Hence, 𝑥 = (11.25) / (6) = 1.88 m from A. For the rectangular loading of height 0.5 kN/m and width 3 m, F R = 0.5 kN/m  3 m = 1.5 kN

x 1 = 1.5 m from A

For the triangular loading of height 2 kN/m and width 3 m, F R = (0.5) (2 kN/m) (3 m) = 3 kN and its line of action is at 𝑥 = 1 m from A