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Therefore, the magnitude of the resultant force is equal to the
total area A under the loading diagram, Fig. 1-5,c.
Location of Resultant Force. Applying Eq. M R 6M the
location of the line of action of F can be determined by equating the
R
moments of the force resultant and the parallel force distribution about
point O (the y axis). Since dF produces a moment of xdF xw ()x dx
about O, Fig. 1-5,b, then for the entire length, Fig. 1-5,c,

(M RO M ; xF R ³ xw ()x dx . 1-8
) 6
O
L
Solving for x using Eq. 1-7, we have

³ xw ()x dx ³ xdA
x L A . 1-9
³ w ()x dx ³ dA
L A
This coordinate x , locates the geometric center or centroid of
the area under the distributed loading. In other words, the resultant
force has a line of action which passes through the centroid C
(geometric center) of the area under the loading diagram, Fig. 1-5,c.
Detailed treatment of the integration techniques for finding the
location of the centroid for areas is given in Chapter Center of Mass.
In many cases, however, the distributed-loading diagram is in the
shape of a rectangle, triangle, or some other simple geometric form.
The centroid location for such common shapes does not have to be
determined from the above equation but can be obtained directly from
the tabulation given on the inside back cover.
Once x is determined, F by symmetry passes through point
R
( x ,0) on the surface of the beam, Fig. 1-5,a. Therefore, in this case
the resultant force has a magnitude equal to the volume under the
loading curve p px
()and a line of action which passes through the
centroid (geometric center) of this volume.
For example, Fig. 1-6, the beam supporting this stack of lumber
is subjected to a uniform loading of w . The resultant force is
O
therefore equal to the area under the loading diagram F wb. It acts
O
R
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