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sECTIONB Special Topics
s6 | Beams—External Effects
Beams are structural members which offer resistance to bending due to applied loads. Most beams are long prismatic bars, and the loads are usually applied normal
to the axes of the bars.
Beams are undoubtedly the most important of all structural members, so it is important to understand the basic theory underlying their design. To analyze the load-carrying capacities of a beam, we must first establish the equilibrium re- quirements of the beam as a whole and any portion of it considered separately. Second, we must establish the relations between the resulting forces and the ac- companying internal resistance of the beam to support these forces. The first part of this analysis requires the application of the principles of statics. The second part involves the strength characteristics of the material and is usually treated in studies of the mechanics of solids or the mechanics of materials. This article is concerned with the external loading and reactions acting on a beam. In Art. 5/7 we calculate the distribution along the beam of the internal force
and moment.
Types of Beams
Beams supported so that their external support reactions can be calculated by the methods of statics alone are called statically determinate beams. A beam which has more supports than needed to provide equilibrium is statically indeterminate. To determine the support reactions for such a beam we must consider its load-deformation properties in addition to the equations of static equilibrium. Figure 5/18 shows examples of both types of beams. In this article we will analyze statically determi- nate beams only. Beams may also be identified by the type of external loading they support. The beams in Fig. 5/18 are supporting concentrated loads, whereas the beam in Fig. 5/19 is supporting a distributed load. The intensity w of a distributed load may be
Cantilever End-supported cantilever
Combination Fixed Statically determinate beams Statically indeterminate beams FIGURE 5/
126 CHAPTER 5 Distributed Forces
a [o] D A B FIGURE 5/
p—L/2—
Ry =!(wz—w¡)L
(e)
— M —
— — s
FIGURE 5/
-——x
expressed as force per unit length of beam. The intensity may be con- stant or variable, continuous or discontinuous. The intensity of the load-
ing in Fig. 5/19 is constant from C to D and variable from A to C and
from D to B. The intensity is discontinuous at D, where it changes mag-
nitude abruptly. Although the intensity itself is not discontinuous at C,
the rate of change of intensity dw/dx is discontinuous.
Distributed Loads
Loading intensities which are constant or which vary linearly are easily
handled. Figure 5/20 illustrates the three most common cases and the resultants of the distributed loads in each case. In cases a and b of Fig. 5/20, we see that the resultant load R is
represented by the area formed by the intensity w (force per unit length
of beam) and the length Z over which the force is distributed. The resul- tant passes through the centroid of this area. In part c of Fig. 5/20, the trapezoidal area is broken into a rectangu- lar and a triangular area, and the corresponding resultants R; and R, of these subareas are determined separately. Note that a single resultant could be determined by using the composite technique for finding cen- troids, which was discussed in Art. 5/4. Usually, however, the determina-
tion of a single resultant is unnecessary.
For a more general load distribution, Fig. 5/21, we must start with
a differential increment of force dR = w dx. The total load R is then the
sum of the differential forces, or
R=Íwdx
As before, the resultant R is located at the centroid of the area under
consideration. The x-coordinate of this centroid is found by the principle
of moments Rx = f xw dx, or
For the distribution of Fig. 5/21, the vertical coordinate of the centroid
need not be found.
Once the distributed loads have been reduced to their equivalent concentrated loads, the external reactions acting on the beam may be found by a straightforward static analysis as developed in Chapter 3.
3.6 - Introduction to Distributed Normal Loads
specified by the function p(x. y), called the load intensity. The units of load intensity are N/m?, 1b/f, and so on. The plane region o is known as the load area, and the surface formed by the plot of the load intensity is called the load surface. The region that lies between the load area 4 and the load surface is
labeled Y.
As shown in Fig. 3.9(a), we let dA represent a differential (infinitesimal) area element of 4. The force applied to dA is dR = p dA. The distributed surface load can thus be represented mathematically as an infinite number of forces dR that are parallel to the z-axis. Therefore, the resultant can be determined by employing the methods explained previously for parallel forces. However, because the force system here consists of an infinite number of differential forces, the summations must be replaced by an integrations over the load area 4. The resultant force is obtained from R = £F., which becomes
R=LdR=ÁpM (3.16)
where the range of integration is the load area. The coordinates X and y that locate the line of action of R, shown in Fig. 3.9(b), are determined by Eqs. (3.11): ¥ = — E M,/R and y = E My/R. After
replacing E M; by [, py dA and E M, by — [, px dA, these equations become
and >= — (3.17)
Let us now consider Eqs. (3.16) and (3.17) from a geometrical viewpoint. By inspection of Fig. 3.9 we observe that dR = p dA represents a differential volume of the region Y in Fig. 3.8. This volume has been denoted dV in Fig. 3.9(a). Therefore, the resultant force R in Eq. (3.16) can also be written as
R:Ádv:v (3.18)
where V is the total volume of the region Y.
— Load area sl
(a)
Fig. 3.
(b)
129
130 CHAPTER3 Resultants of Force Systems
¢/ Load diagram w(s)
(a)
w(y)
(b)
Fig.3.
Replacing p dA with dV in Eqs. (3.17), we get
[y x dV _ [ x dV
dV v G.19)
dV:fv_vdV
[ dV 12
As will be explained in Chapter 8. Eqs. (3.19) define the coordinates of a point known as the centroid of the volume that occupies the region Y. This point is labeled C in Fig. 3.9(b). The z-coordinate of the centroid is of no concern here because + and y are sufficient to define the line of action of the resultant force. The determination of the resultant force of a normal loading distributed over a plane area may thus be summarized as follows:
- The magnitude of the resultant force is equal to the volume of the region between the load area and the load surface.
- The line of action of the resultant force passes through the centroid ofthe volume bounded by the load area and the load surface.
b. Line loads
Whenever the width of the loading area is negligible compared with its length, a distributed load can be represented as a line load. Loadings distributed along a plane curve and along a straight line are shown in Figs. 3.10(a) and (b), respec- tively. Line loads are characterized by the load intensity w. a function of the distance measured along the line of distribution. The plot of w is called the load diagram. The units of w are N/m. Ib/ft, and so on. In this article, we consider only straight-line loads. Loads distributed along plane curves will be discussed in Chapter 8. As shown in Fig. 3.11(a), a straight-line load is equivalent to an infinite num- ber of differential forces, each of magnitude dR = w dx. Because these forces are parallel, their resultant is R = E F., or
R= fx:‘dR= /oLwdx (3.20)
directed parallel to the z-axis, as indicated in Fig. 3.11(b). The line of action of R can be determined by equating the moments about point O for the two systems in Figs. 3.11(a) and (b):
L L
3 EM(,:Í de=Í wr dx= R
x=0 o
132 CHAPTER 3 Resultants of Force Systems
load surface or the load diagram has a simple shape, then tables of centroids, such as Table 3.1, can be used to determine the resultant as illustrated in the following sample problems.
A. Volumes B. Areas
Rectangular solid Rectangle
Right-triangular solid Right triangle
3 V=14 bhL
Table 3.1 Centroids of Some Common Geometric Shapes (Additional tables are found in Chapter 8.)
