Stresses in Beams
To calculate stresses in beams, one must ?rst model the beam correctly
in terms of its supports and loading (such as simply supported, with
distributed loading), determine the appropriate unknown external
reactions, and establish the corresponding shear and moment diagrams
using a consistent sign convention. Both normal and shear stresses may
have to be calculated, but typically the normal stresses are the most
signi?cant.
Flexure Formula
The normal stresses at a particular cross section in a beam are caused
by the bending moment that acts at that cross section, and are distributed
by magnitude and sign (both tension and compression) so that the beam is
in equilibrium. The basic concept for calculating the stresses is that there
is a neutral axis n-n of ? = ? = 0 in the beam, and that the longitudinal normal
strain varies linearly with distance y from the neutral axis.
If the beam is behaving entirely elastically, the stress distribution is
also linear, as in Figure 1.5.16. In this case, the stress at a distance y from
the neutral axis is calculated from M = ??(y)y dA and results in
FIGURE 1.5.16 Internal normal stresses in a beam caused
by bending.
where I = moment of inertia of the cross-sectional area about the
neutral axis. The maximum stress, with the appropriate sign, is
There are several special cases of bending that require additional
considerations and analysis as outlined below.
Inelastic Bending
A beam may plastically deform under an increasing moment, yielding
?rst in its outer layers and ultimately throughout its depth. Such a beam is
analyzed by assuming that the normal strains are still linearly varying from
zero at the neutral axis to maximum values at the outer layers. Thus, the
stress distributions depend on the stress-strain curve of the material.
With the stress distribution established, the neutral axis can be
determined from ??(y) dA = 0, and the resultant moment from
M = ?y?(y) dA. A fully plastic beam of rectangular cross section and
?at-top yielding supports 50% more bending moment than its maximum
elastic moment.
Neutral Axis of Semisymmetric Area
If the cross-sectional area is semisymmetric, such as a T-shape, and the
loading is in a centroidal plane of symmetry, the neutral axis for elastic
deformations is at the centroid C of the area as shown in Figure
1.5.17, and Equation 1.5.25 can be used. Note that the magnitudes of the
maximum tensile and com- pressive stresses are not the same in this case.
FIGURE 1.5.17 Neutral axis of a semisymmetric area.
Unsymmetric Bending
In the general case, the cross-sectional area has an arbitrary shape and
the loading is arbitrarily applied. The problem of an arbitrary area is
handled by choosing the centroidal xy coordinate system such that
the axes are principal axes of inertia for the area. The principal axes can
be determined by using inertia transformation equations or Mohrs circle of
inertia. Having an axis of symmetry is a simple special case because the
principal axes are the axis of symmetry and the axis perpendicular to it.
The ?exure formula can be applied directly if the principal axes of inertia
are known, and the bending moment is applied about one of these centroidal
principal axes. A more complex case is if the moment is not about a
principal axis as shown in Figure 1.5.18.
FIGURE 1.5.18 Schematic of arbitrary bending moment.
Different texts may present different formulas for calculating the
bending stresses in such situations, depending on the choice of a
coordinate system and the sign convention adopted. It is better not
to rely on a cookbook formula, but to break down the problem into
simple, easily visualized parts, and then reason out an algebraic
superposition of the stress components. To illustrate this approach
schematically, consider the stresses at points A and B in Figure 1.5.18.
Instead of working directly with the applied moment M, resolve M
into its components Mx and My. Mx causes a tensile stress ?zA1 at A
and a compressive stress ??zB1 at B. My causes tensile stresses at
both A and B, ?zA2 and ?zB2 . The magnitudes of these stress
components are readily calculated from the ?exure formula with the
appropriate dimensions and inertias for each. The resultant
stresses are
The neutral axis at angle ? in the general case is not coincident with
the direction of M (Figure 1.5.18). In the present case ? is de?ned by
Composite Beams
Nonhomogeneous beams are often designed to take advantage of the
properties of two different materials. The approach for analyzing these
is to imagine a transformation of the beams cross section to an
equivalent cross section of a different shape but of a single material,
so that the ?exure formula is usable. This is illustrated for a beam A
with reinforcing plates B, as in Figure 1.5.19.
FIGURE 1.5.19 Equivalent area method for a symmetric
composite beam.
The transformation factor n is obtained from
Note that in a composite beam the strains vary linearly with distance
from the neutral axis, but the stresses do not, because of the different
elastic moduli of the components. The actual stress ? in the transformed
area is determined by ?rst calculating the pretend stress ?? for the
uniform transformed area and then multiplying it by n,
? = n??
Nonsymmetric composite beams (such as having only one reinforcing
plate B in Figure 1.5.19) are analyzed similarly, but ?rst require the
location of the neutral axis. Reinforced concrete beams are important
special cases of composite beams. The stress analysis of these is
in?uenced by the fact that concrete is much weaker in tension than in
compression. Empirical approaches are particularly useful in this area.
Curved Beams
The stress analysis of curved beams requires some additional
considerations. For example, the ?exure formula is about 7% in error
(the calculated stresses are too low) when the beams radius of curvature
is ?ve times its depth (hooks, chain links). The curved-beam formula
provides realistic values in such cases.
Shear Stresses in Beams
Transverse loads on beams are common, and they cause transverse
and complementary longitudinal shear stresses in the beams.
Schematically, the transverse shear stresses are distributed on a
rectangular cross section as shown in Figure 1.5.20. The shear stress
is zero at free surfaces by de?nition.
FIGURE 1.5.20 Transverse shear stress distribution.
The internal shear stress is calculated according to Figure 1.5.20 from
where ? = shear stress value at any point on the line 
at a distance y? from the neutral axis
V = total shear force on cross-sectional area A
Q = ?A?;A = area above line ; ? = distance from neutral
axis to centroid of A?
I = moment of inertia of entire area A about neutral axis
t = width of cross section where ? is to be determined
This shear formula gives ?max = 1.5 V/A if t is constant for the
whole section (rectangle). Note that the magnitude of the shear stress
distribution changes sharply where there is an abrupt
change in width t, such as in an I-beam, Figure 1.5.21.
FIGURE 1.5.21 Shear stress distribution for I-beam.
Shear Flow
In the analysis of built-up members, such as welded, bolted, nailed,
or glued box beams and channels, a useful quantity is the shear ?ow q
measured in force per unit length along the beam,
where all quantities are de?ned as for Equation 1.5.30. Care must
be taken to use the appropriate value for Q. For example, consider
a channel section of three ?at pieces glued together as in Figure 1.5.22.
There are two critical joint regions B here, and the area A? is between them.
The shear ?ow is carried by the two joints together, so the actual force
per unit length on one joint is q/2 here.
FIGURE 1.5.22 Critical joint regions of a built-up beam.
Shear Flow in Thin-Walled Beams
The shear-?ow distribution over the cross section of a thin-walled member
is governed by equilibrium requirements. Schematic examples of this are
given in Figure 1.5.23. Note the special case of unsym-metrical loading
in part (c), which causes a bending and a twisting of the beam. The twisting
is prevented if the vertical force V is applied at the shear center C, de?ned
by the quantity e,
where d is the centroidal distance between the two horizontal ?anges
and H is the shear force in the ?anges (qave times width of ?ange).
FIGURE 1.5.23 Shear ?ow distributions.
