Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

283
C
M = Pe
F
+ =
bENdINg duE TO AN
ECCENTRIC AxIAL LOAd
Internal
axial force
and bending
moment
Uniform
stress due
to axial
force
Stress caused
by bending
moment
FIGURE 8.14 Normal stresses caused by an eccentric axial load.
Complete
stress
distribution
Neutral-Axis Location
Whenever an internal axial force F acts simultaneously with an internal bending moment M,
the neutral axis is no longer located at the centroid of the cross section. In fact, depending
upon the magnitude of the force F, there may be no neutral axis at all. All normal stresses
on the cross section may be either tension stresses or compression stresses. The location of
the neutral axis can be determined by setting σ x = 0 in Equation (8.19) and solving for the
distance y measured from the centroid of the cross section.
Limitations
The stresses determined by this approach assume that the internal bending moment in the
flexural member can be accurately calculated from the original undeformed dimensions. In
other words, the deflections caused by the internal bending moment must be relatively
small. If the flexural member is relatively long and slender, the lateral deflections caused by
the eccentric axial load may significantly increase the eccentricity e, thereby amplifying
the bending moment.
The use of Equations (8.19) and (8.20) should be consistent with Saint-Venant’s
principle. In practice, this condition means that stresses cannot be accurately calculated
near points H and K in Figure 8.13a.
ExAmpLE 8.8
A structural member with a rectangular cross section 10 in. wide by 6 in. deep supports
a 30 kip concentrated load as shown. Determine the distribution of normal
stresses on section a–a of the member.
y
13 in.
30 kips
Plan the Solution
The internal forces acting on section a–a must be determined at the outset. The
principle of equivalent force systems will be used to determine a force and a moment
acting at the section of interest that together are equivalent to the single 30 kip
concentrated load acting on the top of the structural member. Once the equivalent
force and moment have been determined, the stresses produced at section a–a can
be computed.
a
10 in.
a
SolutioN
Equivalent Force and Moment
The cross section of the structural member is rectangular; therefore, by symmetry,
the centroid must be located 5 in. from the left side of the structural
member. The 30 kip concentrated load is located 13 in. from the left side of
z
x