EXAMPLE 6.13
The beam shown in Fig. 6-27a has a cross-sectional area in the shape
of a channel, Fig. 6-27b. Determine the maximum bending stress that
occurs in the beam at section a-a.
SOLUTION
Internal Moment. Here the beam's support reactions do not have
to be determined. Instead, by the method of sections, the segment to
the left of section a-a can be used, Fig. 6-27c. In particular, note that
the resultant internal axial force N passes through the centroid of the
cross section. Also, realize that the resultant internal moment must be
calculated about the beam's neutral axis at section a-a.
To find the location of the neutral axis, the cross-sectional area
is subdivided into three composite parts as shown in Fig. 6-27b.
Using Eq. A-2 of Appendix A, we have
$$y = \frac{\sum yA}{\sum A} = \frac{2[0.100 \ m](0.200 \ m)(0.015 \ m) + [0.010 \ m](0.02 \ m)(0.250 \ m)}{2(0.200 \ m)(0.015 \ m) + 0.020 \ m(0.250 \ m)}$$
= 0.05909 m = 59.09 mm
This dimension is shown in Fig. 6-27c.
Applying the moment equation of equilibrium about the neutral
axis, we have
$$(+ \sum M_{NA} = 0; 2.4 \ kN(2 \ m) + 1.0 \ kN(0.05909 \ m) - M = 0$$
$$M = 4.859 \ kN \cdot m$$
Section Property. The moment of inertia about the neutral axis
is determined using $$I = \sum (I + Ad^2)$$ applied to each of the three
composite parts of the cross-sectional area. Working in meters, we have
$$I = \frac{1}{12}(0.250 \ m)(0.020 \ m)^3 + (0.250 \ m)(0.020 \ m)(0.05909 \ m - 0.010 \ m)^2$$
$$+ 2[\frac{1}{12}(0.015 \ m)(0.200 \ m)^3 + (0.015 \ m)(0.200 \ m)(0.100 \ m - 0.05909 \ m)^2]$$
$$= 42.26(10^{-6}) \ m^4$$
Maximum Bending Stress. The maximum bending stress occurs at
points farthest away from the neutral axis. This is at the bottom of the
beam, c = 0.200 m - 0.05909 m = 0.1409 m. Thus,
$$\sigma_{max} = \frac{Mc}{I} = \frac{4.859(10^3) \ N \cdot m(0.1409 \ m)}{42.26(10^{-6}) \ m^4} = 16.2 \ MPa$$
Ans.
Show that at the top of the beam the bending stress is σ' = 6.79 MPa.
NOTE: The normal force of N = 1 kN and shear force V = 2.4 kN will
also contribute additional stress on the cross section. The superposition
of all these effects will be discussed in Chapter 8.
2.6 kN
13/12
2 m
-1 m
(a)
-250 mm
20 mm
200 mm
15 mm
15 mm
(b)
2.4 kN
V
1.0 kN
0.05909 m
6
M
N
4
C
2 m
(c)
Fig. 6-27
