EXAMPLE $6\mathrm{.}13$

The beam shown in Fig. $6-27$a has a cross$-$sectional area in the shape

of a channel, Fig. $6-27$b$.$ 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. $6-27$c$.$ 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. $6-27$b$.$

Using Eq$.$ A$-2$ of Appendix A$,$ we have

This dimension is shown in Fig. $6-27$c$.$

Applying the moment equation of equilibrium about the neutral

axis, we have

$+M_{NA}=0$;$,2\mathrm{.}4kN(2m)+1\mathrm{.}0kN(0\mathrm{.}05909m)-M=0$

$M=4\mathrm{.}859kN*m$

Section Property. The moment of inertia about the neutral axis

is determined using $I=((\frac{?}{\text{b}\text{a}\text{r}}(I))+A{d}^2)$ applied to each of the three

composite parts of the cross$-$sectional area. Working in meters, we have

$I=[\frac{1}{12}(0\mathrm{.}250(m))(0\mathrm{.}020(m){)}^3+(0\mathrm{.}250(m))(0\mathrm{.}020(m))(0\mathrm{.}05909(m)-0\mathrm{.}010(m){)}^2]$

$+2[\frac{1}{12}(0\mathrm{.}015(m))(0\mathrm{.}200(m){)}^3+(0\mathrm{.}015(m))(0\mathrm{.}200(m))(0\mathrm{.}100(m)-0\mathrm{.}05909(m){)}^2]$

$=42\mathrm{.}26({10}^{-6})m^4$

Maximum Bending Stress. The maximum bending stress occurs at

points farthest away from the neutral axis. This is at the bottom of the

beam, $c=0\mathrm{.}200m-0\mathrm{.}05909m=0\mathrm{.}1409m.$ Thus,

${}_{\text{m}\text{a}\text{x}}=\frac{Mc}{\text{I}}=\frac{4\mathrm{.}859({10}^3)N*m(0\mathrm{.}1409(m))}{42\mathrm{.}26({10}^{-6})m^4}=16\mathrm{.}2$MPa, Ans.

Show that at the top of the beam the bending stress is ${}^{\text{'}}=6\mathrm{.}79$MPa.

NOTE: The normal force of $N=1kN$ and shear force $V=2\mathrm{.}4kN$ will

also contribute additional stress on the cross section. The superposition

of all these effects will be discussed in Chapter $8.$