A beam has a rectangular cross section and is subjected to the stress

distribution shown in Fig, $6-25$a$.$ Determine the internal moment M at

the section caused by the stress distribution $($a$)$ using the flexure

formula, $($b$)$ by finding the resultant of the stress distribution using

basic principles.

SOLUTION

Part $($a$).$ The flexure formula is ${}_{\text{m}\text{a}\text{x}}=M\frac{c}{l}.$ From Fig. $6-25a,$

$c=6in.$ and ${}_{\text{m}\text{a}\text{x}}=2$ksi. The neutral axis is defined as line $NA,$

because the stress is zero along this line. Since the cross section has a

rectangular shape, the moment of inertia for the area about NA is

determined from the formula for a rectangle given on the inside front

cover;i$.$e$.,$

$I=\frac{1}{12}b{h}^3=\frac{1}{12}(6in.)(12in.{)}^3=864i{n}^4$

Therefore,

${}_{\text{m}\text{a}\text{s}}=\frac{Mc}{l},2ki\frac{p}{i}{n}^2=\frac{M(6in.)}{864i{n}^4}$

$M=288$kip$*in.=24$kip$*f$

Part $($b$).$ The resultant force for each of the two triangular stress

distributions in Fig. $6-25b$ is graphically equivalent to the volume

contained within each stress distribution. Thus, each volume is

$F=\frac{1}{2}(6in)(2ki\frac{p}{i}{n}^2)(6in.)=36$kip

These forces, which form a couple, act in the same direction as the

stresses within each distribution. Fig, $6-25$b$.$ Furthermore, they act

through the centroid of each volume, i$.$c$.,J(6in)=\mathrm{.}4in.$ from the

neutral axis of the beam. Hence the distance between them is $8$ in$.$ as

shown. The moment of the couple is therefore

$M=36$kip$(8in)=288$kip$*in.=24$kip$*ft,$Ant

NOTE: This result can also be obtained by choosing a horizontal strip

of area $.)dy$ and using integration by applying Eq$.6-11.$