Question 4.10: A curved rectangular bar has a mean radius r = 6 in. and a c...

We first derive the expression for the radius R of the neutral surface. Denoting by r_{1} and r_{2}, respectively, the inner and outer radius of the bar (Fig. 4.75), we use Eq. (4.66) and write

R=\frac{A}{\int \frac{d A}{r}}      Eq. (4.66)

R=\frac{A}{\int_{r_{1}}^{r_{2}} \frac{d A}{r}}=\frac{b h}{\int_{r_{1}}^{r_{2}} \frac{b d r}{r}}=\frac{h}{\int_{r_{1}}^{r_{2}} \frac{d r}{r}}

R=\frac{h}{\ln \frac{r_{2}}{r_{1}}}    (4.73)

For the given data, we have

r_{1}=\bar{r}-\frac{1}{2} h=6-0.75=5.25  in . \\ r_{2}=\bar{r}+\frac{1}{2} h=6+0.75=6.75  in .

Substituting for h, r_{1}, and r_{2} into Eq. (4.73), we have

R=\frac{h}{\ln \frac{r_{2}}{r_{1}}}=\frac{1.5  in .}{\ln \frac{6.75}{5.25}}=5.9686  in.

The distance between the centroid and the neutral axis of the cross section (Fig. 4.76) is thus

e=\bar{r}-R=6-5.9686=0.0314  in .

We note that it was necessary to calculate R with five significant figures in order to obtain e with the usual degree of accuracy.