Question 4.CA.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 \text { and } r_2 , respectively, the inner and outer radius of the bar (Fig. 4.62b), use Eq. (4.66) to write

R=\frac{A}{\int \frac{d A}{r}}              (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,

\begin{aligned} & r_1=\bar{r}-\frac{1}{2} h=6-0.75=5.25 \text {  in. } \\ & r_2=\bar{r}+\frac{1}{2} h=6+0.75=6.75 \text {  in. } \end{aligned}

Substituting for h, r_1 \text {, and } r_2 into Eq. (4.73),

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. 4.62c) is thus

e=\bar{r}-R=6-5.9686=0.0314 \text {  in. }

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