Chapter 4.10, Problem 187P

To determine

The expression of R for a circular cross section.

Answer to Problem 187P

The expression of R for a circular cross section is $\underset{\_}{\frac{1}{2}\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)}$.

Explanation of Solution

Consider w be the width as a function of $\beta$, $w=2c\sin \beta$

Sketch the cross section for the polar coordinate as shown in Figure 1.

Refer to Figure 1.

The distance $r=\overline{r}-c\cos \beta$

Differentiate both sides with respect to $\beta$ as shown below.

$\begin{array}{l}\frac{dr}{d\beta }=0-c\left(-\sin \beta \right)\\ dr=c\sin \beta d\beta \end{array}$

Calculate the area of the strip $\left(dA\right)$ as shown below.

$dA=wdr$

Substitute $2c\sin \beta$ for w and $c\sin \beta d\beta$ for dr.

$\begin{array}{c}dA=2c\sin \beta \times c\sin \beta d\beta \\ \frac{dA}{r}=\frac{2c^2{\sin }^2\beta d\beta }{r}\end{array}$

Substitute $\overline{r}-c\cos \beta$ for r.

$\frac{dA}{r}=\frac{2c^2{\sin }^2\beta d\beta }{\overline{r}-c\cos \beta }$

Integrate both sides of the Equation.

$\begin{array}{c}{\displaystyle \int \frac{dA}{r}}={\displaystyle {\int }_0^{\pi }\frac{2c^2{\sin }^2\beta d\beta }{\overline{r}-c\cos \beta }}\\ =2{\displaystyle {\int }_0^{\pi }\frac{c^2\left(1-{\cos }^2\beta \right)d\beta }{\overline{r}-c\cos \beta }}\\ =2{\displaystyle {\int }_0^{\pi }\frac{\left(c^2-c^2{\cos }^2\beta +{\overline{r}}^2-{\overline{r}}^2\right)d\beta }{\overline{r}-c\cos \beta }}\\ =2{\displaystyle {\int }_0^{\pi }\frac{\left({\overline{r}}^2-c^2{\cos }^2\beta -\left({\overline{r}}^2-c^2\right)\right)d\beta }{\overline{r}-c\cos \beta }}\end{array}$

$\begin{array}{c}=2{\displaystyle {\int }_0^{\pi }\frac{\left({\overline{r}}^2-c^2{\cos }^2\beta \right)d\beta }{\overline{r}-c\cos \beta }}-2{\displaystyle {\int }_0^{\pi }\frac{\left({\overline{r}}^2-c^2\right)d\beta }{\overline{r}-c\cos \beta }}\\ =2{\displaystyle {\int }_0^{\pi }\frac{\left(\overline{r}-c\cos \beta \right)\left(\overline{r}+c\cos \beta \right)d\beta }{\overline{r}-c\cos \beta }}-2\left({\overline{r}}^2-c^2\right){\displaystyle {\int }_0^{\pi }\frac{d\beta }{\overline{r}-c\cos \beta }}\\ =2{\displaystyle {\int }_0^{\pi }\left(\overline{r}+c\cos \beta \right)d\beta }-2\left({\overline{r}}^2-c^2\right){\displaystyle {\int }_0^{\pi }\frac{d\beta }{\overline{r}-c\cos \beta }}\\ =2{\left(\overline{r}\beta +c\sin \beta \right)}_0^{\pi }-2\left({\overline{r}}^2-c^2\right){\left(\frac{2}{\sqrt{{\overline{r}}^2-c^2}}{\tan }^{-1}\frac{\sqrt{{\overline{r}}^2-c^2}\tan \frac{1}{2}\beta }{\overline{r}+c}\right)}_0^{\pi }\end{array}$

$\begin{array}{l}=\left\{2\left(\overline{r}\pi +c\sin \pi -0\right)-4\sqrt{{\overline{r}}^2-c^2}\left\{\begin{array}{l}\left({\tan }^{-1}\frac{\sqrt{{\overline{r}}^2-c^2}\tan \frac{\pi }{2}}{\overline{r}+c}\right)\\ -\left({\tan }^{-1}\frac{\sqrt{{\overline{r}}^2-c^2}\tan 0}{\overline{r}+c}\right)\end{array}\right\}\right\}\\ =\left\{2\left(\overline{r}\pi \right)-4\sqrt{{\overline{r}}^2-c^2}\left(\frac{\pi }{2}-0\right)\right\}\\ =2\pi \overline{r}-2\pi \sqrt{{\overline{r}}^2-c^2}\end{array}$

Calculate the area of the circle $\left(A\right)$ as shown below.

$A=\pi c^2$

Calculate the expression of $R$ as shown below.

$R=\frac{A}{{\displaystyle \int \frac{dA}{r}}}$

Substitute $\pi c^2$ for $A$ and $2\pi \overline{r}-2\pi \sqrt{{\overline{r}}^2-c^2}$ for ${\displaystyle \int \frac{dA}{r}}$.

$\begin{array}{c}R=\frac{\pi c^2}{2\pi \overline{r}-2\pi \sqrt{{\overline{r}}^2-c^2}}\\ =\frac{1}{2\pi }\times \frac{\pi c^2}{\overline{r}-\sqrt{{\overline{r}}^2-c^2}}\\ =\frac{1}{2}\times \frac{c^2}{\overline{r}-\sqrt{{\overline{r}}^2-c^2}}\times \frac{\overline{r}+\sqrt{{\overline{r}}^2-c^2}}{\overline{r}+\sqrt{{\overline{r}}^2-c^2}}\\ =\frac{1}{2}\times \frac{c^2\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)}{\left({\overline{r}}^2-{\left(\sqrt{{\overline{r}}^2-c^2}\right)}^2\right)}\end{array}$

$\begin{array}{c}=\frac{1}{2}\times \frac{c^2\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)}{\left({\overline{r}}^2-{\overline{r}}^2+c^2\right)}\\ =\frac{1}{2}\times \frac{c^2\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)}{c^2}\\ =\frac{1}{2}\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)\end{array}$

Therefore, the expression of R for a circular cross section is $\underset{\_}{\frac{1}{2}\left(\overline{r}+\sqrt{{\overline{r}}^2-c^2}\right)}$.