Chapter 9, Problem 9.185RP

Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes.

Fig. P9.185

To determine

Find the moment of inertia of the shaded area with respect to x and y axes.

Answer to Problem 9.185RP

The moment of inertia of the shaded area with respect to x axes is $\underset{\_}{\frac{16}{105}a{h}^3}$.

The moment of inertia of the shaded area with respect to y axes is $\underset{\_}{\frac{1}{5}h{a}^3}$

Explanation of Solution

Given information:

The curve Equation is $y=4h\left(\frac{x}{a}-\frac{x^2}{a^2}\right)$

Calculation:

Sketch the shaded portion with vertical strip as shown in Figure 1.

Refer to Figure 1.

Write the curve Equation as shown below:

$y=4h\left(\frac{x}{a}-\frac{x^2}{a^2}\right)$ (1)

Determine the moment of inertia $\left(d{I}_x\right)$ with respect to x axis of a rectangular strip:

$d{I}_x=\frac{1}{3}{y}^{3}dx$ (2)

Substitute $4h\left(\frac{x}{a}-\frac{x^2}{a^2}\right)$ for y in Equation (2).

$\begin{array}{c}d{I}_x=\frac{1}{3}{y}^{3}dx\\ =\frac{1}{3}{\left[4h\left(\frac{x}{a}-\frac{x^2}{a^2}\right)\right]}^{3}dx\end{array}$ (3)

Integrate Equation (3) with respect to x.

$\begin{array}{c}{I}_x={\displaystyle \int d{I}_x}\\ =\frac{64h^3}{3}{\displaystyle \underset{0}{\overset{a}{\int }}\left(\frac{x^3}{a^3}-3\left(\frac{x^4}{a^4}\right)+3\left(\frac{x^5}{a^5}\right)-\frac{x^6}{a^6}\right)}\\ =\frac{64h^3}{3}{\left[\left(\frac{x^4}{4a^3}-3\left(\frac{x^5}{5a^4}\right)+3\left(\frac{x^6}{6a^5}\right)-\frac{x^7}{7a^6}\right)\right]}_0^a\\ =\frac{64h^3}{3}{\left[\left(\frac{1}{4}\frac{x^4}{a^3}-\frac{3}{5}\left(\frac{x^5}{a^4}\right)+\frac{1}{2}\left(\frac{x^6}{a^5}\right)-\frac{1}{7}\frac{x^7}{a^6}\right)\right]}_0^a\end{array}$

$\begin{array}{c}=\frac{64h^3}{3}\left[\left(\frac{1}{4}\frac{a^4}{a^3}-\frac{3}{5}\left(\frac{a^5}{a^4}\right)+\frac{1}{2}\left(\frac{a^6}{a^5}\right)-\frac{1}{7}\frac{a^7}{a^6}\right)\right]\\ =\frac{64h^3}{3}a\left(\frac{1}{4}-\frac{3}{5}+\frac{1}{2}-\frac{1}{7}\right)\\ =\frac{64h^3}{3}a\left(\frac{1}{140}\right)\\ I_x=\frac{16}{105}a{h}^3\end{array}$

Thus, the moment of inertia of the shaded area with respect to x axes is $\underset{\_}{\frac{16}{105}a{h}^3}$.

Determine the area of the strip element $dA$ as shown in below:

$dA=ydx$

Determine the moment of inertia $\left(d{I}_y\right)$ with respect to y axis of a rectangular strip:

$\begin{array}{c}d{I}_y=x^{2}dA\\ =4h{x}^2\left(\frac{x}{a}-\frac{x^2}{a^2}\right)dx\end{array}$ (4)

Integrate Equation (4) with respect to y.

$\begin{array}{c}{I}_y=4h{\displaystyle \underset{0}{\overset{a}{\int }}\left(\frac{x^3}{a}-\frac{x^4}{a^2}\right)dx}\\ =4h{\left[\frac{x^4}{4a}-\frac{x^5}{5a^2}\right]}_0^a\\ =4h\left[\frac{a^4}{4a}-\frac{a^5}{5a^2}\right]\\ =4h\left[\frac{a^3}{4}-\frac{a^3}{5}\right]\end{array}$

$\begin{array}{c}=4h{a}^3\left[\frac{1}{4}-\frac{1}{5}\right]\\ =4h{a}^3\left[\frac{1}{20}\right]\\ I_y=\frac{1}{5}h{a}^3\end{array}$

Thus, the moment of inertia of the shaded area with respect to y axes is $\underset{\_}{\frac{1}{5}h{a}^3}$.