Chapter 10, Problem 1FP

Determine the moment of inertia of the shaded area about the x axis.

To determine

The moment of inertia for the shaded area about the $x$ axis.

Answer to Problem 1FP

The moment of inertia for the shaded area about the $x$ axis is $\underset{\_}{0.1112{\text{m}}^4}$.

Explanation of Solution

Given:

The height of the shaded area is $\text{1}\text{m}$.

The width of the shaded area is $\text{1}\text{m}$.

Show the area of the differential element parallel to the $y$ axis as in Figure 1.

From Figure 1,

Compute the area of the differential element parallel to the $y$ axis as shown with the shaded area from Figure 1.

$dA=ydx$

Here, the area of the differential element is $dA$ and the centroid of the differential element along the $y$ axis is $y$.

Express the moment of inertia of the differential element parallel to the $x$ axis.

$d{I}_x=d{\overline{I}}_{x^{\prime }}+dA{\tilde{y}}^2$ (I)

Here, the first integral of the moment of inertia of the area about the centroidal axis is ${\overline{I}}_{x^{\prime }}$ and the centroid of the differential element along the y axis is $\tilde{y}$.

Substitute $\frac{1}{12}\left(dx\right)y^3$ for $d{\overline{I}}_{x^{\prime }}$, $ydx$ for $dA$, and $y/2$ for $\tilde{y}$ in Equation (I).

$\begin{array}{c}d{I}_x=\frac{1}{12}\left(dx\right)y^3+ydx{\left(y/2\right)}^2\\ =\left[\frac{y^3}{12}+y\frac{y^2}{4}\right]\left(dx\right)\\ =\left[\frac{y^3}{12}+\frac{y^3}{4}\right]\left(dx\right)\end{array}$

$\begin{array}{c}=\frac{4y^3}{12}dx\\ =\frac{y^3}{3}dx\end{array}$ (II)

Substitute $x^2$ for $y^3$ in Equation (II).

$d{I}_x=\frac{x^2}{3}dx$

Conclusion:

Express the moment of inertia for the shaded area about the $x$ axis.

$I_x={\displaystyle {\int }_0^{1\text{m}}d{I}_x}$ (III)

Substitute $\frac{x^2}{3}dx$ for $d{I}_x$ in Equation (III).

$\begin{array}{c}{I}_x={\displaystyle {\int }_0^{1\text{m}}\frac{x^2}{3}dx}\\ =\frac{1}{3}{\displaystyle {\int }_0^{1\text{m}}{x}^2}dx\\ =\frac{1}{3}{\left[\frac{x^3}{3}\right]}_0^1\end{array}$

$\begin{array}{c}=\frac{1}{9}{\left(1\right)}^3-\frac{1}{9}{\left(0\right)}^3\\ =\frac{1}{9}{\text{m}}^{\text{4}}\\ =0.1112{\text{m}}^{\text{4}}\end{array}$

Hence, the moment of inertia for the shaded area about the $x$ axis is $\underset{\_}{0.1112{\text{m}}^4}$.