Chapter 9.1, Problem 9.20P

To determine

Find the moment of inertia $\left(I_y\right)$ and radius of gyration $\left(k_y\right)$ of the shaded area with respect to y axis.

Answer to Problem 9.20P

The moment of inertia $\left(I_y\right)$ and radius of gyration $\left(k_y\right)$ of the shaded area with respect to y axis is $\underset{\_}{0.61345a^{3}h}$ and $\underset{\_}{1.299a}$ respectively.

Explanation of Solution

Given information:

The equation of the upper segment is $y=mx+b$.

The equation of the lower curve is $y=c\sin kx$.

Calculation:

Show the subject area with vertical differential strip element as Figure 1.

Consider the curve equation as $y=c\sin kx$.

Apply the boundary condition at $y_1$.

Condition 1:

At $x=2a$ and $y=0$.

$0=c\sin k\left(2a\right)$

Since $\sin \theta$ will be zero when angle at $180°\left(\pi \right)$.

$\begin{array}{c}2ak=\pi \\ k=\frac{\pi }{2a}\end{array}$

Condition 2:

At $x=a$ and $y=h$.

$h=c\sin k\left(a\right)$

Substitute $\frac{\pi }{2a}$ for k.

$\begin{array}{c}h=c\sin \frac{\pi }{2a}\left(a\right)\\ h=c\sin \frac{\pi }{2}\\ h=c\left(1\right)\left(\because \sin \frac{\pi }{2}=1\right)\\ c=h\end{array}$

Consider the upper curve equation $y=mx+b$.

Apply the boundary condition at $y_2$.

Condition 1:

At $x=a$ and $y=2h$.

$2h=ma+b$ (1)

Condition 2:

At $x=2a$ and $y=0$.

$\begin{array}{c}0=m2a+b\\ 2ma+b=0\end{array}$ (2)

Subtract the equation (1) and (2).

$\begin{array}{c}2h-0=ma+b-2ma-b\\ 2h=-ma\\ m=-\frac{2h}{a}\end{array}$

Substitute $-\frac{2h}{a}$ for m in the equation (2).

$\begin{array}{c}2\left(-\frac{2h}{a}\right)a+b=0\\ -4h+b=0\\ b=4h\end{array}$

Rewrite the lower equation.

$y=c\sin kx$

Substitute $y_1$ for y, h for c and $\frac{\pi }{2a}$ for k.

$y_1=h\sin \frac{\pi }{2a}x$

Rewrite the upper equation.

$y=mx+b$

Substitute $y_2$ for y, $-\frac{2h}{a}$ for m and 4h for b.

$\begin{array}{c}{y}_2=-\frac{2h}{a}x+4h\\ =\frac{2h}{a}\left(-x+2a\right)\end{array}$

Calculate the area of the given figure (A) using the formula:

$\begin{array}{c}A={\displaystyle \int dA}\\ ={\displaystyle \int \left(y_2-y_1\right)dx}\end{array}$

Substitute $h\sin \frac{\pi }{2a}x$ for $y_1$ and $\frac{2h}{a}\left(-x+2a\right)$ for $y_2$.

$\begin{array}{c}A={\displaystyle \int \left[\frac{2h}{a}\left(-x+2a\right)-\left(h\sin \frac{\pi }{2a}x\right)\right]dx}\\ ={\displaystyle \underset{a}{\overset{2a}{\int }}h\left[\frac{2}{a}\left(-x+2a\right)-\left(\sin \frac{\pi }{2a}x\right)\right]dx}\\ =h{\left[\left(-\frac{2}{a}\frac{{\left(-x+2a\right)}^2}{2\left(1\right)}\right)-\left(-\frac{\cos \frac{\pi }{2a}x}{\frac{\pi }{2a}}\right)\right]}_a^{2a}\end{array}$

$\begin{array}{c}=h{\left[\left(-\frac{1}{a}{\left(-x+2a\right)}^2\right)+\frac{2a}{\pi }\cos \frac{\pi }{2a}x\right]}_a^{2a}\\ =h\left[\left(-\frac{1}{a}{\left(-2a+2a\right)}^2+\frac{1}{a}{\left(-a+2a\right)}^2\right)+\left(\frac{2a}{\pi }\cos \frac{\pi }{2a}2a-\left(\frac{2a}{\pi }\cos \frac{\pi }{2a}a\right)\right)\right]\\ =h\left[\left(-\frac{1}{a}{\left(-a+2a\right)}^2\right)+\frac{2a}{\pi }\left(-1\right)-0\right]\end{array}$

$\begin{array}{c}=h\left[\left(-\frac{1}{a}{\left(-a+2a\right)}^2\right)-\frac{2a}{\pi }\right]\\ =ah\left[\left(-\frac{1}{1}{\left(-1+2\right)}^2-\frac{2}{\pi }\right)\right]\\ =ah\left[1-\frac{2}{\pi }\right]\\ =0.36338ah\end{array}$

Calculate the moment of inertia about y axis using the formula:

$\begin{array}{c}{I}_y={\displaystyle \int d{I}_y}\\ ={\displaystyle \int x^{2}dA}\\ ={\displaystyle \underset{a}{\overset{2a}{\int }}{x}^2\left(y_2-y_1\right)dx}\end{array}$

Substitute $h\sin \frac{\pi }{2a}x$ for $y_1$ and $\frac{2h}{a}\left(-x+2a\right)$ for $y_2$.

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

Take $u=x^2$ from the equation (3).

Differentiate the u equation with respect to x.

$du=2xdx$

Take $dv=\sin \frac{\pi }{2a}xdx$ from the equation (3).

Integrate the above equation with respect to x.

$\begin{array}{c}{\displaystyle \int dv}={\displaystyle \int \sin \frac{\pi }{2a}xdx}\\ v=-\frac{\cos \frac{\pi }{2a}x}{\frac{\pi }{2a}}\\ =-\frac{2a}{\pi }\cos \frac{\pi }{2a}x\end{array}$

Consider the second term from the equation (3).

${\displaystyle \int x^2\sin \frac{\pi }{2a}x}dx$

Integrate the above equation by UV method with respect to x.

Substitute $-\frac{2a}{\pi }\cos \frac{\pi }{2a}x$ for $\sin \frac{\pi }{2a}x$.

Since $uv=udv-vdu$.

$\begin{array}{c}{\displaystyle \int x^2\sin \frac{\pi }{2a}x}dx=x^2\left(-\frac{2a}{\pi }\cos \frac{\pi }{2a}x\right)-{\displaystyle \int \left(-\frac{2a}{\pi }\cos \frac{\pi }{2a}x\right)\left(2xdx\right)}\\ =-\frac{2a{x}^2}{\pi }\cos \frac{\pi }{2a}x+\frac{4a}{\pi }\left[x\left(\frac{2a}{\pi }\sin \frac{\pi }{2a}x\right)+{\displaystyle \int \frac{2a}{\pi }\sin \frac{\pi }{2a}x}\left(1\right)\right]\\ =-\frac{2a{x}^2}{\pi }\cos \frac{\pi }{2a}x+\frac{8a^2}{{\pi }^2}x\sin \frac{\pi }{2a}x+\frac{4a}{\pi }\left(\frac{2a}{\pi }\times \frac{2a}{\pi }\cos \frac{\pi }{2a}x\right)\\ =-\frac{2a{x}^2}{\pi }\cos \frac{\pi }{2a}x+\frac{8a^2}{{\pi }^2}x\sin \frac{\pi }{2a}x+\frac{16a^3}{{\pi }^3}\cos \frac{\pi }{2a}x\end{array}$

Substitute $-\frac{2a{x}^2}{\pi }\cos \frac{\pi }{2a}x+\frac{8a^2}{{\pi }^2}x\sin \frac{\pi }{2a}x+\frac{16a^3}{{\pi }^3}\cos \frac{\pi }{2a}x$ for ${\displaystyle \int x^2\sin \frac{\pi }{2a}x}dx$ in the equation (1).

$\begin{array}{c}{I}_y=h{\left[\frac{2}{a}\left(\frac{-x^4}{4}+\frac{2a{x}^3}{3}\right)-\left(-\frac{2a{x}^2}{\pi }\cos \frac{\pi }{2a}x+\frac{8a^2}{{\pi }^2}x\sin \frac{\pi }{2a}x+\frac{16a^3}{{\pi }^3}\cos \frac{\pi }{2a}x\right)\right]}_a^{2a}\\ =\left\{\begin{array}{c}h\left[\frac{2}{a}\left(\frac{-{\left(2a\right)}^4}{4}+\frac{2a{\left(2a\right)}^3}{3}\right)-\left(-\frac{2a{\left(2a\right)}^2}{\pi }\cos \frac{\pi }{2a}2a+\frac{8a^2}{{\pi }^2}2a\sin \frac{\pi }{2a}2a+\frac{16a^3}{{\pi }^3}\cos \frac{\pi }{2a}2a\right)\right]-\\ h\left[\frac{2}{a}\left(\frac{-a^4}{4}+\frac{2a{a}^3}{3}\right)-\left(-\frac{2a{a}^2}{\pi }\cos \frac{\pi }{2a}a+\frac{8a^2}{{\pi }^2}a\sin \frac{\pi }{2a}a+\frac{16a^3}{{\pi }^3}\cos \frac{\pi }{2a}a\right)\right]\end{array}\right\}\\ =h\left\{\left[\frac{2}{a}\left(-\frac{16a^4}{4}+\frac{16a^4}{3}\right)-\left(-\frac{8a^3}{\pi }\left(-1\right)+\frac{16a^3}{{\pi }^3}\right)\right]\right\}-h\left[\left(\frac{2}{a}\left(-\frac{a^4}{4}+\frac{2a^4}{3}\right)\right)-\frac{8a^3}{{\pi }^2}\right]\\ =0.61345a^{3}h\end{array}$

Calculate the radius of gyration $\left(k_y\right)$ along y axis using the formula:

$k_y=\sqrt{\frac{I_y}{A}}$

Substitute $0.36338ah$ for A and $0.61345a^{3}h$ for $I_x$.

$\begin{array}{c}{k}_y=\sqrt{\frac{0.61345a^{3}h}{0.36338ah}}\\ =1.299a\end{array}$

Therefore, the moment of inertia $\left(I_y\right)$ and radius of gyration $\left(k_y\right)$ of the shaded area with respect to y axis are $\underset{\_}{0.61345a^{3}h}$ and $\underset{\_}{1.299a}$ respectively.