Chapter 9, Problem 9.192RP

To determine

Find the moment and product of inertia of the area with respect to x and y axes about through $30°$ by using Mohr circle method.

Answer to Problem 9.192RP

The moment of inertia of the area with respect to x about through $30°$ using Mohr circle method is $\underset{\_}{5.27{\text{in}}^{\text{4}}}$

The moment of inertia of the area with respect to y about through $30°$ using Mohr circle method is $\underset{\_}{6.71{\text{in}}^{\text{4}}}$

The product of inertia of the area with respect to x and y axes about through $30°$ by using Mohr circle method is $\underset{\_}{-4.39{\text{in}}^{\text{4}}}$.

Explanation of Solution

Sketch the cross section as shown in Figure 1.

Refer to Figure 9.13.

The moment of inertia ${\overline{I}}_x$ about x axis is $9.43{\text{in}}^{\text{4}}$.

The moment of inertia ${\overline{I}}_y$ about y axis is $2.55{\text{in}}^{\text{4}}$.

Refer to Problem 9.191.

Sketch the cross section as shown in Figure 2.

Express the product of inertia as shown below:

${\overline{I}}_{xy}={\left(I_{xy}\right)}_1+{\left({\overline{I}}_{xy}\right)}_2$

Here, $\left[{\left(I_{xy}\right)}_1,{\left(I_{xy}\right)}_2\right]$ is product of inertia section 1 and 2.

Applying parallel axis theorem,

$I_{xy}=I_{x^{\prime }{y}^{\prime }}+\overline{x}\overline{y}A$

When the x and y axis is symmetry.

$I_{x^{\prime }{y}^{\prime }}=0$

Refer to Figure 1.

Find the area of semicircle section 1 as shown below:

$A_1=\left(b_1{h}_1\right)$ (1)

Here, $b_1$ is width of the section 1 and $h_1$ is height of the section 1.

Substitute $3\text{in}\text{.}$ for $b_1$ and $0.5\text{in}\text{.}$ for $h_1$ in Equation (1).

$\begin{array}{c}{A}_1=3\times 0.5\\ =1.5{\text{in}}^{\text{2}}\end{array}$

Find the area of rectangle section 2 as shown below:

$A_2=\left(b_2{h}_2\right)$ (2)

Here, $b_2$ is width of the section 2 and $h_2$ is height of the section 2.

Substitute $4.5\text{in}\text{.}$ for $b_2$ and $0.5\text{in}\text{.}$ for $h_2$ in Equation (2).

$\begin{array}{c}{A}_2=4.5\times 0.5\\ =2.25{\text{in}}^{\text{2}}\end{array}$

Find the centroid for section 1 about x axis $\left(x_1\right)$ as shown below:

$\begin{array}{c}{x}_1=\frac{1}{2}\left(3\right)-0.746\\ =1.5-0.746\\ =0.754\text{in}\text{.}\end{array}$

Find the centroid for section 1 about x axis $\left(x_2\right)$ as shown below:

$\begin{array}{c}{x}_2=-\left(0.746-\left(0.5\times 0.5\right)\right)\\ =0.746-0.25\\ =-0.496\end{array}$

Find the centroid for section 1 about y axis $\left(y_1\right)$ as shown below:

$\begin{array}{c}{y}_1=-\left(1.74-\frac{1}{2}\left(\frac{1}{2}\right)\right)\\ =-1.49\text{in}\text{.}\end{array}$

Find the centroid for section 2 about x axis $\left(y_2\right)$ as shown below:

$\begin{array}{c}{y}_2=\frac{1}{2}\left(4.5\right)-\left(1.74-\frac{1}{2}\right)\\ =2.25-1.24\\ =1.01\text{in}\text{.}\end{array}$

Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem as shown below:

$\begin{array}{c}{\overline{I}}_{xy}={\left(\overline{x}\overline{y}A\right)}_1+{\left(\overline{x}\overline{y}A\right)}_2\\ =\left(x_1{y}_{1}A\right)+\left({\overline{x}}_2{\overline{y}}_{2}A\right)\end{array}$ (3)

Substitute $0.754\text{in}\text{.}$ for $x_1$, $-0.496$ for $x_2$, $-1.49\text{in}\text{.}$ for $\left(y_1\right)$, $1.01\text{in}\text{.}$ for $y_2$, $1.5{\text{in}}^{\text{2}}$ for $A_1$, and $2.25{\text{in}}^{\text{2}}$ for $A_3$ in Equation (3).

$\begin{array}{c}{\overline{I}}_{xy}=\left(0.754\times \left(-1.49\right)\times 1.5\right)+\left(-0.496\times \left(1.01\right)\times 2.25\right)\\ =-1.68519-1.12716\\ =-2.81{\text{in}}^{\text{4}}\end{array}$

The Mohr circle is defined by the diameter XY, where $X\left(9.43-2.81235\right)$ and $Y\left(2.55,2.81235\right)$.

Find the average moment of inertia $\left(I_{ave}\right)$ using the relation as shown below:

$I_{ave}=\frac{1}{2}\left({\overline{I}}_x+{\overline{I}}_y\right)$ (4)

Here, ${\overline{I}}_x$ is moment of inertia about x axis and ${\overline{I}}_y$ is moment of inertia about y axis.

Substitute $9.43{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$ and $2.55{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$ in Equation (4).

$\begin{array}{c}{I}_{\text{ave}}=\frac{1}{2}\left(9.43+2.55\right)\\ =\frac{1}{2}\left(11.98\right)\\ =5.99{\text{in}}^{\text{4}}\end{array}$

Find the radius (R) using the relation as shown below:

$R=\sqrt{{\left(\frac{{\overline{I}}_x-{\overline{I}}_y}{2}\right)}^2+{\overline{I}}_{xy}^2}$ (5)

Here, R is radius and ${\overline{I}}_{xy}$ is product of inertia.

Substitute $9.43{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$ and $2.55{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$,$-2.81{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$, in Equation (5).

$\begin{array}{c}R=\sqrt{{\left(\frac{9.43-2.55}{2}\right)}^2+{\left(-2.81235\right)}^2}\\ =\sqrt{11.8336+7.9093}\\ =\sqrt{19.7429}\\ =4.4433{\text{in}}^{\text{4}}\end{array}$

Sketch the Mohr circle as shown in Figure 3.

Refer to Figure 2.

$\tan 2{\theta }_m=-\frac{2I_{xy}}{I_x-I_y}$ (6)

Substitute $9.43{\text{in}}^{\text{4}}$ for ${\overline{I}}_x$ and $2.55{\text{in}}^{\text{4}}$ for ${\overline{I}}_y$,$-2.81{\text{in}}^{\text{4}}$ for ${\overline{I}}_{xy}$, in Equation (6).

$\begin{array}{c}\tan 2{\theta }_m=-\frac{2\left(-2.81\right)}{9.43-2.55}\\ =-\frac{-5.62}{6.88}\\ \tan 2{\theta }_m=0.817\\ 2{\theta }_m=39.267°\end{array}$

Find the angle $\left(\alpha \right)$ value using the relation:

$\begin{array}{c}\alpha =180°-\left(39.267+60°\right)\\ =80.733°\end{array}$

Find the moment of inertia of the area with respect to x about through $30°$ using Mohr circle method as shown below:

$I_{x^{\prime }}=I_{\text{ave}}-R\cos \alpha$ (7)

Here, $I_{\text{ave}}$ is average value of moment of inertia.

Substitute $5.99{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$, $4.4433{\text{in}}^{\text{4}}$ for R, and $80.733°$ for $\alpha$ in Equation (7).

$\begin{array}{c}{I}_{x^{\prime }}=\left(5.99-4.4433\cos 80.733°\right)\\ =\left(5.99-0.7155\right)\\ =5.2744{\text{in}}^{\text{4}}\end{array}$

Thus, the moment of inertia of the area with respect to x about through $30°$ using Mohr circle method is $\underset{\_}{5.27{\text{in}}^{\text{4}}}$

Find the moment of inertia of the area with respect to y about through $30°$ using Mohr circle method as shown below:

$I_{y^{\prime }}=I_{\text{ave}}+R\cos \alpha$ (8)

Substitute $5.99{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$, $4.4433{\text{in}}^{\text{4}}$ for R, and $80.733°$ for $\alpha$ in Equation (8).

$\begin{array}{c}{I}_{y^{\prime }}=\left(5.99+4.4433\cos 80.733°\right)\\ =\left(5.99+0.7155\right)\\ =6.71{\text{mm}}^{\text{4}}\end{array}$

Thus, the moment of inertia of the area with respect to y about through $30°$ using Mohr circle method is $\underset{\_}{6.71{\text{in}}^{\text{4}}}$

Find the product of inertia of the area with respect to y about through $45°$ using Mohr circle method as shown below:

$I_{x^{\prime }{y}^{\prime }}=-R\sin \alpha$ (9)

Substitute $4.4433{\text{in}}^{\text{4}}$ for R and $80.733°$ for $\alpha$ in Equation (9).

$\begin{array}{c}{I}_{x^{\prime }{y}^{\prime }}=-4.4433\times \sin \left(80.733°\right)\\ =-4.385\\ =-4.39{\text{in}}^{\text{4}}\end{array}$

Thus, the product of inertia of the area with respect to x and y axes about through $30°$ by using Mohr circle method is $\underset{\_}{-4.39{\text{in}}^{\text{4}}}$.

(b)

To determine

Find the orientation of the principal axes through the centroid and corresponding values.

Answer to Problem 9.192RP

The orientation of the principal axes at the origin is $\underset{\_}{19.63°\text{counter}\text{clockwise}}$.

The maximum moment of inertia is $\underset{\_}{10.43{\text{in}}^4}$.

The minimum moment of inertia is $\underset{\_}{1.547{\text{in}}^4}$

Explanation of Solution

Calculation:

Find the orientation of the principal axes through at origin as shown below.

Refer part a.

$\begin{array}{c}\tan 2{\theta }_m=39.267°\\ {\theta }_m=19.63°\end{array}$

Thus, the orientation of the principal axes at the origin is $\underset{\_}{19.63°\text{counter}\text{clockwise}}$.

Sketch the orientation axis as shown in Figure 4.

Find the maximum moment $I_{\max }$ of inertia using the relation:

$I_{\max }=I_{\text{ave}}+R$ (10)

Substitute $5.99{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $4.4433{\text{in}}^{\text{4}}$ for R, in Equation (10).

$\begin{array}{c}{I}_{\max }=5.99+4.4433\\ =10.4333{\text{in}}^{\text{4}}\end{array}$

Thus, the maximum moment of inertia is $\underset{\_}{10.4333{\text{in}}^{\text{4}}}$.

Find the minimum moment $I_{\min }$ of inertia using the relation:

$I_{\min }=I_{\text{ave}}-R$ (11)

Substitute $5.99{\text{in}}^{\text{4}}$ for $I_{\text{ave}}$ and $4.4433{\text{in}}^{\text{4}}$ for R, in Equation (11).

$\begin{array}{c}{I}_{\min }=5.99-4.4433\\ =1.547{\text{in}}^{\text{4}}\end{array}$

Thus, the minimum moment of inertia is $\underset{\_}{1.547{\text{in}}^4}$.