Chapter 9.2, Problem 9.41P

To determine

Find the moment of inertia ${\overline{I}}_x$ and ${\overline{I}}_y$ with respect to centroidal axes.

Answer to Problem 9.41P

The moment of inertia about x axis is $\underset{\_}{46.8\times {10}^6{\text{mm}}^{\text{4}}}$.

The moment of inertia about y axis is $\underset{\_}{13.89\times {10}^6{\text{mm}}^{\text{4}}}$.

Explanation of Solution

Given information:

The width $\left(b_1\right)$ of flange is $120\text{mm}$.

The height $\left(h_1\right)$ of flange is $30\text{mm}$.

The width $\left(b_2\right)$ of web is $120\text{mm}$.

The height $\left(h_2\right)$ of web is $40\text{mm}$.

Calculation:

Sketch the cross section as shown in Figure 1.

Refer to Figure 1.

Here, area of flange 1 is equal to area of flange 3.

Find the Area $\left(A_1\right)$ of flange using the relation:

$A_1=2\left(b_1{h}_1\right)$ (1)

Substitute $120\text{mm}$ for $\left(b_1\right)$, $30\text{mm}$  for $h_1$ in Equation (1).

$\begin{array}{c}{A}_1=2\left(120\times 30\right)\\ =7,200{\text{mm}}^{\text{2}}\end{array}$

Find the Area $\left(A_2\right)$ of flange using the relation:

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

Substitute $40\text{mm}$ for $\left(b_2\right)$, $120\text{mm}$  for $h_2$ in Equation (2).

$\begin{array}{c}{A}_2=\left(40\times 120\right)\\ =4,800{\text{mm}}^{\text{2}}\end{array}$

Find the total area (A) of section using the relation:

$A=A_1+A_2$ (3)

Substitute $7,200{\text{mm}}^{\text{2}}$ for $A_1$ and $4,800{\text{mm}}^{\text{2}}$ for $A_2$ in Equation (3).

$\begin{array}{c}A=7,200+4,800\\ =12,000{\text{mm}}^{\text{2}}\end{array}$

Find the centroid section $\left(x_1\right)$ as shown below:

$\begin{array}{c}{x}_1=\frac{120}{2}\\ =60\text{mm}\end{array}$

Find the centroid section $\left(x_2\right)$ as shown below:

$\begin{array}{c}{x}_2=\frac{40}{2}\\ =20\text{mm}\end{array}$

Find the centroid $\left(\overline{x}\right)$ using the relation:

$\overline{x}=\frac{2\left(A_1{x}_1\right)+A_2{x}_2}{A_1+A_2}$ (4)

Substitute $7,200{\text{mm}}^{\text{2}}$ for $A_1$, $4,800{\text{mm}}^{\text{2}}$ for $A_2$, $60\text{mm}$ for $x_1$, $20\text{mm}$ for $x_2$ in Equation (4).

$\begin{array}{c}\overline{x}=\frac{\left(7,200\times 60\right)+\left(4,800\times 20\right)}{7,200+4,800}\\ =\frac{432,000+96,000}{12,000}\\ =44\text{mm}\end{array}$

Find the moment of inertia ${\left(I{x}_{}\right)}_F$ of flange section using the relation:

${\left(I_x\right)}_F=\frac{1}{12}{b}_1{h}_1^3+\left(A_1\right){\left(\overline{h}\right)}^2$ (5)

Here, $A_1$ is area of individual section and $\overline{h}$ is the vertical distance from the centroid of the segment to the neutral axis.

Substitute $120\text{mm}$ for $\left(b_1\right)$, $30\text{mm}$  for $h_1$, $3,600{\text{mm}}^{\text{2}}$ for $A_1$, $\left(90-15\right)\text{mm}$ for $\overline{h}$ in Equation (5).

$\begin{array}{c}{\left(I_x\right)}_F=\frac{1}{12}\left(120\right)\left({30}^3\right)+\left(3,600\right){\left(90-15\right)}^2\\ =270,000+20,250,000\\ =20,520,000\\ =20.520\times {10}^6{\text{mm}}^4\end{array}$

Find the moment of inertia ${\left(I{x}_{}\right)}_W$ of web section using the relation:

${\left(I_x\right)}_W=\frac{1}{12}\left(b_2{h}_2^3\right)$ (6)

Substitute $40\text{mm}$ for $\left(b_2\right)$ and $120\text{mm}$  for $h_2$, in Equation (6).

$\begin{array}{c}{\left(I_x\right)}_W=\frac{1}{12}\left(40\right){\left(120\right)}^3\\ =5,760,000\\ =5.760\times {10}^6{\text{mm}}^{\text{4}}\end{array}$

Find the total moment of inertia $\left({\overline{I}}_x\right)$ using the relation as follows:

$\left({\overline{I}}_x\right)={\left(I{x}_{}\right)}_W+{\left(I{x}_{}\right)}_F$ (7)

Substitute $20.520\times {10}^6{\text{mm}}^4$ for ${\left(I{x}_{}\right)}_F$ and $5.760\times {10}^6{\text{mm}}^{\text{4}}$ for ${\left(I{x}_{}\right)}_W$ in Equation (7).

$\begin{array}{c}\left({\overline{I}}_x\right)=20.520\times {10}^6+5.760\times {10}^6+20.520\times {10}^6\\ =46,800,00\\ =46.8\times {10}^6{\text{mm}}^{\text{4}}\end{array}$

Thus, the moment of inertia about x axis is $\underset{\_}{46.8\times {10}^6{\text{mm}}^{\text{4}}}$.

Find the moment of inertia ${\left(I{y}_{}\right)}_F$ of about y axis using the relation:

${\left(I_y\right)}_F=\frac{1}{12}{b}_1^3{h}_1+\left(A_1\right){\left(\overline{h}\right)}^2$ (8)

Substitute $120\text{mm}$ for $\left(b_1\right)$, $30\text{mm}$  for $h_1$, $3,600{\text{mm}}^{\text{2}}$ for $A_1$, $\left(16\right)\text{mm}$ for $\overline{h}$ in Equation (8).

$\begin{array}{c}{\left(I_y\right)}_F=\left[2\times \frac{1}{12}{\left(120\right)}^3\left(30\right)+\left(3,600\right){\left(16\right)}^2\right]\\ =2\times \left(4,320,000+921,600\right)\\ =10,483,200\\ =10.483\times {10}^6{\text{mm}}^4\end{array}$

Find the moment of inertia ${\left(I{y}_{}\right)}_W$ of about y axis using the relation:

${\left(I_y\right)}_W=\frac{1}{12}{b}_1^3{h}_1+\left(A_1\right){\left(\overline{h}\right)}^2$ (9)

Substitute $40\text{mm}$ for $\left(b_1\right)$, $120\text{mm}$  for $h_1$, $4,800{\text{mm}}^{\text{2}}$ for $A_2$, $\left(24\right)\text{mm}$ for $\overline{h}$ in Equation (9).

$\begin{array}{c}{\left(I_y\right)}_W=\left[\frac{1}{12}{\left(40\right)}^3\left(120\right)+\left(4,800\right){\left(24\right)}^2\right]\\ =\left(640,000+2,764,800\right)\\ =3,404,800{\text{mm}}^{\text{4}}\end{array}$

Find the total moment of inertia $\left(\overline{I}{y}_{}\right)$ of about y axis using the relation:

$\left(\overline{I}{y}_{}\right)={\left(I{y}_{}\right)}_F+{\left(I{y}_{}\right)}_w$ (10)

Substitute $3,404,800{\text{mm}}^{\text{4}}$ for ${\left(I{y}_{}\right)}_w$ and $10.483\times {10}^6{\text{mm}}^4$ for $\left(\overline{I}{y}_{}\right)$ in Equation (10).

$\begin{array}{c}\left({\overline{I}}_y\right)=3,404,800+10.483\times {10}^6\\ =13.89\times {10}^6{\text{mm}}^4\end{array}$

Thus, the moment of inertia about y axis is $\underset{\_}{13.89\times {10}^6{\text{mm}}^{\text{4}}}$.