Step 1

We are given the various dimensions of the system.

We are asked to determine the distance $\bar y$ to the centroid of the beam's cross-sectional area and the moment of inertia about the centroidal x' axis.

 
Step 2

The three possible sections of the system are made and shown as:

We have the width of part 1 is ${b_1} = 1\;{\rm{in}}$.

We have the height of part 1 is ${h_1} = 4\;{\rm{in}}$.

We have the width of part 2 is ${b_2} = 6\;{\rm{in}}$.

We have the height of part 2 is ${h_2} = 1\;{\rm{in}}$.

We have the width of part 3 is ${b_3} = 1\;{\rm{in}}$.

We have the height of part 3 is ${h_3} = 4\;{\rm{in}}$.

 
Step 3

The formula to calculate the $\bar y$ coordinate of the centroid for the total area is,

\[\begin{array}{l} \bar y = \frac{{{A_1}{y_1} + {A_2}{y_2} + {A_3}{y_3}}}{{{A_1} + {A_2} + {A_3}}}\\ \bar y = \frac{{\left( {{b_1}{h_1}} \right)\left( {\frac{{{h_1}}}{2}} \right) + \left( {{b_2}{h_2}} \right)\left( {\frac{{{h_2}}}{2}} \right) + \left( {{b_3}{h_3}} \right)\left( {\frac{{{h_3}}}{2}} \right)}}{{{b_1}{h_1} + {b_2}{h_2} + {b_3}{h_3}}} \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} \bar y = \frac{{\left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right)\left( {\frac{{4\;{\rm{in}}}}{2}} \right) + \left( {6\;{\rm{in}} \times 1\;{\rm{in}}} \right)\left( {\frac{{1\;{\rm{in}}}}{2}} \right) + \left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right)\left( {\frac{{4\;{\rm{in}}}}{2}} \right)}}{{\left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right) + \left( {6\;{\rm{in}} \times 1\;{\rm{in}}} \right) + \left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right)}}\\ \bar y = 1.36\;{\rm{in}} \end{array}\]
 
Step 5

The formula to calculate the moment of inertia of part 1 about the centroidal axis is,

\[\begin{array}{l} {\left( {{I_x}} \right)_1} = {I_1} + {A_1}{\left( {{y_1} - \bar y} \right)^2}\\ {\left( {{I_x}} \right)_1} = \frac{{{b_1}{{\left( {{h_1}} \right)}^3}}}{{12}} + \left( {{b_1}{h_1}} \right){\left( {\frac{{{h_1}}}{2} - \bar y} \right)^2} \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{I_x}} \right)_1} = \frac{{\left( {1\;{\rm{in}}} \right){{\left( {4\;{\rm{in}}} \right)}^3}}}{{12}} + \left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right){\left( {\frac{{4\;{\rm{in}}}}{2} - 1.36} \right)^2}\\ {\left( {{I_x}} \right)_1} = 6.97\;{\rm{i}}{{\rm{n}}^4} \end{array}\]
 
Step 7

The formula to calculate the moment of inertia of part 2 about the centroidal axis is,

\[\begin{array}{l} {\left( {{I_x}} \right)_2} = {I_2} + {A_2}{\left( {{y_2} - \bar y} \right)^2}\\ {\left( {{I_x}} \right)_2} = \frac{{{b_2}{{\left( {{h_2}} \right)}^3}}}{{12}} + \left( {{b_2}{h_2}} \right){\left( {\frac{{{h_2}}}{2} - \bar y} \right)^2} \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{I_x}} \right)_2} = \frac{{\left( {6\;{\rm{in}}} \right){{\left( {1\;{\rm{in}}} \right)}^3}}}{{12}} + \left( {6\;{\rm{in}} \times 1\;{\rm{in}}} \right){\left( {\frac{{1\;{\rm{in}}}}{2} - 1.36} \right)^2}\\ {\left( {{I_x}} \right)_2} = 4.94\;{\rm{i}}{{\rm{n}}^4} \end{array}\]
 
Step 9

The formula to calculate the moment of inertia of part 3 about the centroidal axis is,

\[\begin{array}{l} {\left( {{I_x}} \right)_3} = {I_3} + {A_3}{\left( {{y_3} - \bar y} \right)^2}\\ {\left( {{I_x}} \right)_3} = \frac{{{b_3}{{\left( {{h_3}} \right)}^3}}}{{12}} + \left( {{b_3}{h_3}} \right){\left( {\frac{{{h_3}}}{2} - \bar y} \right)^2} \end{array}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{I_x}} \right)_3} = \frac{{\left( {1\;{\rm{in}}} \right){{\left( {4\;{\rm{in}}} \right)}^3}}}{{12}} + \left( {1\;{\rm{in}} \times 4\;{\rm{in}}} \right){\left( {\frac{{4\;{\rm{in}}}}{2} - 1.36} \right)^2}\\ {\left( {{I_x}} \right)_3} = 6.97\;{\rm{i}}{{\rm{n}}^4} \end{array}\]
 
Step 11

The formula to calculate the moment of inertia of the shaded area about the centroidal $x'$ axis is,

\[{I_{x'}} = {\left( {{I_x}} \right)_1} + {\left( {{I_x}} \right)_2} + {\left( {{I_x}} \right)_3}\]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{l} {I_{x'}} = 6.97\;{\rm{i}}{{\rm{n}}^4} + 4.94\;{\rm{i}}{{\rm{n}}^4} + 6.97\;{\rm{i}}{{\rm{n}}^4}\\ {I_{x'}} = 18.88\;{\rm{i}}{{\rm{n}}^4} \end{array}\]