Step 1

We are given a shaded area of a composite section in the diagram showing all of its dimensions.

We are required to determine the moment of inertia for the shaded area about the $x$ axis.

 
Step 2

Consider the given composite section and divide it into four sections represented by 1, 2, 3 and 4 in the diagram. The diagrammatic representation of the shaded composite area is shown below:

 
Step 3

For solid section 1:

Using the parallel axis theorem, the expression for moment of inertia of the solid rectangular section 1 about the $x$ axis is given by,

\[\begin{array}{l} {\left( {{{\overline I }_x}} \right)_1} = {I_1} + {A_1}{y_1}^2\\ {\left( {{{\overline I }_x}} \right)_1} = \frac{{{\rm{6 in}} \times {{\left( {{\rm{6 in}}} \right)}^3}}}{{12}} + \left( {6 \times 6} \right){\rm{ i}}{{\rm{n}}^2} \times {\left( {\frac{{{\rm{6 in}}}}{2}} \right)^2}\\ {\left( {{{\overline I }_x}} \right)_1} = 108{\rm{ i}}{{\rm{n}}^4} + 324{\rm{ i}}{{\rm{n}}^4}\\ {\left( {{{\overline I }_x}} \right)_1} = 432{\rm{ i}}{{\rm{n}}^4} \end{array}\]
 
Step 4

For solid section 2:

Similarly, the expression for the moment of inertia of the solid triangular section 2 about the $x$ axis is given by,

\[\begin{array}{l} {\left( {{{\overline I }_x}} \right)_2} = {I_2} + {A_2}{y_2}^2\\ {\left( {{{\overline I }_x}} \right)_2} = \frac{{{\rm{6 in}} \times {{\left( {{\rm{3 in}}} \right)}^3}}}{{36}} + \left( {\frac{1}{2} \times 6 \times 3} \right){\rm{ i}}{{\rm{n}}^2} \times {\left( {6{\rm{ in}} + \frac{{{\rm{3 in}}}}{3}} \right)^2}\\ {\left( {{{\overline I }_x}} \right)_2} = 4.5{\rm{ i}}{{\rm{n}}^4} + 441{\rm{ i}}{{\rm{n}}^4}\\ {\left( {{{\overline I }_x}} \right)_2} = 445.5{\rm{ i}}{{\rm{n}}^4} \end{array}\]
 
Step 5

For solid section 3:

The expression for the moment of inertia of the solid triangular section 3 about the $x$ axis is given by,

\[\begin{array}{l} {\left( {{{\overline I }_x}} \right)_3} = {I_3} + {A_3}{y_3}^2\\ {\left( {{{\overline I }_x}} \right)_3} = \frac{{{\rm{6 in}} \times {{\left( {{\rm{9 in}}} \right)}^3}}}{{36}} + \left( {\frac{1}{2} \times 6 \times 9} \right){\rm{ i}}{{\rm{n}}^2} \times {\left( {\frac{{{\rm{2}} \times 9{\rm{ in}}}}{3}} \right)^2}\\ {\left( {{{\overline I }_x}} \right)_3} = 121.5{\rm{ i}}{{\rm{n}}^4} + 972{\rm{ i}}{{\rm{n}}^4}\\ {\left( {{{\overline I }_x}} \right)_3} = 1093.5{\rm{ i}}{{\rm{n}}^4} \end{array}\]
 
Step 6

For hollow section 4:

The expression for the moment of inertia of the circular hollow section 4 about the $x$ axis is given by,

\[\begin{array}{l} {\left( {{{\overline I }_x}} \right)_4} = {I_4} + {A_4}{y_4}^2\\ {\left( {{{\overline I }_x}} \right)_4} = \frac{{\pi \times {{\left( {{\rm{2 in}}} \right)}^4}}}{4} + \left( {\pi \times {{\left( {{\rm{2 in}}} \right)}^2}} \right){\rm{ i}}{{\rm{n}}^2} \times {\left( {{\rm{3 in}}} \right)^2}\\ {\left( {{{\overline I }_x}} \right)_4} = 12.57{\rm{ i}}{{\rm{n}}^4} + 113.097{\rm{ i}}{{\rm{n}}^4}\\ {\left( {{{\overline I }_x}} \right)_4} = 125.66{\rm{ i}}{{\rm{n}}^4} \end{array}\]
 
Step 7

The expression for the moment of inertia ${\overline I _x}$ of the beam about the $x$ axis is given by,

\[\begin{array}{l} {\overline I _x} = {\left( {{{\overline I }_x}} \right)_1} + {\left( {{{\overline I }_x}} \right)_2} + {\left( {{{\overline I }_x}} \right)_3} - {\left( {{{\overline I }_x}} \right)_4}\\ {\overline I _x} = 432{\rm{ i}}{{\rm{n}}^4} + 445.5{\rm{ i}}{{\rm{n}}^4} + 1093.5{\rm{ i}}{{\rm{n}}^4} - 125.66{\rm{ i}}{{\rm{n}}^4}\\ {\overline I _x} = 1845.34{\rm{ i}}{{\rm{n}}^4} \end{array}\]