Step 1

We are given a composite area shown in the diagram with its dimensions.

We are required to locate the centroid $\overline y $ of the cross section and the moment of inertia of the section about the $x'$ axis.

 
Step 2

The diagram of the cross-section showing all the dimensions are given as,

Consider section a horizontal axis $XX$ located at the lowest point and the centroid of the sections are measured from this axis.

 
Step 3

Using the formula for the centroid of the cross-section, the expression for the centroid $\overline y $ of the cross section about the $x'$ axis is given by,

\[\begin{array}{l} \overline y = \frac{{\sum {\overline y A} }}{{\sum A }}\\ \overline y = \frac{{{{\overline y }_1}{A_1} + {{\overline y }_2}{A_2} + {{\overline y }_3}{A_3} + {{\overline y }_4}{A_4}}}{{{A_1} + {A_2} + {A_3} + {A_4}}}\\ \overline y = \frac{{\left\{ \begin{array}{l} \left( {0.05 + \frac{{0.4}}{2}} \right){\rm{ m}} \times \left( {0.3 \times 0.4} \right){\rm{ }}{{\rm{m}}^2} + \left( {0.05 + \frac{{0.4}}{3}} \right){\rm{ m}} \times \left( {\frac{1}{2} \times 0.4 \times 0.4} \right){\rm{ }}{{\rm{m}}^2}\\ + \left( {\frac{{0.05}}{2}} \right){\rm{ m}} \times \left( {1.1 \times 0.05} \right){\rm{ }}{{\rm{m}}^2} \end{array} \right\}}}{{\left( {0.3 \times 0.4} \right){\rm{ }}{{\rm{m}}^2} + \left( {\frac{1}{2} \times 0.4 \times 0.4} \right){\rm{ }}{{\rm{m}}^2} + \left( {1.1 \times 0.05} \right){\rm{ }}{{\rm{m}}^2}}} \end{array}\]

Solve further as,

\[\begin{array}{l} \overline y = \frac{{\left\{ \begin{array}{l} 0.25{\rm{ m}} \times \left( {0.12} \right){\rm{ }}{{\rm{m}}^2} + 0.1833{\rm{ m}} \times \left( {0.08} \right){\rm{ }}{{\rm{m}}^2}\\ + 0.025{\rm{ m}} \times \left( {0.055} \right){\rm{ }}{{\rm{m}}^2} \end{array} \right\}}}{{\left( {0.12} \right){\rm{ }}{{\rm{m}}^2} + \left( {0.08} \right){\rm{ }}{{\rm{m}}^2} + \left( {0.055} \right){\rm{ }}{{\rm{m}}^2}}}\\ \overline y = \frac{{0.046{\rm{ }}{{\rm{m}}^3}}}{{0.255{\rm{ }}{{\rm{m}}^2}}}\\ \overline y = 0.181{\rm{ m}} \end{array}\]
 
Step 4

Using the parallel-axis theorem, the moment of inertia about the $x'$ axis for section 1 can be given as,

\[\begin{array}{l} {\left( {{I_{x'}}} \right)_1} = {\overline I _1} + {A_1}{\left( {{{\overline y }_1} - \overline y } \right)^2}\\ {\left( {{I_{x'}}} \right)_1} = \frac{{0.3{\rm{ m}} \times {{\left( {0.4{\rm{ m}}} \right)}^3}}}{{12}} + \left( {0.3{\rm{ m}} \times {\rm{0}}{\rm{.4 m}}} \right) \times {\left( {{\rm{0}}{\rm{.25 m}} - 0.181{\rm{ m}}} \right)^2}\\ {\left( {{I_{x'}}} \right)_1} = 2.171 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} \end{array}\]

Similarly, the moment of inertia about the $x'$ axis for section 2 can be given as,

\[\begin{array}{l} {\left( {{I_{x'}}} \right)_2} = {\overline I _2} + {A_2}{\left( {{{\overline y }_2} - \overline y } \right)^2}\\ {\left( {{I_{x'}}} \right)_2} = \frac{{0.4{\rm{ m}} \times {{\left( {0.4{\rm{ m}}} \right)}^3}}}{{36}} + \left( {\frac{1}{2} \times 0.4{\rm{ m}} \times {\rm{0}}{\rm{.4 m}}} \right) \times {\left( {{\rm{0}}{\rm{.1833 m}} - 0.181{\rm{ m}}} \right)^2}\\ {\left( {{I_{x'}}} \right)_2} = 0.7115 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} \end{array}\]

The moment of inertia about the $x'$ axis for section 3 can be given as,

\[\begin{array}{l} {\left( {{I_{x'}}} \right)_3} = {\overline I _3} + {A_3}{\left( {{{\overline y }_3} - \overline y } \right)^2}\\ {\left( {{I_{x'}}} \right)_3} = \frac{{{\rm{1}}{\rm{.1 m}} \times {{\left( {0.05{\rm{ m}}} \right)}^3}}}{{12}} + \left( {{\rm{1}}{\rm{.1 m}} \times {\rm{0}}{\rm{.05 m}}} \right) \times {\left( {{\rm{0}}{\rm{.025 m}} - 0.181{\rm{ m}}} \right)^2}\\ {\left( {{I_{x'}}} \right)_3} = 1.35 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} \end{array}\]
 
Step 5

The expression for the moment of inertia of the whole section about the $x'$ axis is given by,

\[\begin{array}{l} {I_{x'}} = {\left( {{I_{x'}}} \right)_1} + {\left( {{I_{x'}}} \right)_2} + {\left( {{I_{x'}}} \right)_3}\\ {I_{x'}} = 2.171 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} + 0.7115 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} + 1.35 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4}\\ {I_{x'}} = 4.2325 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^4} \end{array}\]