Step 1

We are given a cross-sectional area of a beam which is symmetrical in geometry about $y$ axis.

We are required to locate the centroid $\overline y $ of the beam’s cross-sectional area.

 
Step 2

The diagram showing all the dimensions of the beam’s cross-section is given below,

Here, point $C$ represents the centroid of the beam’s cross-section and $\overline y $ is the centroidal distance from horizontal $x$ axis.

The area of the top section 1 is given by,

\[\begin{array}{l} {A_1} = 300{\rm{ mm}} \times {\rm{25 mm}}\\ {A_1} = 7500{\rm{ m}}{{\rm{m}}^2} \end{array}\]

The centroidal distance of the top section 1 from the $x$ axis is given by,

\[\begin{array}{l} {\overline y _1} = 100{\rm{ mm}} + \frac{{25{\rm{ mm}}}}{2}\\ {\overline y _1} = 112.5{\rm{ mm}} \end{array}\]
 
Step 3

The area of the bottom section 2 is given by,

\[\begin{array}{l} {A_2} = 25{\rm{ mm}} \times 100{\rm{ mm}}\\ {A_2} = 2500{\rm{ m}}{{\rm{m}}^2} \end{array}\]

Since, the beams cross section is symmetric about the central vertical axis, it means that the area of section 2 would be equal to section 3.

So, the area of the bottom section 3 is given by,

\[\begin{array}{l} {A_3} = {A_2}\\ {A_3} = 2500{\rm{ m}}{{\rm{m}}^2} \end{array}\]

The centroidal distance of the bottom section 2 from the $x$ axis is given by,

\[\begin{array}{l} {\overline y _2} = \frac{{{\rm{100 mm}}}}{2}\\ {\overline y _2} = 50{\rm{ mm}} \end{array}\]

Similarly, for the symmetric section 3, the centroidal distance of the bottom section 3 from the $x$ axis is given by,

\[\begin{array}{l} {\overline y _3} = {\overline y _2}\\ {\overline y _3} = 50{\rm{ mm}} \end{array}\]
 
Step 4

Using the formula for the centroid, the expression for the centroid $\overline y $ of the beam’s cross-sectional area is given by,

\[\begin{array}{l} \overline y = \frac{{\sum {\overline y A} }}{{\sum A }}\\ \overline y = \frac{{{{\overline y }_1}{A_1} + {{\overline y }_2}{A_2} + {{\overline y }_3}{A_3}}}{{{A_1} + {A_2} + {A_3}}}\\ \overline y = \frac{{\left\{ {\left( {112.5 \times 7500} \right){\rm{ m}}{{\rm{m}}^3}} \right\} + \left\{ {\left( {50 \times 2500} \right){\rm{ m}}{{\rm{m}}^3}} \right\} + \left\{ {\left( {50 \times 2500{\rm{ }}} \right){\rm{m}}{{\rm{m}}^3}} \right\}}}{{\left( {7500{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {2500{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {2500{\rm{ m}}{{\rm{m}}^2}} \right)}}\\ \overline y = 87.5{\rm{ mm}} \end{array}\]