Step 1

We are given the T-beam with their dimensions.

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

 
Step 2

To calculate the area and the centroid of each segment, we need to divide the beam’s cross-sectional area into two rectangles as:

To calculate the area of segment 1, we have:

\[{A_1} = {b_1} \times {d_1}\]

Here, ${b_1}$ is width of the segment 1 and ${d_1}$ is the depth of the segment 1.

Substitute the values in the above expression, we get:

\begin{array}{c} {A_1} = \left( {50\;{\rm{mm}}} \right) \times \left( {300\;{\rm{mm}}} \right)\\ {A_1} = 15000\;{\rm{m}}{{\rm{m}}^2} \end{array}

Similarly, to calculate the area of segment 2, we have:

\[{A_2} = {b_2} \times {d_2}\]

Here, ${b_2}$ is width of the segment 2 and ${d_2}$ is the depth of the segment 2.

Substitute the values in the above expression, we get:

\begin{array}{c} {A_2} = \left( {300\;{\rm{mm}}} \right) \times \left( {50\;{\rm{mm}}} \right)\\ {A_2} = 15000\;{\rm{m}}{{\rm{m}}^2} \end{array}
 
Step 3

To calculate the centroid of segment 1, we have:

\[{y'_1} = \frac{{{y_1}}}{2}\]

Substitute the values in the above expression, we get:

\begin{array}{c} {{y'}_1} = \frac{{\left( {300\;{\rm{mm}}} \right)}}{2}\\ {{y'}_1} = 150\;{\rm{mm}} \end{array}

To calculate the centroid of segment 2, we have:

\[{y'_2} = \left( {\frac{{{y_2}}}{2} + {y_1}} \right)\]

Substitute the values in the above expression, we get:

\begin{array}{c} {{y'}_2} = \left( {\frac{{50\;{\rm{mm}}}}{2} + 300\;{\rm{mm}}} \right)\\ {{y'}_2} = 325\;{\rm{mm}} \end{array}
 
Step 4

To calculate the centroid of the cross-sectional area of the beam about y-axis, we have:

\begin{array}{c} \bar y = \frac{{\sum y'A}}{{\sum A}}\\ \bar y = \frac{{\left( {{{y'}_1}{A_1} + {{y'}_2}{A_2}} \right)}}{{\left( {{A_1} + {A_2}} \right)}} \end{array}

Substitute the values in the above expression, we get:

\begin{array}{c} \bar y = \frac{{\left[ {\left( {150\;{\rm{mm}}} \right) \times \left( {15000\;{\rm{m}}{{\rm{m}}^3}} \right)} \right] + \left[ {\left( {325\;{\rm{mm}}} \right) \times \left( {15000\;{\rm{m}}{{\rm{m}}^2}} \right)} \right]}}{{\left( {15000\;{\rm{m}}{{\rm{m}}^2}} \right) + \left( {15000\;{\rm{m}}{{\rm{m}}^2}} \right)}}\\ \bar y = \frac{{7125000\;{\rm{m}}{{\rm{m}}^3}}}{{30000\;{\rm{m}}{{\rm{m}}^2}}}\\ \bar y = 237.5\;{\rm{mm}} \end{array}