[1] $\bar{y} = \frac{10 \times 150 \times 5 + 280 \times 10 \times (10 + 140) + 150 \times 10 \times (280 + 10 + 5)}{10 \times 150 + 280 \times 10 + 150 \times 10}$

$\bar{y} = 150 \ mm$ OR

$\bar{y_1} = \frac{300}{2} = 150 \ mm$ (Symmetrical about X – axis)

Moment Inertia about XX – axis and YY – axis is $I_{xx}$ & $I_{yy}$

[2] $I_{xx} = \frac{150 \times 300^3}{12} - \frac{(150 – 10) \times 280^3}{12} = 81.39 \times 10^6 \ mm^4$

$I_{yy} = 2[ 10 \times \frac{150^3}{12}] + \frac{280 \times 10^3}{12} = 5.64 \times 10^6 \ mm^4$

[3] $\bar{y_2} = \frac{150}{2} = 75 \ mm$ (Symmetrical about Y – axis)

[4] Now, Let ex and ey be the maximum eccentricities about the XX – axis and YY – axis in order to avoid tensile stress. Let P be the longitudinal load on section.

Hence,

$\frac{P}{A} = \frac{P_x \ e_x}{Z_{xx}} $ & $\frac{P}{A} = \frac{P_{ey}}{Z_{yy}}$

$(A = 10 \times 150 \times 2 + 280 \times 10)$

i.e. $A = 5800 \ mm^2$

i.e. $e_x = \frac{Zxx}{A}$ & $e_y = \frac{Z_{yy}}{A}$

But $Z_{xx} = \frac{I_{xx}}{y_1}$ & $Z_{yy} = \frac{I_{yy}}{y^2}$

$Z_{xx} = \frac{81.39 \times 10^6}{150}$ & $Z_{yy} = \frac{5.64 \times 10^6}{75}$

$Z_{xx} = 542.6 \times 10^3 \ mm^3$ & $Z_{yy} = 75.2 \times 10^3 \ mm^3$

$\therefore$ $e_x = \frac{542.6 \times 10^3}{5800}$ & $e_y = \frac{75.2 \times 10^3} {5800}$

$\therefore$ $e_x = 933.55 \ mm$ & $e_y = 12.96 \ mm$

The shaded region in I – Section shows core/kernel of section.