[1] Area of section = $400 \times 600 = 24 \times 10^4 \ mm^2$

[2] Section modulus about XX – axis = $\frac{bd^2}{6} = \frac{400 \times 600^2}{6} = 24 \times 10^6 \ mm^3$

[3] Section modulus about YY – axis = $\frac{db^2}{6} = \frac{600 \times 400^2}{6} = 16 \times 10^6 \ mm^3$

[4] 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, $\frac{P}{A} = \frac{Pex}{Zxx}$ & $\frac{P}{A} = \frac{Pey}{Zyy}$

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

i.e. $e_x = \frac{ 24 \times 10^6}{24 \times 100^4}$ & $e_y = \frac{16 \times 10^6}{24 \times 10^4}$

$e_x = 100 \ mm$ and $e_y = 66.67 \ mm$

$\therefore$ The shaded portion of section is core or kernel of the solid rectangular section.