Data: P = 1500 kN e = 1 m

To find $\sigma_{max} = ?$

$Direct stress, \sigma_{0}=\frac{P}{A}=\frac{1500 \times 10^{3}}{2\left(\frac{1}{2} \times 3000 \times 3000\right)}=0.1667 \mathrm{N} / \mathrm{mm}^{2}$

$Bending Moment, \mathrm{M}=\operatorname{Pxe}=\left(1500 \times 10^{3}\right) \mathrm{x}\left(1 \times 10^{3}\right)=1.5 \times 10^{9} \mathrm{N}-\mathrm{mm}$

$As the eccentricity is about Centroidal \mathrm{X}-\mathrm{X} axis, therefore$

$\mathrm{I}_{\mathrm{xx}}=\mathrm{I}_{\mathrm{DB}}=\mathrm{I}_{\text {base }}=2 \mathrm{x}\left(\frac{b h^{3}}{12}\right)=2 \times\left(\frac{3000 \times 3000^{3}}{12}\right)=13.5 \times 10^{12} \mathrm{mm}^{4}$

$Distance of extreme layer (i.e. point A or C) from \mathrm{X}-\mathrm{X} axis - \mathrm{Y}=6000 / 2=3000 \mathrm{mm}$

Section modulus: $$ \begin{aligned} Z_{\mathrm{xx}}=\frac{I_{\mathrm{xx}}}{Y}=& \frac{13.5 \times 10^{12}}{3000}=4.5 \times 10^{9} \mathrm{mm}^{3} \\ \text { Bending Stress: } \end{aligned} $$ $$ \sigma_{\mathrm{b}}=\frac{M}{Z_{\mathrm{xx}}}=\frac{1.5 \times 10^{9}}{4.5 \times 10^{9}}=0.333 \mathrm{N} / \mathrm{mm}^{2} $$ Maximum Bending Stress: $$ \sigma_{\max }=\sigma_{0}+\sigma_{\mathrm{b}}=0.1667+0.333=0.5 \mathrm{N} / \mathrm{mm}^{2} \text { (Compressive) } $$