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 …