Corner stresses of 450 x 750 column (150 kN)

Data

Section 450 mm $\times$ 750 mm, $P=150$ kN. The eccentricity of the load from the centroid is 50 mm along the 750 mm side and 45 mm along the 450 mm side.

Eccentricities

$$\begin{aligned} e_{1} &= 45\ \text{mm (across the 450 mm side)}\\ e_{2} &= 50\ \text{mm (along the 750 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 450\times750 = 337500\ \text{mm}^2\\ Z_1 &= \frac{750\times450^2}{6} = 25312500\ \text{mm}^3\\ Z_2 &= \frac{450\times750^2}{6} = 42187500\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{150\times10^3}{337500} = 0.4444\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{150\times10^3\times45}{25312500} = 0.2667\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{150\times10^3\times50}{42187500} = 0.1778\ \text{N/mm}^2 \end{aligned}$$

Corner stresses $\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2}$ (positive = compression, N/mm²):

Check: $\sigma_0$ = average of the four corners = 0.4444. Corner D has zero stress because the load lies exactly on the edge of the kernel ($6e_1/450+6e_2/750 = 0.6+0.4 = 1.000$), so no tension develops.

Answer: A = 0.889, B = 0.533, C = 0.356, D = 0.000 N/mm² (compression positive); maximum 0.889 N/mm² at A.