Corner stresses of 500 x 900 column (300 kN)

Distances are measured from the faces. The 170 mm distance is across the 500 mm side (from the 900 mm face) and the 360 mm distance is along the 900 mm side (from the 500 mm face). Hence $e_1 = 250-170 = 80$ mm and $e_2 = 450-360 = 90$ mm.

Data

Section 500 mm $\times$ 900 mm, $P=300$ kN. The load acts 170 mm from the 900 mm face and 360 mm from the 500 mm face.

Eccentricities

$$\begin{aligned} e_{1} &= 80\ \text{mm (across the 500 mm side)}\\ e_{2} &= 90\ \text{mm (along the 900 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 500\times900 = 450000\ \text{mm}^2\\ Z_1 &= \frac{900\times500^2}{6} = 37500000\ \text{mm}^3\\ Z_2 &= \frac{500\times900^2}{6} = 67500000\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{300\times10^3}{450000} = 0.6667\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{300\times10^3\times80}{37500000} = 0.6400\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{300\times10^3\times90}{67500000} = 0.4000\ \text{N/mm}^2 \end{aligned}$$

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

Check: $\sigma_0$ = average of the four corners = 0.6667. Corner D is in tension (-0.3733 N/mm²) because $6e_1/500+6e_2/900 = 1.560 > 1$, i.e. the load lies outside the kernel (middle-third) of the section.

Answer: A = 1.707, B = 0.907, C = 0.427, D = -0.373 N/mm² (compression positive); maximum 1.707 N/mm² at A.