Corner stresses of 400 x 800 column (200 kN)

Distances are measured from the faces. The 160 mm distance is across the 400 mm side (from the 800 mm face) and the 350 mm distance is along the 800 mm side (from the 400 mm face). Hence $e_1 = 200-160 = 40$ mm and $e_2 = 400-350 = 50$ mm.

Data

Section 400 mm $\times$ 800 mm, $P=200$ kN. The load acts 160 mm from the 800 mm face and 350 mm from the 400 mm face.

Eccentricities

$$\begin{aligned} e_{1} &= 40\ \text{mm (across the 400 mm side)}\\ e_{2} &= 50\ \text{mm (along the 800 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 400\times800 = 320000\ \text{mm}^2\\ Z_1 &= \frac{800\times400^2}{6} = 21333333\ \text{mm}^3\\ Z_2 &= \frac{400\times800^2}{6} = 42666667\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{200\times10^3}{320000} = 0.6250\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{200\times10^3\times40}{21333333} = 0.3750\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{200\times10^3\times50}{42666667} = 0.2344\ \text{N/mm}^2 \end{aligned}$$

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

Check: $\sigma_0$ = average of the four corners = 0.6250. All four corners are in compression, because the load lies inside the kernel ($6e_1/400+6e_2/800 = 0.975 < 1$).

Answer: A = 1.234, B = 0.766, C = 0.484, D = 0.016 N/mm² (compression positive); maximum 1.234 N/mm² at A.