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Column 500 x 900: stress intensities, critical corner

Short answer

Eccentricities80 mm,90 mm; corner stresses: A =1.707, B =0.907, C =0.427, D =−0.373 N/mm². Corner A has the highest compression (1.707 N/mm²), corner D is in tension (−0.373 N/mm²). Both are critical because masonry cannot resist tension and compression must not exceed permissible stress.

A column section 500 mm ×\times 900 mm carries a load of 300 kN acting at 170 mm from the 900 mm face and 360 mm from the 500 mm face. Determine the stress intensities. Which one is critical?

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 e1=250−170=80e_1 = 250-170 = 80 mm and e2=450−360=90e_2 = 450-360 = 90 mm.

Data

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

Eccentricities

e1=80 mm (across the 500 mm side)e2=90 mm (along the 900 mm side)\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

A=500×900=450000 mm2Z1=900×50026=37500000 mm3Z2=500×90026=67500000 mm3\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

σ0=PA=300×103450000=0.6667 N/mm2σb1=Pe1Z1=300×103×8037500000=0.6400 N/mm2σb2=Pe2Z2=300×103×9067500000=0.4000 N/mm2\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 σ=σ0±σb1±σb2\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2} (positive = compression, N/mm²):

CornerPositionWorkingStress
Anear the 900 face and near the 500 face0.6667+0.6400+0.40000.6667 + 0.6400 + 0.40001.7067
Bnear the 900 face, far from the 500 face0.6667+0.6400−0.40000.6667 + 0.6400 - 0.40000.9067
Cfar from the 900 face, near the 500 face0.6667−0.6400+0.40000.6667 - 0.6400 + 0.40000.4267
Dfar from both faces0.6667−0.6400−0.40000.6667 - 0.6400 - 0.4000-0.3733

Check: σ0\sigma_0 = average of the four corners = 0.6667. Corner D is in tension (-0.3733 N/mm²) because 6e1/500+6e2/900=1.560>16e_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.

Critical stress

  • The maximum compressive stress is at corner A = 1.707 N/mm2^2, the corner nearest the load. This must be compared with the permissible compressive stress of the masonry (fbkskakpf_b k_s k_a k_p).
  • Corner D has a tension of 0.373 N/mm2^2 because the load is outside the kernel of the section. Masonry and plain concrete have little tensile strength (IS 1905 cl. 5.4.2 permits none for vertical loads), so the corner would crack and the compression at A would be even higher than 1.707 N/mm2^2.

Therefore corner A (compression) and corner D (tension) both govern; the column must be enlarged or the eccentricity reduced.

✓ Answer checked by Claude Opus 5.5 (high effort)

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