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Ring footing 8 m / 4 m, stress at 8 m

Short answer

Stress = q[1/(1+(r/z)²)^(3/2) − 1/(1+(R/z)²)^(3/2)]. For R=4 m, r=2 m, z=8 m, q=150 kN/m²: outer factor=0.2845, inner=0.0869, net=0.1976 × 150 = 29.6 kN/m².

A ring footing of external diameter 8 m and internal diameter 4 m rests at a depth 2 m below the ground surface. It carries a load intensity of 150 kN/m². Find the vertical stress at a depth of 8 m along the axis of the footing below the footing base. Neglect the effect of the excavation on the stress.

Given

Outer D=8D = 8 m (R=4R = 4 m), inner d=4d = 4 m (r=2r = 2 m), q=150q = 150 kN/m². Depth of interest is 8 m below the base of the footing, so z=8z = 8 m. The effect of excavation is neglected.

Calculation

Δσz=q[(1−1(1+(R/z)2)3/2)−(1−1(1+(r/z)2)3/2)]\Delta\sigma_z = q\left[\left(1 - \frac{1}{(1+(R/z)^2)^{3/2}}\right) - \left(1 - \frac{1}{(1+(r/z)^2)^{3/2}}\right)\right]
  • Outer: R/z=0.5R/z = 0.5, (1.25)1.5=1.3975(1.25)^{1.5} = 1.3975, factor =1−0.7155=0.2845= 1 - 0.7155 = 0.2845
  • Inner: r/z=0.25r/z = 0.25, (1.0625)1.5=1.0952(1.0625)^{1.5} = 1.0952, factor =1−0.9131=0.0869= 1 - 0.9131 = 0.0869
Δσz=150 (0.2845−0.0869)=150×0.1976=29.6 kN/m2\Delta\sigma_z = 150\,(0.2845 - 0.0869) = 150\times0.1976 = 29.6\ \text{kN/m}^2

Answer: Vertical stress at 8 m below the footing base along the axis ≈29.6\approx 29.6 kN/m².

✓ Answer checked by Claude Opus 5.5 (high effort)

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