Ring foundation 3.60 m stress

Given

Outer $D = 3.60$ m ($R = 1.80$ m), inner $d = 2.40$ m ($r = 1.20$ m), $q = 135$ kN/m², $z = 1.80$ m, point on the axis.

Stress on the axis of a ring = stress of full circle (radius $R$) minus stress of the inner circle (radius $r$):

$$\Delta\sigma_z = q\left[\frac{1}{\left(1+(r/z)^2\right)^{3/2}} - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right]$$

Calculation

• Outer circle: $R/z = 1.0$, $(1+1)^{3/2} = 2.828$, factor $= 1 - 0.3536 = 0.6464$
• Inner circle: $r/z = 0.667$, $(1+0.444)^{3/2} = 1.736$, factor $= 1 - 0.576 = 0.4240$

$$\Delta\sigma_z = 135\,(0.6464 - 0.4240) = 135\times0.2224 = 30.0\ \text{kN/m}^2$$

Answer: Vertical stress at 1.80 m depth below the centre $\approx 30.0$ kN/m².