Question 4 of 6: Vertical stress increase by superposition

Question 4: Vertical stress increase by superposition (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A square ring (frame) footing: outer plan 50 m × 50 m with a central 30 m × 30 m unloaded opening (frame width 10 m). Uniform contact pressure $q = 5 \times 10 = 50\ \text{kPa}$. Point A is at the outer top corner of the footing.

Find. $\Delta\sigma_z$ below the outer corner A at $z=2\ \text{m}$ and $z=5\ \text{m}$.

Approach. Use the Boussinesq corner influence factor $I(m,n)$ for a uniformly loaded rectangle, and superpose: the frame load = the full outer square (cornered at A) minus the central opening (built from four rectangles about A).

1. Corner influence factor. For a rectangle $B\times L$ with the corner over the point, at depth $z$, with $m=B/z$, $n=L/z$, $\Delta\sigma = qI(m,n)$ where $I$ is the standard Boussinesq/Newmark rectangular-corner factor (from the supplied chart).
2. Set up the superposition about A. Point A is the corner of the outer 50×50. The opening spans 10–40 m in both directions from A, so its influence at A is $I(40,40)-2\,I(40,10)+I(10,10)$. Hence $$\Delta\sigma_z = q\big[\,I(50,50) - \big(I(40,40)-2I(40,10)+I(10,10)\big)\big].$$
3. At $z = 2$ m. The outer square subtends a near-quarter-plane, so $I(50,50)\approx 0.250$, while the opening is far from A and contributes only $\approx 0.000$. Thus $\Delta\sigma_z = 50\times(0.250-0.000) = \boxed{12.5\ \text{kPa}}$.
4. At $z = 5$ m. Now $I(50,50)\approx 0.250$ and the opening contributes $\approx 0.002$, so $\Delta\sigma_z = 50\times(0.250-0.002) = \boxed{12.4\ \text{kPa}}$.
5. Newmark cross-check. $N = \Delta\sigma/(0.005\,q) = 12.5/(0.005\times50) \approx 50$ chart blocks at 2 m — consistent with counting elements on the influence chart.

Comment. Directly below the corner of this large footing the stress increase is close to $q/4 \approx 12.5\ \text{kPa}$ and is almost constant over the shallow 2–5 m zone, because the loaded plan dimensions (50 m) are large relative to these depths. The central opening removes almost nothing near the surface but its effect grows with depth, so $\Delta\sigma_z$ decreases slightly from 2 m to 5 m. For a geotechnical engineer these $\Delta\sigma_z$ values are the load input for settlement (multiplied by the compressibility of each sublayer), for locating the depth of significant influence (the “stress bulb”, usually taken where $\Delta\sigma \approx 0.1q$), and for checking consolidation of deeper compressible strata.