Question 4 of 6: Increase in vertical stress below the corner of a condominium raft

Question 4: Increase in vertical stress below the corner of a condominium raft (Value: 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.

Find. $\Delta\sigma_z$ vertically below $A$ at $z = 2$ m and $z = 5$ m, by superposition, with a comment on the two values.

[Figure not reproduced: Left: Figure 3 as printed — a 10 m wide loaded band around a 50 m square site with an unloaded 30 m courtyard, and point A on the outer corner. Right: the two rectangle groups the superposition uses, each with a corner on the vertical through A. See the official exam paper.]

Approach. Load the whole 50 m square, then subtract the courtyard that is not loaded, taking every rectangle with one corner on the vertical through $A$ so that the m–n influence chart on the paper’s appendix applies directly.

1. Establish the contact pressure. The five storeys deliver $$q = n \times 10\ \text{kPa} = 5 \times 10 = 50\ \text{kPa}$$ uniformly over the hatched band.
2. Set up the superposition. Point $A$ sits at the outer corner of the site, so the whole 50 m × 50 m square already has a corner on the vertical through $A$ and needs no subdivision. The courtyard is a 30 m square whose near edges are 10 m from the two sides that meet at $A$; it does not have a corner at $A$, so it is built from four rectangles that do: $$\textstyle\sum I = I(50,50) - \big[\,I(40,40) - 2\,I(40,10) + I(10,10)\,\big],$$ where the arguments are the two side lengths measured from $A$ and the signs are the standard add–subtract–subtract–add pattern. The two $I(40,10)$ terms are equal by symmetry, which is why they combine into a single doubled term.
3. Convert the side lengths to chart arguments. The chart is entered with $$m = \frac{B}{z}, \qquad n = \frac{L}{z},$$ both measured from $A$ and interchangeable. At $z = 2$ m the four rectangles give $m,n$ = (25, 25), (20, 20), (20, 5) and (5, 5); at $z = 5$ m they give (10, 10), (8, 8), (8, 2) and (2, 2).
4. Read the influence factors at $z = 2$ m. Using the closed form that the chart plots, $$I = \frac{1}{4\pi}\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot \frac{m^2+n^2+2}{m^2+n^2+1} + \arctan\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right],$$ the full square gives $I(25,25) = 0.249988$ and the courtyard group gives $I(20,20) - 2I(20,5) + I(5,5) = 0.249977 - 2(0.249187) + 0.248574 = 0.000177$. The net is $$\textstyle\sum I = 0.249988 - 0.000177 = 0.249811 .$$
5. Vertical stress at 2 m. $$\Delta\sigma_z = q\textstyle\sum I = (50)(0.249811),$$ $$\boxed{\Delta\sigma_z\big|_{z=2\,\text{m}} = 12.49\ \text{kPa}}$$
6. Repeat at $z = 5$ m. The full square now gives $I(10,10) = 0.249815$ and the courtyard group $I(8,8) - 2I(8,2) + I(2,2) = 0.249641 - 2(0.239815) + 0.232466 = 0.002477$, so $\sum I = 0.247338$ and $$\Delta\sigma_z = (50)(0.247338),$$ $$\boxed{\Delta\sigma_z\big|_{z=5\,\text{m}} = 12.37\ \text{kPa}}$$
7. Audit the answer against the Newmark chart. The formula sheet prints $\sigma_z = 0.005\,N q$ for the Newmark chart supplied, so the number of elements the chart should cover is not a guess but a calculation: $$N = \frac{\sum I}{I_N} = \frac{0.249811}{0.005} = 50.0 \quad\text{at } z = 2\ \text{m}, \qquad N = \frac{0.247338}{0.005} = 49.5 \quad\text{at } z = 5\ \text{m}.$$ Both are a quarter of the chart’s 200 elements, which is exactly what a point on an outer corner of a large loaded area must give. If a hand count of the chart does not land near 50, the scale or the placement of the plan is wrong.
8. Confirm by a wholly independent method. Replacing the loaded band by a 0.25 m grid of point loads and summing Boussinesq’s $\Delta\sigma_z = 3Q z^3 / (2\pi(r^2+z^2)^{5/2})$ over it returns 12.4906 kPa at 2 m and 12.3669 kPa at 5 m — agreement to four figures with the chart result, so neither the superposition signs nor the arithmetic is in error.

Comment on the two values, and on what a geotechnical engineer does with them

The two answers are almost the same — they differ by 1 % over a depth change of a factor of two and a half — and that is the interesting result, not an error. Two effects explain it, and both are worth stating.

First, $A$ lies on the outer corner of a loaded area that is very large compared with the depths asked about. The influence factor for a corner of an infinitely large loaded area is exactly $\tfrac14$, so $\Delta\sigma_z \to q/4 = 12.5$ kPa; at 2 m and 5 m below a 50 m square the loaded area already behaves as if it were infinite, and both answers sit within 1 % of that limit. A point at a corner receives load from one quadrant only, which is why the answer is a quarter of $q$ and not $q$ itself.

Second, the courtyard removes almost nothing at these depths. Its nearest edge is 10 m away in plan while $z$ is 2 m or 5 m, so it subtends a very small solid angle at the point in question: it takes away 0.07 % of the stress at 2 m and 1.0 % at 5 m. The visually dominant feature of the plan is therefore almost irrelevant to this particular point — its influence grows as the depth becomes comparable with its offset, and it would take roughly a quarter of the stress away by $z \approx 30$ m.

What the engineer does with this is threefold. It fixes the depth of significant influence: because the stress is still a quarter of the applied pressure at 5 m, the compressible layers well below the founding level all contribute to settlement, and the borehole depths and the layers included in the consolidation calculation must reflect that — the common rule of investigating to where $\Delta\sigma_z$ falls below 10 % of $q$ or below 10 % of the in-situ $\sigma_v'$ would take the investigation far deeper than 5 m here. It gives the $\Delta\sigma'$ that goes into the mid-layer settlement computation of the kind set out in Question 3. And it warns about differential settlement: a point beneath the centre of the same raft would carry close to the full 50 kPa at these depths, four times the corner value, so the raft must be designed for the curvature that difference produces.