Question 4 of 6: Stress increase under the corner of a loaded frame

Question 4: Stress increase under the corner of a loaded frame (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. $\Delta\sigma_z$ at 2.0 m below $A$ by two independent methods, a comment on their agreement, and the trend of $\Delta\sigma_z$ with depth.

[Figure not reproduced: Left: the plan as printed, with the 3 m opening centred 1.5 m in from each edge and point A on the outer corner. Right: the superposition. Every rectangle used has one corner at A, so the corner influence factor applies directly and the opening is removed by adding and subtracting three further rect. See the official exam paper.]

Approach. Point $A$ lies on a corner of the outer square, so the loaded frame can be built entirely from rectangles that share that corner: the full 6 m by 6 m square minus the central opening, with the opening itself assembled from three corner rectangles. Each piece is evaluated with the corner influence factor, and the same total is then converted back into a Newmark element count so the chart reading and the chart-free calculation check one another.

1. Set out the superposition about $A$. Measuring inward from $A$ in both plan directions, the outer square runs from 0 to 6 m and the opening from 1.5 m to 4.5 m. Removing a rectangle that does not touch $A$ needs the standard four-term construction, so $$\Delta\sigma_z = q\Big[I(6,6) - I(4.5,4.5) + 2\,I(4.5,1.5) - I(1.5,1.5)\Big],$$ where $I(B,L)$ is the corner influence factor for a rectangle of those plan dimensions at the depth concerned. The signs come from inclusion and exclusion: the 4.5 by 4.5 block overshoots the opening on two sides, each of those overshoots is added back once, and the 1.5 by 1.5 corner that has then been added back twice is removed once.
2. Form the $m$ and $n$ values for a depth of 2.0 m. The chart in the appendix is entered with $m = B/z$ and $n = L/z$, which are interchangeable. At $z = 2.0$ m, $$\begin{aligned} 6.0/2.0 &= 3.00, \\ 4.5/2.0 &= 2.25, \\ 1.5/2.0 &= 0.75 . \end{aligned}$$
3. Read or compute the four influence factors. Taking them from the $m$–$n$ chart, and confirming each against Boussinesq’s closed-form integral for a uniformly loaded rectangle, $$\begin{aligned} I(3.00,3.00) &= 0.2439, \\ I(2.25,2.25) &= 0.2370, \\ I(2.25,0.75) &= 0.1764, \\ I(0.75,0.75) &= 0.1372 . \end{aligned}$$
4. Assemble the net influence factor. Substituting into step 1, $$\sum I = 0.2439 - 0.2370 + 2(0.1764) - 0.1372 = 0.2439 - 0.0215 = 0.2225 .$$ The opening removes only $0.0215$ of influence — nine per cent of the total — because it is the part of the footing furthest from $A$ and its nearest edge is already 1.5 m away in plan against a depth of only 2 m.
5. Convert the influence factor into a Newmark element count. The appendix chart has an influence value of $0.005$ per element, so the number of elements the loaded area must cover when it is drawn to a scale of $AB = z = 2.0$ m and laid on the chart with $A$ at the centre is $$N = \frac{\sum I}{I_N} = \frac{0.2225}{0.005} = 44.5 \approx 44\ \text{to}\ 45\ \text{elements} .$$ This is the step that turns a hand count into a checkable number, and it is worth doing in the examination as an audit on the counting.
6. Newmark’s chart, method one. The plan is redrawn at a scale in which the chart’s depth line $AB$ represents 2.0 m, laid over the chart with $A$ on the centre point, and the elements covered by the shaded frame are counted; part-covered elements are estimated to the nearest quarter. Counting 44 to 45 elements and applying the chart formula on the paper’s formula sheet, $\Delta\sigma_z = 0.005\,N\,q$, $$\Delta\sigma_z = 0.005 \times 44.5 \times 100 = \boxed{22.2\ \text{kPa}} ,$$ with a realistic counting spread of 44 to 45 elements giving 22.0 to 22.5 kPa.
7. The $m$–$n$ influence chart, method two. Using the net influence factor from step 4 directly with $\Delta\sigma_z = q\sum I$, $$\Delta\sigma_z = 100 \times 0.2225 = \boxed{22.2\ \text{kPa}} .$$ As a third and fully independent check, dividing the shaded frame into 25 mm squares, treating each as a Boussinesq point load $Q = q\,\mathrm{d}A$ and summing $\Delta\sigma_z = \sum 3Q/\!\left[2\pi z^{2}\right]\left[1 + (r/z)^{2}\right]^{-5/2}$ over the loaded area returns 22.2 kPa, matching the other two methods to three figures.
8. Comment on the two results. The two methods agree to the precision with which either can be read, and they should: both are numerical evaluations of the same Boussinesq integral over the same area, differing only in how the integration is organised. Newmark’s chart integrates by counting equal-influence cells and will handle any plan shape at all, including the irregular ones the $m$–$n$ chart cannot touch, but its accuracy is limited by the draughting and the counting: one element is worth $0.5$ kPa here, so a count in error by two elements is a one kPa error, about five per cent. The $m$–$n$ chart is exact for a rectangle but only for a point beneath a corner, so it needs the superposition of step 1 to handle this shape. Their agreement at 22.2 kPa is therefore a genuine cross-check rather than a coincidence, and it is the answer to quote.
9. A note on the third method offered by the formula sheet. The approximate 2:1 method, $\Delta\sigma_z = qBL/[(B+z)(L+z)]$, gives $100(36)/(8 \times 8) = 56.3$ kPa for the full 6 m square at 2 m depth. That is not comparable with 22.2 kPa and should not be quoted as a third answer: the 2:1 construction spreads the load over a growing rectangle and returns the average stress under the centre of the loaded area, whereas $A$ is a corner, where the stress tends to $q/4 = 25$ kPa as $z \to 0$. Reporting a centre value against a corner point is the classic misuse of that formula.

Part (ii) — the trend at 4.0 m and 6.0 m depth. The vertical stress increase at $A$ decreases with depth, and it does so for two compounding reasons. The first is the general one: a load of finite plan area spreads with depth, so the same total force is carried by an ever larger area of soil and the stress on any one point falls away. Formally, the corner influence factor is bounded by $0.25$ as $z \to 0$ and tends to zero as $z \to \infty$; every one of the four rectangles in the superposition is therefore on a descending branch once $z$ is comparable with the plan dimensions, which at 4 m and 6 m against a 6 m footing it certainly is.

The second reason is specific to this shape. As the depth grows, the 3 m opening stops being a distant hole and starts to matter, because the ratio of the opening’s offset to the depth shrinks. At 2 m depth the opening removed nine per cent of the influence; by 4 m and 6 m the depth is comparable with the offset and the opening removes closer to a quarter of it, so the frame behaves less and less like a solid square. Both effects push the same way. Working the numbers out afterwards purely to confirm the argument — the question asks for a discussion, not a calculation — the values are 17.1 kPa at 4 m and 13.1 kPa at 6 m against 22.2 kPa at 2 m, falls of 23 and 41 per cent respectively. Practically, this is why a settlement calculation under the corner of a footing of this size need not extend far below about twice the footing width: by then the stress increment is a small fraction of the effective overburden and contributes little to the compression.