Question 4 of 6: Vertical stress increase beneath a footing with an opening

Question 4: Vertical stress increase beneath a footing with an opening (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. The vertical stress increase $\Delta\sigma_z$ on the vertical through A at $z = 1.5\ \text{m}$, computed twice by independent methods and compared, and the direction in which that stress changes when the depth is doubled to 3.0 m.

[Figure not reproduced: Figure 3 (redrawn to scale). The loaded area is the hatched square annulus; the central 3 m by 3 m opening carries no load. Point A is the outer corner. See the official exam paper.]

Approach. The loaded shape is not a rectangle, so it is built by superposition: take the full 6 m by 6 m square loaded at 80 kPa and subtract the central 3 m by 3 m patch, every rectangle being referred to a corner at A so that the standard corner influence factor applies. The same net influence factor is then re-expressed as a Newmark element count, which gives the two independent routes the question asks for.

1. Set out the superposition about point A. Place the origin at A with axes along the two sides of the square. Because A is the outer corner, the full square already has a corner at A, so its influence factor is read directly. The opening does not, so it is constructed from four rectangles that all have a corner at A: the large 4.5 m by 4.5 m rectangle reaching to the far side of the opening, less the two 4.5 m by 1.5 m rectangles that overshoot it, plus the 1.5 m by 1.5 m rectangle subtracted twice. In symbols $$\Delta\sigma_z = q\left[I_{6\times 6} - \left(I_{4.5\times 4.5} - 2 I_{4.5\times 1.5} + I_{1.5\times 1.5}\right)\right].$$
2. Convert each rectangle to its $m$ and $n$ values at $z = 1.5\ \text{m}$. With $m = B/z$ and $n = L/z$ as defined on the formula sheet, and the two interchangeable, the four rectangles become $6/1.5 = 4$, $4.5/1.5 = 3$ and $1.5/1.5 = 1$. The influence factors read from the appendix m–n chart, and checked against the closed-form Boussinesq expression for a uniformly loaded rectangle, are $I_{(4,4)} = 0.2473$, $I_{(3,3)} = 0.2439$, $I_{(3,1)} = 0.2034$ and $I_{(1,1)} = 0.1752$.
3. Assemble the influence factor of the opening. Substituting into the four-rectangle construction, $$I_{\text{opening}} = 0.2439 - 2(0.2034) + 0.1752 = 0.0123 .$$ The opening contributes almost nothing. That is expected rather than suspicious: its nearest corner is 1.5 m away from A in both plan directions, which at $z = 1.5\ \text{m}$ is already a full depth away, and the stress bulb beneath a corner point falls off quickly with horizontal offset.
4. Combine to the net influence factor and the stress. Subtracting the opening from the full square, $$I_{\text{net}} = 0.2473 - 0.0123 = 0.2349 ,$$ so that $$\Delta\sigma_z = q\,I_{\text{net}} = 80 \times 0.2349 = \boxed{18.8\ \text{kPa}} .$$ Notice how close this is to $q/4 = 20\ \text{kPa}$. A point on the outer corner of a large loaded area sees only one quadrant of the load, and the corner influence factor tends to the limiting value 0.25 as $m$ and $n$ grow; recognising that limit is the quickest check that the superposition has been set up the right way round.
5. Now do the same calculation with Newmark’s chart. The chart is used by drawing the plan of the loaded area to the scale set by the depth — here the chart line marked depth scale represents $z = 1.5\ \text{m}$ — laying the plan on the chart with point A over the centre, and counting the elements the loaded area covers. With $I_N = 0.005$ the stress is $\Delta\sigma_z = I_N N q$. Counting the hatched annulus gives $N = 47$ elements: about 49.5 elements are covered by the full square, which is a quarter of the chart’s 200 elements as a corner point demands, and about 2.5 elements are removed by the opening. Hence $$\Delta\sigma_z = I_N N q = 0.005 \times 47 \times 80 = \boxed{18.8\ \text{kPa}} .$$ The element counts here are audited rather than guessed: dividing each influence factor from step 4 by $I_N$ back-figures $0.2473/0.005 = 49.5$ elements for the square and $0.0123/0.005 = 2.5$ for the opening, which is what a careful count on the chart should reproduce.
6. Comment on the two results. The two methods agree to better than one per cent, 18.80 kPa against 18.8 kPa, and the agreement is not accidental: Newmark’s chart is nothing more than a graphical integration of the same Boussinesq point-load solution that underlies the m–n influence chart, so the two must converge on the same answer whenever the counting is done accurately. What differs is the error each carries. The chart method is limited by the resolution of hand counting, and two miscounted elements move the answer by $0.005 \times 2 \times 80 = 0.8\ \text{kPa}$, roughly four per cent; against that it handles any plan shape at all, which is exactly why it is worth using on an annulus. The influence-factor method is exact for the shape it can represent but is restricted to rectangles and their superpositions. On this plan both are applicable, so the chart is best used as the independent check and the analytical result quoted as the answer. It is worth adding that the approximate 2:1 method also printed on the formula sheet, $\Delta\sigma_z = qBL/[(B+z)(L+z)]$, gives 51.2 kPa here — nearly three times the correct value — because it is a rule for the point beneath the centre of the loaded area and says nothing about a corner. Choosing the wrong tool costs far more than counting the chart badly.
7. Part (ii): the trend with depth at 3.0 m. The vertical stress increase decreases. Three arguments converge on this, none needing a calculation. First, Boussinesq’s point-load solution has $\Delta\sigma_z \propto Q/z^{2}$, so for any finite load the stress must fall away with depth once the depth is comparable to the plan dimensions. Second, in influence-factor terms the parameters $m = B/z$ and $n = L/z$ both halve when the depth doubles, and $I$ increases monotonically with $m$ and $n$, so the influence factor of the governing 6 m by 6 m square must fall. Third, physically, the fixed total load $q \times \text{area}$ spreads over an ever-widening horizontal plane as it is carried deeper, so the average pressure on that plane, and with it the pressure on the vertical through A, diminishes. The opening reinforces the trend rather than opposing it: at greater depth the opening subtends a larger angle at A, so the load it removes becomes relatively more important and takes a bigger bite out of the total. Taken together the stress at $z = 3.0\ \text{m}$ is appreciably smaller than at 1.5 m — a calculation, which the question does not ask for, puts it near 15.6 kPa, a drop of about 17 per cent.