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

Question 4: Vertical stress increase beneath a footing with a central opening (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. Loaded area (shaded) $=$ a 6 m $\times$ 6 m square with a concentric 3 m $\times$ 3 m unloaded opening (1.5 m margin on every side); uniform contact pressure $q = 80\text{ kPa}$. Point A is at the outer top-right corner of the loaded square. Depth of interest $z = 1.5\text{ m}$ (part i); $z = 3.0\text{ m}$ (part ii).

Find. $\Delta\sigma_z$ below A at 1.5 m by Newmark’s chart and by an independent method, with a comparison; then a qualitative discussion of the trend at 3.0 m.

Approach. Treat the shaded load as the solid 6×6 square minus the 3×3 opening, and superpose corner-rectangle influence factors with A at the corner (this is the “other suitable method”); then reproduce the same value by counting influence blocks on Newmark’s chart.

1. Corner influence factor for the solid square. The rectangular corner factor is $I(m,n)$ with $m=B/z,\ n=L/z$. For the full 6×6 square from corner A at $z=1.5\text{ m}$, $m=n=6/1.5=4$: $$I_{\text{solid}} = I(4,4) = 0.2473.$$
2. Remove the opening by superposition about A. The opening spans $1.5\text{ m}$ to $4.5\text{ m}$ from A in both directions, so its corner-rectangle combination is $$I_{\text{hole}} = I(3,3) - 2\,I(3,1) + I(1,1) = 0.01235,$$ using $m,n = 4.5/1.5=3$ and $1.5/1.5=1$.
3. Net influence factor and stress (independent method). $$\Delta\sigma_z = q\,\big(I_{\text{solid}} - I_{\text{hole}}\big) = 80\,(0.2473-0.0123)=80(0.2349)=\boxed{18.8\text{ kPa}}.$$
4. Newmark’s chart. On the provided influence chart, $\Delta\sigma_z = 0.005\,N\,q$. Drawing the loaded plan to the depth scale ($AB = z = 1.5\text{ m}$) with A at the centre and counting the influence squares covered by the shaded (solid-minus-opening) area gives $$N \approx \frac{I_{\text{solid}}-I_{\text{hole}}}{0.005}=\frac{0.2349}{0.005}\approx 47\ \text{blocks},$$ $$\Delta\sigma_z = 0.005\,(47)\,(80) = 18.8\text{ kPa}.$$
5. Comment. The two methods agree to within the block-counting resolution of the chart ($\pm$ a few blocks, i.e. $\pm$5%). Newmark’s chart is general — it handles the irregular, holed shape directly by counting — whereas the influence-factor superposition is exact for the rectangular pieces but requires the area to be decomposed into rectangles sharing corner A. Their agreement confirms the reading.

(ii) At a depth of 3.0 m the vertical stress increase will decrease. Point A lies at the corner of the loaded area, so the stress it feels comes only from load lying to one side of it. The rectangular influence factor depends on $m=B/z$ and $n=L/z$: doubling the depth from 1.5 m to 3.0 m halves both $m$ and $n$, which lowers $I(m,n)$ and hence $\Delta\sigma_z$. Physically, the load is finite and offset from A, so as the point moves deeper the stress bulb spreads and dilutes (Boussinesq dissipation) and less of the applied pressure reaches directly below A. The central opening reduces the stress at both depths, but the dominant effect — loss of intensity with depth beneath a corner — means $\Delta\sigma_z$ at 3.0 m is smaller than at 1.5 m.