Question 4 of 6: Vertical stress increase below a corner by superposition

Question 4: Vertical stress increase below a corner by superposition (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. The mat is a $50\ \text{m} \times 50\ \text{m}$ square uniformly loaded at $q = 50$ kPa, with an unloaded interior court $30\ \text{m}\times 30\ \text{m}$ (spanning 10–40 m in each direction). Point A is at the outer corner of the mat.

Find. The vertical stress increase $\Delta\sigma_z$ on the vertical line through point A at $z = 2$ m and $z = 5$ m, and an interpretation of the two values.

Approach. Use the Boussinesq corner-influence factor $I(m,n)$ for a uniformly loaded rectangle. Point A already sits at the corner of the full $50\times50$ square, so its stress is the full-square value minus the contribution of the unloaded central court, the court being assembled from rectangles that all share corner A.

1. State the corner-stress tool. For a uniformly loaded rectangle with one corner above the point, $\Delta\sigma_z = q\,I(m,n)$ with $m = B/z$, $n = L/z$ (interchangeable), where $I$ is Newmark’s integral of Boussinesq’s solution.
2. Set up the superposition. The loaded area = (full $50\times50$ square, corner at A) − (central court). The court’s corners are $10$ m and $40$ m from A in each direction, so its contribution is the four-rectangle combination $$I_{\text{court}} = I(40,40) - 2\,I(10,40) + I(10,10)\quad(\text{arguments as }a/z),$$ and $\;\Delta\sigma_z = q\big[\,I_{\text{full}} - I_{\text{court}}\,\big]$.
3. Evaluate at $z = 2$ m. Here $I_{\text{full}} = I(25,25) = 0.2500$ (the corner limit) and the court terms give $I_{\text{court}} = 0.00018$, because the court is 10 m away laterally and contributes almost nothing this shallow: $$\Delta\sigma_z = 50\,(0.2500 - 0.0002) = 12.49\ \text{kPa}.$$
4. Evaluate at $z = 5$ m. Now $I_{\text{full}} = I(10,10) = 0.2498$ and $I_{\text{court}} = 0.00248$: $$\boxed{\;\Delta\sigma_z(2\,\text{m}) = 12.49\ \text{kPa},\qquad \Delta\sigma_z(5\,\text{m}) = 12.37\ \text{kPa}\;}$$
5. Comment. Both values are close to $q/4 = 12.5$ kPa and barely change with depth. Two facts explain this: (i) point A lies at the corner of a very large loaded footprint, so near the surface the stress tends to the one-quadrant value $q/4$; and (ii) the unloaded court is 10 m away laterally, so it removes only a fraction of a kPa even at 5 m. Vertical stress from a surface load decays with depth and with horizontal offset; here the point is at the loaded edge, so depth dominates only very slowly over this shallow range.

The information matters to the geotechnical engineer because $\Delta\sigma_z$ at depth is the driver of consolidation settlement ($s_c = \dfrac{C_c H}{1+e_0}\log\dfrac{\sigma'_0 + \Delta\sigma_z}{\sigma'_0}$) and of bearing / factor-of-safety checks. Knowing that the added stress stays near $q/4$ to several metres below the corner tells the engineer where the compressible strata will feel the load and confirms that a distant interior opening is immaterial to the settlement at this corner.