Question 4 of 6: Stress increase below the corner of a courtyard raft
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: PEO / Engineers Canada National Examinations, December 2014, 07-Str-A3 Geotechnical Materials and Analysis. Three hours, closed book, drawing instruments required, one approved Casio or Sharp calculator. All charts and equations are supplied at the back of the paper (the m–n influence chart on page 8, a Newmark chart with $I_N = 0.005$ on page 9, and a two-page formula sheet). Total value 100 marks over six compulsory questions (20 + 10 + 10 + 20 + 20 + 20). Every question and every sub-part is answered in full below.
Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. (Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (Ch. 2 seepage and flow-net construction, Ch. 5 shear strength, Ch. 11 lateral earth pressure); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (compacted-fill fabric and Proctor behaviour by USCS group); Harr, Groundwater and Seepage, Ch. 4 (closed-form conformal solutions for a single cut-off); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for earth pressures, seepage control and settlement; Canadian Dam Association, Dam Safety Guidelines (2013, rev. 2019); ASTM D698 / D1557 (Proctor compaction), ASTM D4767 (consolidated-undrained triaxial with pore-pressure measurement), ASTM D7181 (consolidated-drained triaxial).
Check — two printing slips in the source, carried exactly as printed. (1) The Question 1 header reads “(4 × 5 = 20 marks)” but five statements (i)–(v) are printed; this solution treats the question as 5 × 4 = 20 marks and answers all five parts. (2) Question 5(a) instructs the candidate to “draw on Figure 4”, while the only section supplied is labelled Figure 3 — the blank grid sheets on pages 6 and 7 are the intended drawing space. Neither slip changes the 100-mark total.
Question 4: Stress increase below the corner of a courtyard raft (20 marks)
Given. The site is a 50 m × 50 m square. The foundation is the shaded picture-frame: the whole square less an unloaded courtyard 30 m × 30 m that is set back 10 m from every edge. The contact pressure over the loaded area is $q = 50$ kPa, applied at the ground surface. Point A is the outer corner of the 50 m square, and the stress is required on the vertical through A at depths $z = 2$ m and $z = 5$ m. The paper supplies the m–n influence chart (page 8) and a Newmark chart with $I_N = 0.005$ over 200 elements (page 9); the soil is treated as a homogeneous, isotropic, linear-elastic half-space, which is the basis of both charts.
Find. $\Delta\sigma_z$ at 2 m and at 5 m below A, obtained by superposition of corner rectangles, and a comment on what the two values mean for a geotechnical engineer.
Figure 4.1 — the loaded plan. The hatched frame is loaded at 50 kPa; the central 30 m × 30 m courtyard carries no load. Point A is the outer corner, so every rectangle needed for superposition already has a corner on the vertical through A.
Approach. The influence chart gives $\Delta\sigma_z$ only beneath the corner of a uniformly loaded rectangle, so the loaded frame is written as one large rectangle with a corner at A, minus the courtyard, and the courtyard is itself written as a signed sum of four rectangles that each have a corner at A.
Set up the superposition about A. Measure $x$ from A along the top edge of the plan and $y$ from A down the right edge. The whole site is then the rectangle $0 \le x \le 50$, $0 \le y \le 50$, and the courtyard occupies $10 \le x \le 40$, $10 \le y \le 40$. Writing $I(a,b)$ for the corner influence factor of an $a \times b$ rectangle at the depth in question,
$$\Delta\sigma_z \;=\; q\Big[\,I(50,50) \;-\; \big(I(40,40) - I(40,10) - I(10,40) + I(10,10)\big)\Big]$$
The bracket is the standard four-rectangle expression for an area that does not touch A: the large rectangle out to the far courtyard corner, less the two strips that overhang it, plus the small rectangle that was subtracted twice.
Evaluate the corner factors at $z = 2$ m. The chart is entered with $m = B/z$ and $n = L/z$, which are interchangeable. Every rectangle here is enormous relative to a 2 m depth, so all five factors sit at the flat right-hand end of the chart, close to the limiting value 0.25.
Corner influence factors $I(m,n)$ at $z = 2$ m
Rectangle
$m$
$n$
$I$
Sign
50 × 50
25.0
25.0
0.24999
+
40 × 40
20.0
20.0
0.24998
−
40 × 10
20.0
5.0
0.24919
+
10 × 40
5.0
20.0
0.24919
+
10 × 10
5.0
5.0
0.24857
−
The courtyard therefore contributes $I_{\text{court}} = 0.24998 - 0.24919 - 0.24919 + 0.24857 = 0.000177$, and the net influence factor is $I_{\text{net}} = 0.24999 - 0.00018 = 0.24981$.
Convert to a stress at 2 m. With $\sigma_z = q\,I$,
$$\Delta\sigma_z\big|_{z=2\,\text{m}} = 50 \times 0.24981 = \boxed{12.49\ \text{kPa}}$$
The courtyard has removed only $50 \times 0.000177 = 0.009$ kPa — nine thousandths of a kilopascal — because its nearest edge is 10 m away in plan while the point of interest is only 2 m down.
Repeat at $z = 5$ m. The same five rectangles are re-entered with $z = 5$ m, which halves and quarters the $m$ and $n$ values and starts to move the smaller rectangles off the flat part of the chart:
Corner influence factors $I(m,n)$ at $z = 5$ m
Rectangle
$m$
$n$
$I$
Sign
50 × 50
10.0
10.0
0.24982
+
40 × 40
8.0
8.0
0.24964
−
40 × 10
8.0
2.0
0.23982
+
10 × 40
2.0
8.0
0.23982
+
10 × 10
2.0
2.0
0.23247
−
Now $I_{\text{court}} = 0.24964 - 0.23982 - 0.23982 + 0.23247 = 0.002477$ and $I_{\text{net}} = 0.24982 - 0.00248 = 0.24734$, so
$$\Delta\sigma_z\big|_{z=5\,\text{m}} = 50 \times 0.24734 = \boxed{12.37\ \text{kPa}}$$
The courtyard now removes 0.124 kPa — fourteen times more than at 2 m, but still only one per cent of the answer.
Cross-check on the Newmark chart. The alternative route offered by the paper is $\sigma_z = 0.005\,N\,q$, where $N$ is the number of influence elements covered by the loaded plan drawn to the depth scale. Inverting that relation turns the hand count into a checkable number:
$$N \;=\; \frac{I_{\text{net}}}{I_N} \;=\; \frac{0.24981}{0.005} = 50.0 \quad\text{and}\quad \frac{0.24734}{0.005} = 49.5$$
so a candidate working the Newmark chart should cover about 50 of the 200 elements at both depths — a quarter of the chart, which is precisely what is expected for a point sitting on the corner of a very large loaded area. Agreement of the two independent methods to better than one per cent confirms the superposition.
Comment. The two answers are almost identical: the stress falls by only 1 % between 2 m and 5 m, and both are within 1 % of $q/4 = 12.5$ kPa. Physically, a point on the corner of a loaded area sees one quadrant of an infinitely wide load, and the Boussinesq limit for that geometry is exactly $q/4$; at depths of a few metres beneath a 50 m plan, the loaded area is effectively infinite and the courtyard is too far away in plan to register.
That observation is the useful engineering content of the question. For a small strip or pad footing the vertical stress has decayed to a tenth of the contact pressure within about two footing widths, so a borehole 5 m deep tells the whole story. Under a 50 m raft nothing of the sort happens: the pressure bulb extends to a depth comparable with the plan dimension, so significant stress increase persists to 50 m or more, and the settlement calculation must integrate compressibility over that whole depth. The practical consequences are that (1) the site investigation must reach the depth at which $\Delta\sigma_z$ has fallen below about 10 % of the in-situ vertical effective stress — here tens of metres, not five — because a soft layer at 30 m depth still matters; (2) any consolidation settlement estimate that stops at the bottom of the boreholes will be unconservative; and (3) the near-constancy of $\Delta\sigma_z$ with depth under a wide load is the reason large rafts on layered ground settle slowly for decades. It is also worth noting for differential-settlement purposes that A is the least-loaded point of the whole plan: an interior point of the frame would see something approaching the full $q$, so the raft must be stiff enough, or the structure tolerant enough, to accommodate a stress ratio of about four across the plan.
