07-Str-A3 · Undated paper

Question 3 of 6: Stress increase beneath the centre of a square footing

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, May 2019 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator. All six questions are compulsory and are weighted 10 / 10 / 15 / 20 / 15 / 30. The paper carries its own appendix: two sheets of blank semi-logarithmic and squared graph paper (pages 7–9) for the plotted answers, the rectangular-loading $m$–$n$ influence chart (page 10), a Newmark influence chart (page 11) and a two-page formula sheet (pages 12–13) on which the Newmark influence value is printed as $\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Every chart value quoted below is taken from those appendix sheets.

Sitting. The printed cover reads NATIONAL EXAMINATIONS – May 2019, 07-Str-A3 Geotechnical Materials and Analysis, 3 hours duration, and every interior page repeats it. The sitting is May 2019.

Reference texts.

B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 2–3 (grain size and phase relations), Ch. 6 (compaction), Ch. 7–8 (permeability and seepage, flow nets), Ch. 9 (stresses in a soil mass, the $m$–$n$ chart and Newmark’s chart), Ch. 11 (consolidation, Casagrande’s construction), Ch. 12 (shear strength and the triaxial test).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 2 (seepage and flow nets), Ch. 3 (effective stress), Ch. 4 (consolidation), Ch. 5 (shear strength, the stress-path or “modified” failure envelope and the choice of test), Ch. 6 (stress distribution).
M. E. Harr, Groundwater and Seepage — Ch. 4 (the conformal solution for a single sheet pile in a stratum of finite depth, used here to audit the drawn flow net).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling), Ch. 8 (settlement), Ch. 11 (laboratory strength testing). The Canadian reference for practice and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 14 (one-dimensional consolidation theory and its assumptions), Art. 16–17 (seepage and piping), Art. 19–20 (shear strength).

Check — two readings of the printed paper, carried as stated.

(1) The printed paper has exactly six questions worth 10 + 10 + 15 + 20 + 15 + 30 = 100 marks, and that numbering is used here.

(2) Question 4 lists the void ratio at $\sigma' = 500$ kPa as $e = 0.925$. Taken with the $400$ kPa point that implies $C_c = 0.57$ over the last increment, against $0.27$–$0.30$ over every earlier virgin increment — the last row is inconsistent with the rest of the test. It is carried exactly as printed, plotted, and excluded from the straight-line fit for $C_c$, with the effect of including it stated where it matters.

Question 3: Stress increase beneath the centre of a square footing (15 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.

Quantity	Symbol	Value
Footing plan dimensions	$B \times L$	4 m × 4 m (square)
Uniform contact pressure	$q$	100 kPa
Depth below the footing base	$z$	5 m
Point of interest	A	on the vertical through the centre O of the footing
Newmark chart influence value (formula sheet)	$I_N$	0.005 (200 elements)

Find. The vertical stress increase $\Delta\sigma_z$ at A, computed by all four of the listed procedures, with the results tabulated and compared.

The 4 m × 4 m footing. Because every chart method gives the stress under a corner, the plan is split at O into four identical 2 m × 2 m rectangles and their contributions are added.

Approach. Superpose four corner rectangles for the chart methods, replace the footing by an equivalent point load for the Boussinesq method, and use the 2 : 1 spread for the trapezoidal method; then compare all four against each other.

Set up the superposition. All of the rectangular-loading solutions are written for a point beneath the corner of the loaded area. Splitting the 4 m square along its two centre-lines produces four 2 m × 2 m rectangles that all share the corner O, so $$\Delta\sigma_z = 4\,q\,I(m,n), \qquad m = \frac{B_1}{z} = \frac{2}{5} = 0.400, \qquad n = \frac{L_1}{z} = \frac{2}{5} = 0.400 .$$ Both $m$ and $n$ are 0.400 because the sub-rectangle is square; the formula sheet notes that $m$ and $n$ are interchangeable, which is why only one of the two ratios has to be looked up on the chart.
Method 1 — the $m$ and $n$ coefficients. Reading the appendix chart at $m = n = 0.4$ gives $I \approx 0.060$. The closed form the chart was drawn from, $$I=\frac{1}{4\pi}\left[\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1+m^{2}n^{2}}\cdot\frac{m^{2}+n^{2}+2}{m^{2}+n^{2}+1}+\tan^{-1}\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1-m^{2}n^{2}}\right],$$ evaluates to $I = 0.06024$ at $m = n = 0.4$, confirming the chart reading. Summing the four quadrants, $$\Delta\sigma_z = 4 \times 100 \times 0.06024 = \boxed{24.1\ \text{kPa}}$$ This is the reference value: it is the exact Boussinesq result for a uniformly loaded flexible rectangle, and the remaining three methods are judged against it.
Method 2 — Newmark’s influence chart. The chart is used by drawing the footing to a scale in which the printed line $AB$ represents the depth $z = 5$ m, placing the point of interest (here the centre O) over the chart’s centre, and counting the elements $N$ covered by the plan. With the influence value printed on the formula sheet, $$\Delta\sigma_z = I_N\,N\,q = 0.005\,N\,q .$$ The count that the drawing produces can be audited rather than guessed: the number of elements a plan covers is just its total influence divided by the influence of one element, $N = \sum I / I_N = 0.2409 / 0.005 = 48.2$, so a careful drawing covers 48 of the chart’s 200 elements — a little under a quarter of the chart, which is what a 4 m square centred over a 5 m depth should give. Hence $$\Delta\sigma_z = 0.005 \times 48 \times 100 = \boxed{24.0\ \text{kPa}}$$ within 0.4 % of Method 1, as it must be, since Newmark’s chart is a graphical integration of the same Boussinesq solution.
Method 3 — the point load method. Replacing the whole footing by a single equivalent point load acting at O, $P = q B L = 100 \times 4 \times 4 = 1600$ kN, and applying Boussinesq’s point-load equation from the formula sheet with $r = 0$, $$\Delta\sigma_z = \frac{3P}{2\pi z^{2}}\cdot\frac{1}{\left[1+(r/z)^{2}\right]^{5/2}} = \frac{3 \times 1600}{2\pi (5)^{2}} = \boxed{30.6\ \text{kPa}}$$ which overestimates the true value by 27 %. The reason is geometric: concentrating the load at the centre moves every kilonewton of it closer to the point directly beneath, and the point-load solution falls off as $r^{-5/2}$-weighted distance, so the concentration cannot be undone by averaging. The cure is to subdivide. Dividing the footing into sixteen 1 m × 1 m squares, each carrying $P_i = 100$ kN at its own centroid, and summing the sixteen point-load contributions gives 24.36 kPa — already within 1.1 % of Method 1, and a 40 × 40 subdivision converges on 24.10 kPa. The single-point form is therefore only admissible when $z$ is large compared with the footing, conventionally $z > 3B$; here $z = 1.25 B$.
Method 4 — the trapezoidal (2 : 1) rule. The approximate method on the formula sheet spreads the load outward at two vertical to one horizontal, so at depth $z$ the same total load is assumed to be uniformly carried on a rectangle $(B+z)$ by $(L+z)$: $$\Delta\sigma_z = \frac{qBL}{(B+z)(L+z)} = \frac{100 \times 4 \times 4}{(4+5)(4+5)} = \frac{1600}{81} = \boxed{19.8\ \text{kPa}}$$ 18 % below the true centre value. That sign is systematic and worth remembering: the 2 : 1 rule spreads the load as a uniform block, whereas the real distribution is bell-shaped with its peak on the centre-line, so the rule always understates the centre-line stress and overstates the stress near the edges of the spread rectangle.

Compare. Ranking the four against the exact chart solution:

Procedure	$\Delta\sigma_z$ (kPa)	Difference from Method 1	Comment
$m$–$n$ influence coefficients (4 corners)	24.1	—	exact Boussinesq solution for the rectangle
Newmark’s chart, $N = 48$	24.0	−0.4 %	graphical form of the same integral
Point load, single equivalent $P$	30.6	+27 %	unsafe concentration; needs $z > 3B$
Point load, 16 sub-areas	24.4	+1.1 %	subdivision recovers the true value
Trapezoidal (2 : 1) rule	19.8	−18 %	uniform-block idealisation, always low on the centre-line

The value to carry forward into a settlement calculation is $\Delta\sigma_z \approx 24$ kPa.

Quantity	Value
Influence factor for one 2 m × 2 m quadrant, $I(0.4, 0.4)$	0.0602
$\Delta\sigma_z$ — $m$ and $n$ coefficients	24.1 kPa
$\Delta\sigma_z$ — Newmark’s chart ($N = 48$ of 200 elements)	24.0 kPa
$\Delta\sigma_z$ — point load method (single equivalent load)	30.6 kPa
$\Delta\sigma_z$ — point load method (16 sub-areas)	24.4 kPa
$\Delta\sigma_z$ — trapezoidal (2 : 1) rule	19.8 kPa
Recommended design value at A	24 kPa

Check — reading of the stem. The question asks for “the average increase in stress at the centre of the area and at a depth of 5 m”. It is answered here as the stress increase at the point 5 m below the centre, which is the only reading consistent with the four named procedures (all four return a point value). If instead the average over the 0–5 m depth were wanted, Simpson’s rule $\Delta\sigma_{av} = \tfrac{1}{6}(\Delta\sigma_t + 4\Delta\sigma_m + \Delta\sigma_b)$ applied to the centre-line values at $z = 0$, 2.5 and 5 m gives $\tfrac{1}{6}(100 + 4 \times 58.4 + 24.1) = 59.6$ kPa.

Topic. Elastic stress distribution beneath a uniformly loaded flexible area — corner superposition, Newmark’s graphical integration, the point-load idealisation and the 2 : 1 spread.

Key relations. Boussinesq point load $\Delta\sigma_z = \dfrac{3P}{2\pi z^{2}}\left[1+(r/z)^{2}\right]^{-5/2}$; rectangle corner $\Delta\sigma_z = q\,I(m,n)$ with $m = B/z$, $n = L/z$; Newmark $\Delta\sigma_z = 0.005\,N q$; 2 : 1 spread $\Delta\sigma_z = qBL / [(B+z)(L+z)]$.

Why this works. All four procedures integrate the same Boussinesq kernel over the loaded area; they differ only in how the integration is done and therefore in what they throw away. The $m$–$n$ chart performs the integral analytically for a rectangle and is exact. Newmark’s chart is the same integral drawn as 200 equal-influence tiles, so its accuracy is limited only by the care of the counting. The point-load method truncates the integral to a single sample of the load, which is only acceptable once the depth is large compared with the loaded dimension. The 2 : 1 rule abandons elasticity altogether and simply conserves the total load across a spreading rectangle — useful for a quick check, but never for a centre-line stress that a settlement estimate depends on.

Common pitfall. Using $m = B/z = 4/5$ and $n = L/z = 4/5$ with the full footing dimensions and then reading a single influence factor. That combination gives the stress under a corner of the 4 m square, not under its centre — the plan must be quartered first. The second recurring error is quoting the 2 : 1 value as “another suitable method” without saying that it is 18 % low on the centre-line; the comparison, not the number, is what this question is testing.

Source. Das & Sobhan, Principles of Geotechnical Engineering, 9th ed., §9.2 (point load), §9.6 (rectangular area, the $m$–$n$ chart), §9.9 (Newmark’s chart), §9.11 (approximate 2 : 1 method); Craig / Knappett, 8th ed., §6.2–6.3.

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