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16-Civ-A4 Geotechnical Materials and Analysis · Undated paper

Question 3 of 6: Stress Increase Below a Square Footing

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

Notes on this paper

National Examinations — May 2019  |  16-Civ-A4 Geotechnical Materials and Analysis  |  3 hours, open book  |  100 marks  |  Answer ALL questions (Q1–Q6).

Reference texts: Das & Sobhan, Principles of Geotechnical Engineering, 9th ed. (Cengage); Craig, Craig's Soil Mechanics, 8th ed.; Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. Canadian practice frame (CFEM 4th ed., EGBC).

Source-quality note. Flow-net field counts carry the usual ±½-field hand-sketch tolerance (Exam Note 3).

Question 3: Stress Increase Below 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.

Given data
Footing plan4 m × 4 m, flexible, uniform
Contact pressure, $q$100 kPa
Point of interestbeneath the centre, depth $z = 5$ m

Find. The vertical stress increase $\Delta\sigma_z$ at 5 m below the footing centre, by three independent methods, and a comparison.

O (centre)4 m4 m2 mq = 100 kPa(4 quadrants of 2 m x 2 m,each cornered at O)Stress increment sought at depth z = 5 m below O
Footing plan — for the influence-factor method the 4 m square is split into four 2 m × 2 m rectangles that share the corner O directly above the point of interest.

Approach. Superpose Boussinesq solutions for a uniformly loaded area: the exact m–n corner-influence factor (four quadrants), Newmark's influence chart, and the approximate 2:1 spread — then cross-check against the crude point-load idealisation.

  1. Method 1 — m–n influence factors (four quadrants). Split the footing into four 2 m×2 m rectangles with a common corner at O. Each has $$m=\frac{B}{z}=\frac{2}{5}=0.4,\qquad n=\frac{L}{z}=\frac{2}{5}=0.4.$$ The Boussinesq corner factor is $$I=\frac{1}{4\pi}\!\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot\frac{m^2+n^2+2}{m^2+n^2+1}+\tan^{-1}\!\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right]=0.0602.$$ Superposing the four identical quadrants, $$\Delta\sigma_z = 4\,I\,q = 4(0.0602)(100)=\boxed{24.1\ \text{kPa}}.$$
  2. Method 2 — Newmark's influence chart. Placing the scaled footing plan on the chart with the centre over the plot origin, the loaded area covers about $N\approx48$ influence units. With the chart's influence value $I_N=0.005$, $$\Delta\sigma_z=q\,(0.005\,N)=100(0.005)(48)\approx 24\ \text{kPa},$$ in essentially exact agreement with the influence-factor result (the two methods share the same Boussinesq basis).
  3. Method 3 — 2:1 (trapezoidal) spread. The load is assumed to spread on 2 vertical : 1 horizontal planes, so at depth $z$ it acts over $(B+z)(L+z)$: $$\Delta\sigma_z=\frac{qBL}{(B+z)(L+z)}=\frac{100(4)(4)}{(4+5)(4+5)}=\frac{1600}{81}=\boxed{19.8\ \text{kPa}}.$$
  4. Cross-check — point-load (Boussinesq) idealisation. Lumping the whole load $Q=qBL=1600$ kN at the centre and evaluating directly below it ($r=0$), $$\Delta\sigma_z=\frac{3Q}{2\pi z^2}=\frac{3(1600)}{2\pi(5)^2}=30.6\ \text{kPa}.$$ This overestimates, because concentrating a 4 m footing at a point exaggerates the near-axis stress at a depth only slightly greater than the footing width.
Q3 — comparison of methods ($\Delta\sigma_z$ at centre, $z=5$ m)
Method$\Delta\sigma_z$ (kPa)Comment
m–n influence factors (exact)24.1reference value
Newmark's chart≈24same Boussinesq basis
2:1 trapezoidal spread19.8underestimates (empirical)
point-load idealisation30.6overestimates (footing not a point)

The two rigorous Boussinesq methods agree at $\Delta\sigma_z\approx24$ kPa. The 2:1 rule is about 18% low and the point-load idealisation about 27% high — both expected at $z/B=1.25$, where the footing is neither a point nor deep enough for the 2:1 spread to be accurate.