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16-Civ-A4 Geotechnical Materials and Analysis · December 2017

Question 3 of 6: Vertical Stress Increase Beneath a Frame Footing

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

Notes on this paper

Paper format: National Examinations — 16-Civ-A4 Geotechnical Materials and Analysis, December 2017. Closed book, 3 hours, 100 marks. Answer all six questions. Charts (rectangular-area influence chart, Newmark chart) and a formula sheet are provided at the back of the exam.

Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations.

Question 3: Vertical Stress Increase Beneath a Frame Footing (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. A square frame footing: a 6 m × 6 m loaded (shaded) area with a concentric 3 m × 3 m unloaded void (1.5 m of loaded soil on every side). Point A is at the outer corner of the 6 m square.

Given data
Uniform contact pressure, q100 kPa
Outer footing plan6 m × 6 m
Central void3 m × 3 m (centred)
Depth below Az = 1.5 m
Point Aouter corner of the 6 m square
A 6 m 6 m 3 m 1.5 m Loaded (shaded) area q = 100 kPa; central 3×3 m void unloaded; A at outer corner.
Figure 3 — plan of the frame footing. The shaded ring (6×6 outer, 3×3 void) carries 100 kPa; A is at the outer corner. Stress is sought at z = 1.5 m directly below A.

Find. The increase in vertical stress Δσz at 1.5 m below A, by Newmark’s chart and by a second method, with a comparison.

Approach. Because A sits at the outer corner, the loaded frame is treated by superposition of corner rectangles: the full 6×6 corner rectangle minus the influence of the central void (itself expressed as four corner rectangles about A). The rectangular-area corner influence factor supplies the “other suitable method,” and Newmark’s chart supplies the graphical count.

  1. Corner influence factor. For a uniform rectangle of sides B, L loaded at its corner, with m = B/z and n = L/z (interchangeable), $$I=\frac{1}{4\pi}\left[\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1+m^{2}n^{2}}\cdot\frac{m^{2}+n^{2}+2}{m^{2}+n^{2}+1}+\tan^{-1}\!\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1-m^{2}n^{2}}\right].$$
  2. Whole 6×6 area at corner A. With m = n = 6/1.5 = 4: $I_\text{outer}=0.2473$.
  3. Subtract the central void. The void spans 1.5–4.5 m from A in both directions, so its influence is the four-rectangle superposition $$I_\text{void}=I(3,3)-I(1,3)-I(3,1)+I(1,1)=0.0124\quad(m,n=\text{distance}/z).$$
  4. Net stress (influence-factor method). $$\Delta\sigma_z=q\,(I_\text{outer}-I_\text{void})=100\,(0.2473-0.0124)=\boxed{23.5\ \text{kPa}}.$$
  5. Newmark’s chart. Draw the footing to the chart’s depth scale (length AB = z = 1.5 m, so the 6 m side is 4AB and the void starts 1AB from A). Put A over the chart centre and count the influence elements covered by the loaded ring, leaving out those under the void. The analytical result predicts N = Δσz/(0.005q) ≈ 47 blocks, and a careful count lands within about ±2 blocks of that. With N ≈ 47 and IN = 0.005, $$\Delta\sigma_z=0.005\,N\,q=0.005\times47\times100=23.5\ \text{kPa},$$ which agrees with the analytical value to within the counting tolerance (±2 blocks is ±1 kPa).
  6. Quick approximate check. The 2:1-type formula $\sigma_z=qBL/[(B+z)(L+z)]$ applied, very roughly, as (outer − void) gives $100\cdot\tfrac{36}{56.25}-100\cdot\tfrac{9}{20.25}=19.6$ kPa — about 17% lower. This is only an order-of-magnitude check: the 2:1 formula gives the average stress over the spread area, not the stress under a particular point such as a corner, so its closeness here is partly coincidence.
Increase in vertical stress at 1.5 m below A
MethodΔσz
Newmark’s chart (N ≈ 47)23.5 kPa
Rectangular-area influence factor23.5 kPa
Approximate 2:1 spread19.6 kPa

Comment. The Newmark chart and the influence-factor method are both rigorous Boussinesq solutions and agree almost exactly (23.5 kPa); any small difference in practice comes only from the human error of counting partial blocks on the chart. The approximate 2:1 method is quicker but differs by roughly 17% here, because it spreads the load uniformly over a widening area and cannot represent the stress at a specific point such as corner A — acceptable for a first estimate, not for design at a specific point.