16-Civ-A4 Geotechnical Materials and Analysis · May 2014
Question 4 of 6: Vertical stress increase below a frame footing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — May 2014 · 98-Civ-A4 Geotechnical Materials and Analysis · 3 hours, closed book · 100 marks · answer all six questions. A formula sheet plus m–n influence and Newmark charts are supplied at the back of the paper.
Reference texts. B. M. Das, Principles of Geotechnical Engineering (9th ed.); R. F. Craig / Knappett & Craig, Craig’s Soil Mechanics (8th ed.); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering; M. Budhu, Soil Mechanics and Foundations. Canadian practice: Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed.). Unit weight of water taken as $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
Given. A square 6 m × 6 m footing with a central 3 m × 3 m opening (a 1.5 m wide frame; the shaded area carries the load), uniform contact pressure $q = 100\ \text{kPa}$; point A is at the outer top-right corner; depth of interest $z = 2.0\ \text{m}$.
Figure 4 — plan of the frame footing. The 3 m × 3 m opening is centred, leaving a 1.5 m loaded strip on every side; point A is the outer corner directly above which the stress is required.
Find. The vertical stress increase $\Delta\sigma_z$ at 2.0 m below A, by Newmark’s influence chart and by an independent method, with a comparison.
Approach. Both methods rest on Boussinesq theory. Treat the loaded frame as the outer 6 m × 6 m square (corner at A) minus the central 3 m × 3 m opening, and superpose corner influence factors; the Newmark chart is the graphical form of the same superposition.
Set up the superposition. Point A is a corner of the full outer square, so the outer load contributes $q\,I(m,n)$ directly, with $m = B/z,\ n = L/z$. The opening (which carries no load) is subtracted using the four-rectangle rule about A. Thus $\Delta\sigma_z = q\big[I_{outer} - I_{opening}\big]$.
Outer 6×6 square (corner at A). $m = n = 6/2 = 3$, giving the corner influence factor $I(3,3) = 0.2439$ from the Boussinesq corner integral (Newmark/Fadum chart).
Central opening by four rectangles about A. The centred opening spans 1.5 m to 4.5 m from A in both directions, so $I_{opening} = I(2.25,2.25) - 2\,I(0.75,2.25) + I(0.75,0.75) = 0.2370 - 2(0.1764) + 0.1372 = 0.0215$ (each $I$ evaluated at $m,n = \text{distance}/z$).
Net influence and stress (independent m–n method). $I_{frame} = 0.2439 - 0.0215 = 0.2224$, so $\Delta\sigma_z = q\,I_{frame} = 100(0.2224)$, giving $\boxed{\Delta\sigma_z \approx 22.3\ \text{kPa}}$.
Newmark’s chart. With $\Delta\sigma_z = 0.005\,N\,q$, the equivalent number of influence blocks is $N = \dfrac{\Delta\sigma_z}{0.005\,q} = \dfrac{22.3}{0.005(100)} \approx 45$ blocks. Drawing the frame to the chart’s depth scale ($AB = z = 2\ \text{m}$) with A over the centre and counting the covered blocks (outer square minus the opening) returns $\Delta\sigma_z = 0.005(45)(100) \approx 22\text{–}23\ \text{kPa}$, i.e. the same value.
Vertical stress increase 2.0 m below point A
Method
Influence quantity
$\Delta\sigma_z$
m–n influence factors (superposition)
$I_{frame} = 0.2224$
22.3 kPa
Newmark’s chart
$N \approx 45$ blocks
≈ 22–23 kPa
Comment. The two methods agree to within the accuracy of the graphical block count. Newmark’s chart depends on drawing the footing to scale and counting partial blocks by eye, so it typically carries a $\pm 1$–$2\ \text{kPa}$ reading error, whereas the m–n influence-factor superposition is analytical. Note also that removing the central opening lowers the corner stress by only about $2\ \text{kPa}$ (from 24.4 to 22.3 kPa): the opening sits diagonally far from the corner A, so it contributes little to the stress there.
