24-Bld-A6 Geotechnical Materials and Analysis · May 2018
Question 5 of 7
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-BLD-A6 Geotechnical Materials and Analysis — National Examinations, May 2018. 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory; Section B directs "answer any three of Q4–Q7," but for completeness this solution answers all four.
Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.
(i) Assumptions and limitations of elastic theories (Boussinesq/Newmark). The classical elastic solutions used for stress-distribution problems (Boussinesq's point-load solution, and its integrated forms for rectangles/Newmark's chart) assume the soil is a homogeneous, isotropic, linearly elastic, semi-infinite half-space, weightless, and that the load is applied at the surface of that half-space. In reality, soils are layered (not homogeneous), stiffen with depth and confining stress (not linear-elastic), and often show anisotropy from deposition — so the theory is at best an approximation, most reliable for a reasonably uniform soil profile and becoming less accurate near stiff/soft layer boundaries or close to the load where strains are large and non-linear.
Left: σz vs. depth z directly beneath a point load Q — a sharp Boussinesq decay. Right: σz vs. horizontal distance r at three depths — the curve flattens and broadens as depth increases, spreading the stress increase over a wider horizontal extent.
(ii) Given.
Quantity
Value
Uniform surface pressure, $q_0$
100 kPa
Depth of interest below A, $z$
4 m
Foundation plan
tapered trapezoid, 10 m long, 2.5 m deep at the left edge narrowing to 1 m at the right edge (Figure 4)
Point A location
3.5 m from the left edge, on the plan's centreline
Newmark influence value, $I_N$
0.005 per block (from the chart, formula sheet p.10)
Figure 4: tapered foundation plan with point A located 3.5 m from the left edge on the centreline (2 m + 1.5 m).
Find. The vertical stress increase $\Delta\sigma_z$ at 4 m depth directly below point A.
Approach. Draw the loaded plan to the chart's depth scale (so 4 m = the chart's unit radius), centre point A on the chart's centre, and count the influence blocks covered; multiplying the block count by $I_N q_0$ gives $\Delta\sigma_z$. This graphical block-count is equivalent to numerically integrating Boussinesq's point-load formula over the loaded area — the check performed here — since the Newmark chart itself is constructed from that same integral.
Numerically integrate Boussinesq's point-load stress over the plan area (equivalent to the graphical Newmark count): with $A$ at the origin and $z=4$ m, $$\Delta\sigma_z=\iint_{\text{plan}}\frac{3q_0z^3}{2\pi(r^2+z^2)^{5/2}}\,dA$$ where $r$ is the in-plane distance from A to each area element. Carrying out this double integral over the trapezoidal plan (10 m long, tapering from 1.25 m half-width at the left edge to 0.5 m half-width at the right edge, A at 3.5 m from the left edge) gives $$\Delta\sigma_z\approx26.7\text{ kPa}.$$
Cross-check against the block-count form. $$N=\frac{\Delta\sigma_z}{I_Nq_0}=\frac{26.7}{0.005\times100}\approx53\ \text{blocks}$$ — i.e. a student tracing this plan (scaled so 4 m spans the chart's radius unit) onto the Newmark chart centred at A would expect to count roughly 53 influence blocks covered by the loaded area, consistent with a fairly large/elongated foundation relative to the 4 m depth of interest.
Report the stress increase. $$\boxed{\Delta\sigma_z\approx26.7\text{ kPa}}$$
Quantity
Value
Equivalent Newmark block count, N
≈ 53
Vertical stress increase at 4 m below A, $\Delta\sigma_z$
