24-Bld-A6 Geotechnical Materials and Analysis · December 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, December 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.
Figure 4 (reproduced): plan of the 4 m × 2 m flexible loaded area. A is the bottom-right corner; B is interior (1.0 m from the right edge, 0.5 m from the bottom edge); C is 1.5 m beyond A, on the extension of the bottom edge, outside the loaded area.
Given. $q=100\text{ kN/m}^2$ over a 4 m × 2 m flexible rectangle; A at a corner, B interior, C outside on the bottom edge's extension, 1.5 m past A.
Find. (i) $\Delta\sigma_z$ at C, $z=2\text{ m}$, plus a qualitative A-vs-B comparison; (ii) qualitative trend at $z=5\text{ m}$ vs. $z=2\text{ m}$.
Approach. C lies outside the loaded rectangle but on the extension of one edge, so superpose two fictitious rectangles that both have a corner directly above C — add the "big" rectangle spanning C to the far corner, subtract the rectangle spanning C to A (the part that isn't actually loaded) — and apply Boussinesq's corner-of-rectangle influence factor $I(m,n)$ to each (equivalently, two readings off the Fadum chart supplied with the exam).
Big rectangle (C to far corner). $B_1=2\text{ m}$, $L_1=4+1.5=5.5\text{ m}$; $m=B_1/z=1.00$, $n=L_1/z=2.75$ → $I_1\approx0.203$.
A vs. B at $z=2$ m (no calculation). A sits at a corner of the loaded rectangle, so only one "quadrant" of load contributes beneath it; B sits well inside the rectangle, surrounded by load in every direction, so its influence factor is much closer to the full-coverage limit. Hence $\Delta\sigma_z(B) > \Delta\sigma_z(A)$, and both exceed $\Delta\sigma_z(C)$, which lies entirely outside the loaded area.
Depth $z=5$ m vs. $z=2$ m (no calculation). Boussinesq/Newmark influence values fall off monotonically with depth for any fixed plan location — the same surface load spreads its effect over an ever-wider area and a less-peaked stress bulb. So $\Delta\sigma_z$ decreases at $z=5\text{ m}$, relative to $z=2\text{ m}$, at all three points A, B and C.