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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.

Question 5 (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.

q = 100 kN/m² 4 m 2 m B 1.0 m 0.5 m A C 1.5 m
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).

  1. Big rectangle (C to far corner). $B_1=2\text{ m}$, $L_1=4+1.5=5.5\text{ m}$; $m=B_1/z=1.00$, $n=L_1/z=2.75$ → $I_1\approx0.203$.
  2. Fictitious rectangle (C to A). $B_2=2\text{ m}$, $L_2=1.5\text{ m}$; $m=1.00$, $n=0.75$ → $I_2\approx0.155$.
  3. Superpose. $\Delta\sigma_z(C)=q(I_1-I_2)=100(0.203-0.155)=\boxed{4.8\text{ kPa}}$.
  4. 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.
  5. 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.
Final Results — Question 5
QuantityValue
$I_1$ (C→far corner) / $I_2$ (C→A)0.203 / 0.155
$\Delta\sigma_z(C)$, $z=2$ m4.8 kPa
Ranking at $z=2$ m$\Delta\sigma_z(B) > \Delta\sigma_z(A) > \Delta\sigma_z(C)$
Trend at $z=5$ m vs. $z=2$ mdecreases at all three points