24-Bld-A6 Geotechnical Materials and Analysis, December 2016, Q4

Question 4 (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 4 m × 4 m footing carries a 1200 kN column load, founded at 1.5 m depth in sand; the water table is 2 m below ground surface; sand extends to 6 m depth ($\gamma=18.3$ kN/m³); a 3 m clay layer follows ($\gamma_b=9.2$ kN/m³), over sand and gravel ($\gamma_b=9.0$ kN/m³); on Figure 4(a) the in-situ void ratio is marked on the ordinate by a short tick at $e_0\approx 1.03$; the laboratory curve itself starts at $e\approx 0.98$ at 10 kPa, stays nearly flat to about 100 kPa, then steepens onto a virgin-compression branch, so the preconsolidation pressure lies between the heavy 100 kN/m² ordinate and the $\sigma_p'\approx 150$ kPa obtained by extending the flat and virgin branches to their intersection.

Given data — Question 4
Quantity	Symbol	Value
Column load	Q	1200 kN
Footing size	B × L	4 m × 4 m
Founding depth	D_f	1.5 m
Depth to water table	—	2 m
Sand unit weight (above and, submerged, below WT)	γ	18.3 kN/m³
Depth to top / mid-depth of clay	—	6 m / 7.5 m
Clay thickness, submerged unit weight	H, γ_b	3 m, 9.2 kN/m³
Initial void ratio (Fig. 4a ordinate tick)	e₀	1.03
Preconsolidation pressure (Fig. 4a)	σ′_p	100–150 kPa
Recompression index (Fig. 4a, flat branch)	C_r	0.015
Specific gravity	G	2.7

Find. The consolidation settlement of the clay layer under the centre of the footing.

[Figure not reproduced: Figure 4(b) — Footing and soil profile (redrawn from the exam figure), with the clay mid-depth marked at 7.5 m below ground surface. See the official exam paper.]

Approach. Compute the net pressure the footing adds at founding level, find the resulting stress increase at the mid-depth of the clay by the approximate (2:1-type) method printed on the formula sheet, add it to the existing effective overburden stress, and compare the result with the preconsolidation pressure read from Figure 4(a) to decide whether the settlement is governed by the flat recompression branch (C_r) or the steep virgin branch (C_c) of the curve.

Net foundation pressure. The footing replaces sand it was excavated from, so only the NET pressure increases stress in the ground: $$q_{net} = \frac{Q}{BL}-\gamma D_f = \frac{1200}{16}-18.3(1.5) = 75.0-27.45 = 47.55\ \text{kPa}.$$
Effective overburden stress at clay mid-depth (before loading). Using the submerged sand weight $\gamma' = 18.3-9.81 = 8.49$ kN/m³ below the water table, $$\sigma_0' = \underbrace{18.3(2)}_{\text{sand, above WT}}+\underbrace{8.49(4)}_{\text{sand, below WT}}+\underbrace{9.2(1.5)}_{\text{clay, to mid-depth}} = 36.6+33.96+13.8 = 84.4\ \text{kPa}.$$
Stress increase at clay mid-depth. The footing base to clay mid-depth is $z=7.5-1.5=6.0$ m; with the approximate method on the formula sheet, $$\Delta\sigma_z=\frac{q_{net}\,BL}{(B+z)(L+z)}=\frac{47.55(4)(4)}{(4+6)(4+6)}=7.6\ \text{kPa},$$ consistent with an exact Boussinesq (four-quadrant, corner-at-centre) evaluation of 8.5 kPa — the same order of magnitude, as expected for the two methods.
Stress after loading, and check against preconsolidation pressure. $$\sigma_1' = \sigma_0'+\Delta\sigma_z = 84.4+7.6 = 92.0\ \text{kPa}.$$ Both $\sigma_0'=84.4$ kPa and $\sigma_1'=92.0$ kPa are BELOW the preconsolidation pressure read from Figure 4(a) — 100 kPa on the most conservative reading (the heavy ordinate at which the curve visibly leaves its flat branch), about 150 kPa if the flat and virgin branches are extended to their intersection. On either reading the clay is over-consolidated at this location, and the entire loading increment stays on the flat, RECOMPRESSION branch of the curve. The settlement must therefore use $C_r$, not the much larger virgin compression index $C_c$ that governs beyond $\sigma_p'$.
Settlement. With $C_r=0.015$ (the slope of the flat branch of Figure 4(a) between 10 and 100 kPa) and $e_0=1.03$ from the marked ordinate tick, $$S_c = \frac{C_r\,H}{1+e_0}\log_{10}\!\left(\frac{\sigma_1'}{\sigma_0'}\right) = \frac{0.015(3000\ \text{mm})}{2.03}\log_{10}\!\left(\frac{92.0}{84.4}\right)$$ $$\boxed{S_c \approx 0.83\ \text{mm}}$$ Only the factor $(1+e_0)$ depends on which of the figure's three candidate void ratios is adopted (see the check note below), and the spread is small: 0.83 mm at $e_0=1.03$, 0.85 mm at 0.98, 0.93 mm at 0.81. The engineering answer is the same in every case — the consolidation settlement of this clay under this footing is well under a millimetre, i.e. negligible, because the footing is small, its net pressure is modest, and the clay lies 6 m below its base.

Check — Figure 4(a) offers three different void ratios and they do not agree. (1) The ordinate tick labelled e₀ sits at about 1.03. (2) The laboratory compression curve itself begins at e ≈ 0.98 at 10 kPa. (3) The given specific gravity G = 2.7 together with the clay's printed buoyant unit weight $\gamma_b=9.2$ kN/m³ gives $e_0=(G-1)\gamma_w/\gamma_b-1=1.7(9.81)/9.2-1=0.81$, which lands on the SECOND (lower) curve of the figure rather than on the ticked value. That pattern is the usual textbook illustration of sample disturbance, and the exam does not state which curve represents this clay. The explicitly marked e₀ = 1.03 is adopted here because it is the only void ratio the figure actually labels; the discrepancy is flagged rather than hidden, and the sensitivity is quantified in the step above (0.83–0.93 mm, negligible on any reading). G is therefore not redundant — it is the cross-check that exposes the inconsistency.

Question 4 — final results
Quantity	Value
Net foundation pressure, q_net	47.6 kPa
Effective overburden at clay mid-depth, σ′₀	84.4 kPa
Stress increase at clay mid-depth, Δσ_z	7.6 kPa
Stress after loading, σ′₁	92.0 kPa
Comparison with σ′_p (100–150 kPa)	both below — recompression range
Consolidation settlement, S_c	0.83 mm (0.83–0.93 mm across the figure's candidate e₀ values)

Concept & Theory

Topic. One-dimensional consolidation settlement of an over-consolidated clay beneath a shallow footing.

Key relations. Net foundation pressure $q_{net}=Q/A-\gamma D_f$; approximate (2:1-type) stress-increase formula $\Delta\sigma_z=q\,BL/[(B+z)(L+z)]$; recompression settlement $S_c=[C_r/(1+e_0)]H\log_{10}(\sigma_1'/\sigma_0')$ when $\sigma_1'\le\sigma_p'$, switching to (or splitting with) $C_c$ once the stress path crosses $\sigma_p'$.

Why this works. A clay that has previously carried a higher effective stress than it carries today (over-consolidated) responds to a NEW load with the much stiffer recompression stiffness, because the fabric was already compressed to that denser state once before; only once the stress state exceeds the clay's stress "memory" does it soften onto the virgin compression line. Checking $\sigma_0'$ and $\sigma_1'$ against $\sigma_p'$ before picking $C_r$ or $C_c$ is therefore not optional — it changes the answer by roughly an order of magnitude here (0.83 mm on the recompression branch versus several centimetres had the same increment been read off the virgin line, whose slope on this figure is $C_c\approx0.37$).

Common pitfall. Applying the steep virgin compression index $C_c$ to the whole stress increment without first checking it against $\sigma_p'$, or forgetting to net out the overburden the footing excavation removed when computing $q_{net}$.

Source. Das, Principles of Geotechnical Engineering, Ch. 8 (stress distribution) and Ch. 11 (consolidation, preconsolidation pressure); CFEM Ch. 9.

Question 4 of 5

Question 4 (20 marks)