16-Civ-A4 Geotechnical Materials and Analysis · December 2017

Question 4 of 6: Consolidation Settlement and Swelling

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examinations — 16-Civ-A4 Geotechnical Materials and Analysis, December 2017. Closed book, 3 hours, 100 marks. Answer all six questions. Charts (rectangular-area influence chart, Newmark chart) and a formula sheet are provided at the back of the exam.

Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations.

Question 4: Consolidation Settlement and Swelling (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. The oedometer e–logσ′ curve (loading 27→429 kPa, unloading 429→54 kPa) plus the site profile: 4 m sand over 8 m clay, water table at ground surface, and a 4 m × 20 kN/m³ extensive fill.

Given data

Clay layer thickness, H	8 m (mid-depth 8 m below surface)
Overlying sand	4 m, water table at surface
Saturated unit weight (both)	γsat = 19 kN/m³
Fill surcharge	4 m × 20 = 80 kPa (extensive ⇒ constant with depth)
γw	9.81 kN/m³

Find. (a) The ultimate consolidation settlement of the clay under the fill; (b) the eventual heave if the fill is later removed.

Approach. Compute the initial effective stress at mid-clay (buoyant, since the water table is at the surface), add the constant fill surcharge, read the compression index from the virgin portion of the e–logσ′ curve, and apply the one-dimensional settlement equation; for unloading, use the swelling index over the same stress range.

1. Initial effective stress at mid-clay. Buoyant unit weight γ′ = 19 − 9.81 = 9.19 kN/m³; mid-clay is 8 m below the surface (4 m sand + 4 m clay), all submerged: $$\sigma_0'=\gamma'\times 8 = 9.19\times 8 = 73.5\ \text{kPa}.$$
2. Stress after filling. An extensive fill adds a uniform Δσ′ = 20×4 = 80 kPa at every depth: $$\sigma_1'=\sigma_0'+\Delta\sigma'=73.5+80=153.5\ \text{kPa}.$$
3. Compression index from the virgin curve. Using the straight (normally-consolidated) portion, points (107 kPa, e = 1.144) and (429 kPa, e = 0.994): $$C_c=\frac{e_0-e_1}{\log(\sigma_1'/\sigma_0')}=\frac{1.144-0.994}{\log(429/107)}=0.249.$$ The in-situ void ratio at σ′₀ interpolates to e₀ = 1.184.
4. Consolidation settlement (a). $$s_c=\frac{H\,C_c}{1+e_0}\log\frac{\sigma_1'}{\sigma_0'}=\frac{8\times0.249}{1+1.184}\log\frac{153.5}{73.5}=\boxed{0.291\ \text{m}\ (291\ \text{mm})}.$$ Cross-check with m_v over the increment (m_v ≈ 4.6×10⁻⁴ m²/kN): s = m_vΔσ′H = 292 mm — consistent.
5. Swelling index for unloading. From the rebound branch, points (429 kPa, 0.994) and (54 kPa, 1.024): $$C_s=\frac{1.024-0.994}{\log(429/54)}=0.0333.$$
6. Heave on removing the fill (b). The clay unloads from 153.5 kPa back to 73.5 kPa along the swelling line (void ratio at 153.5 kPa is e ≈ 1.104): $$s_s=\frac{H\,C_s}{1+e_1}\log\frac{\sigma_1'}{\sigma_0'}=\frac{8\times0.0333}{1+1.104}\log\frac{153.5}{73.5}=\boxed{0.041\ \text{m}\ (41\ \text{mm})}.$$

Results

Quantity	Value
σ′0 / σ′1 at mid-clay	73.5 / 153.5 kPa
Compression index Cc	0.249
(a) Consolidation settlement	291 mm
Swelling index Cs	0.0333
(b) Heave on unloading	41 mm

The clay settles about 0.29 m under the fill, but recovers only about 0.04 m (roughly one-seventh) when the fill is removed, because unloading follows the much flatter swelling line (C_s ≈ C_c/7.5). Most of the consolidation is therefore permanent — a key reason surcharge pre-loading is an effective ground-improvement technique.

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