24-Bld-A6 Geotechnical Materials and Analysis, December 2017, Q7

Question 7 (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 three-layer profile: sand (H₁ = 1.5 m, above the water table, $\gamma_{dry}=14.6$ kN/m³), sand (H₂ = 1.5 m, below the water table, $\gamma_{sat}=17.3$ kN/m³), then a normally consolidated clay layer (H₃ = 2 m, $\gamma_{sat}=19.3$ kN/m³, LL = 38, e₀ = 0.75), over sand; a surface surcharge $\Delta\sigma=100$ kN/m² is applied over an area wide enough to be treated as extending to infinity (no attenuation with depth).

Given data — Question 7
Quantity	Symbol	Value
Sand thickness above water table	H₁	1.5 m
Sand thickness below water table	H₂	1.5 m
Clay layer thickness	H₃	2 m
Sand, dry / saturated unit weight	γ_dry / γ_sat	14.6 / 17.3 kN/m³
Clay saturated unit weight	γ_sat	19.3 kN/m³
Clay liquid limit	LL	38
Clay initial void ratio	e₀	0.75
Surface surcharge	Δσ	100 kN/m²

Find. The primary consolidation settlement of the normally consolidated clay layer.

[Figure not reproduced: Figure 3 — soil profile with the three distinct thicknesses read from the exam figure (H 1 , H 2 sand above/below the water table, H 3 the clay layer), and the clay mid-depth used for the settlement calculation. See the official exam paper.]

Approach. Since the surcharge is applied over an extensive area, the stress increase at any depth equals the full 100 kPa with no attenuation; compute the effective overburden stress at the clay's mid-depth before loading, add the (unattenuated) surcharge, obtain the compression index from the LL correlation on the formula sheet, and apply the standard normally-consolidated settlement formula.

Effective overburden stress at clay mid-depth (before loading). Clay mid-depth is $z=1.5+1.5+1.0=4.0$ m below ground surface. With submerged unit weights $\gamma_{sand}'=17.3-9.81=7.49$ and $\gamma_{clay}'=19.3-9.81=9.49$ kN/m³ below the water table, $$\sigma_0' = \underbrace{14.6(1.5)}_{\text{sand, above WT}}+\underbrace{7.49(1.5)}_{\text{sand, below WT}}+\underbrace{9.49(1.0)}_{\text{clay, to mid-depth}} = 21.9+11.2+9.5 = 42.6\ \text{kPa}.$$
Stress after loading. The surcharge is extensive (uniform over a large area), so it reaches the clay mid-depth undiminished: $$\sigma_1' = \sigma_0'+\Delta\sigma = 42.6+100 = 142.6\ \text{kPa}.$$
Compression index from the LL correlation. From the formula sheet, $$C_c = 0.009(LL-10) = 0.009(38-10) = 0.252.$$
Settlement. The clay is stated to be normally consolidated, so the entire loading increment lies on the virgin compression line: $$S_c = \frac{C_c\,H_3}{1+e_0}\log_{10}\!\left(\frac{\sigma_1'}{\sigma_0'}\right) = \frac{0.252(2000\ \text{mm})}{1.75}\log_{10}\!\left(\frac{142.6}{42.6}\right)$$ $$\boxed{S_c \approx 151\ \text{mm}}.$$

Question 7 — final results
Quantity	Value
Effective overburden at clay mid-depth, σ′₀	42.6 kPa
Stress after loading, σ′₁	142.6 kPa
Compression index, C_c (from LL = 38)	0.252
Primary consolidation settlement, S_c	≈ 151 mm

Concept & Theory

Topic. One-dimensional primary consolidation settlement of a normally consolidated clay under an extensive surface surcharge.

Key relations. $C_c=0.009(LL-10)$ (Skempton's correlation, formula sheet); $S_c=[C_c/(1+e_0)]H\log_{10}(\sigma_1'/\sigma_0')$ for a normally consolidated clay.

Why this works. For a load applied over an area large compared with the depth to the compressible layer, the vertical stress increase does not attenuate with depth (the loaded area effectively looks infinite in plan from the clay's point of view), so the full surface surcharge can be added directly to the pre-existing effective stress without a Boussinesq/Newmark reduction — a useful simplification that also removes any dependence on the (here, unspecified) plan size of the loaded area.

Common pitfall. Applying a Boussinesq stress-reduction factor when the load is actually described as extensive/uniform over the whole ground surface; and confusing the three distinct layer thicknesses read from the figure (for example by using only two values, H₁ and H₂, and silently merging H₂ and H₃).

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

Question 7 of 7

Question 7 (20 marks)