24-Bld-A6 Geotechnical Materials and Analysis · May 2018

Question 7 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, May 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 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.

Quantity	Value
Surface load, $\Delta\sigma$	200 kN/m²
Sand layer 1 (above GWT), $H_1$	2 m, $\gamma_{dry}=14.6$ kN/m³
Sand layer 2 (below GWT), $H_2$	2 m, $\gamma_{sat}=17.3$ kN/m³
Clay layer, $H_3$	4 m, $\gamma_{sat}=19.3$ kN/m³, $e_0=0.9$
Liquid limit, LL	50%
Specific gravity of clay, $G_s$	2.7

Figure 5: layered profile — sand (dry, $H_1$), sand (submerged, $H_2$) with the groundwater table at the top of $H_2$, clay ($H_3$), on a lower sand layer. The uniform surcharge Δσ is applied at the surface; settlement is estimated at clay mid-depth.

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

Approach. Compute the initial effective overburden stress at the clay's mid-height, use $C_c=0.009(LL-10)$ (formula sheet) since no oedometer $C_c$ is given directly, then apply the standard NC-clay settlement formula with the full $\Delta\sigma=200$ kPa (the load is described as uniformly distributed over the ground surface, so it does not attenuate with depth in this 1-D consolidation idealisation).

Effective overburden stress at clay mid-depth (before loading). $$\gamma'_{sand}=17.3-9.81=7.49\ \text{kN/m}^3,\quad\gamma'_{clay}=19.3-9.81=9.49\ \text{kN/m}^3$$ $$\sigma'_0=\gamma_{dry}H_1+\gamma'_{sand}H_2+\gamma'_{clay}\left(\frac{H_3}{2}\right)=14.6(2)+7.49(2)+9.49(2)$$ $$\boxed{\sigma'_0\approx63.2\text{ kPa}}$$
Compression index from the Liquid Limit correlation. $$C_c=0.009(LL-10)=0.009(50-10)=\boxed{0.36}$$
Primary consolidation settlement (normally consolidated clay, single virgin branch). $$S_c=H_3\frac{C_c}{1+e_0}\log_{10}\!\left(\frac{\sigma'_0+\Delta\sigma}{\sigma'_0}\right)=4\times\frac{0.36}{1.9}\log_{10}\!\left(\frac{63.2+200}{63.2}\right)$$ $$=4\times0.1895\times\log_{10}(4.166)=4\times0.1895\times0.6197$$ $$\boxed{S_c\approx0.470\text{ m}\approx470\text{ mm}}$$

Quantity	Value
Effective overburden at clay mid-depth, $\sigma'_0$	≈ 63.2 kPa
Compression index, $C_c$	0.36
Primary consolidation settlement, $S_c$	≈ 470 mm

Topic: One-dimensional primary consolidation settlement of a normally consolidated clay.

Key relations: $S_c=H\dfrac{C_c}{1+e_0}\log_{10}\!\left(\dfrac{\sigma'_0+\Delta\sigma}{\sigma'_0}\right)$; $C_c=0.009(LL-10)$.

Why this works: "Normally consolidated" means the current effective stress is the highest the clay has ever experienced, so the entire stress increase moves along the single virgin compression line in $e$–$\log\sigma'$ space — no separate recompression-index branch is needed, unlike an over-consolidated clay.

Common pitfall: Using the buoyant unit weight for the sand layer *above* the water table (it should be $\gamma_{dry}$), or forgetting that a widespread ("uniformly distributed over the ground surface") load does not attenuate with depth the way a footing load would — no Boussinesq/Newmark reduction is needed here.

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 11 (Consolidation — settlement calculations, LL–Cc correlation).

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