16-Civ-A4 Geotechnical Materials and Analysis · December 2016
Question 4 of 5: Consolidation settlement of the clay layer
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 98-Civ-A4 Geotechnical Materials and Analysis — December 2016. Closed book; 3 hours; total 100 marks; answer ALL five questions. Newmark / m–n influence charts and a formula sheet are supplied with the paper.
Reference texts: B.M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. (Cengage); R.F. Craig, Craig's Soil Mechanics, 8th ed. (Spon Press); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (Pearson).
Question 4: Consolidation settlement of the clay layer (20 marks)
Given. Square footing 4 m × 4 m carrying 1200 kN with its base at 1.5 m depth (Figure 4(b)); water table at 2 m; 2 m + 4 m of sand ($\gamma=18.3$ kN/m³) over a 3 m compressible clay ($\gamma_b=9.2$ kN/m³, $G_s=2.7$) over sand and gravel; $e$–$\log p$ curve in Figure 4(a).
Given data
Quantity
Value
Footing / load
4 m × 4 m, 1200 kN, founded at 1.5 m
Gross / net contact pressure
$75$ / $47.6$ kPa
Clay layer
3 m thick, 6–9 m depth, $\gamma_b=9.2$ kN/m³
Compressibility (read from Fig. 4a)
$e_0\approx1.03$ (ticked); reloading slope $C_r\approx0.02$; curve breaks at about 150–200 kPa
Find. The consolidation settlement under the centre of the footing.
[Figure not reproduced: Figure 4(b) (redrawn): footing base at 1.5 m, water table at 2 m; the 3 m clay (6–9 m) is the compressible layer. See the official exam paper.]
Approach. Compute the net contact pressure, the initial effective stress at mid-clay, the stress increase there from the footing (centre = four corner rectangles), then place that stress range on Figure 4(a) to choose the correct compression index before computing the one-dimensional settlement.
Net applied pressure. Gross $q = 1200/(4\times4) = 75$ kPa; removing the 1.5 m of soil above founding level (the water table is below the base), $q_{net} = 75 - (1.5)(18.3) = 47.6$ kPa.
Initial effective stress at mid-clay ($z=7.5$ m). Above WT (0–2 m) sand is moist, below WT it is buoyant, and the clay is given as a buoyant unit weight:
$$\sigma'_0 = (2)(18.3) + (4)(18.3-9.81) + (1.5)(9.2) = 36.6 + 34.0 + 13.8 = 84.4\ \text{kPa}.$$
Stress increase at mid-clay. The centre of the 4 m footing is the common corner of four $2\times2$ rectangles at depth $z=7.5-1.5=6.0$ m below the base: $m=n=2/6=0.333$, $I(0.333,0.333)=0.0447$.
$$\Delta\sigma' = q_{net}\,[4\,I] = 47.55(4\times0.0447) = \boxed{8.5\ \text{kPa}}.$$ (The 2:1 method gives $47.55\times16/10^2 = 7.6$ kPa, which agrees.)
Read Figure 4(a) and place the stress range on it. The ticked in-situ void ratio is $e_0\approx1.03$. The laboratory curve is almost flat from 10 to 100 kPa ($e$: 0.978 → 0.958), so the reloading slope is $C_r\approx0.020$. It breaks sharply only above about 150–200 kPa (the preconsolidation pressure), where the virgin slope is $C_c\approx0.33$. The clay goes from $\sigma'_0=84.4$ to $\sigma'_0+\Delta\sigma'=92.9$ kPa, entirely on the flat recompression branch, so $C_r$ applies, not $C_c$.
Check: $e_0$, $C_r$ and the break of the curve are graph readings from Figure 4(a). A flatter read ($C_r=0.015$) gives 0.9 mm, so the settlement is about 1 mm either way. The decisive reading is that $\sigma'_0+\Delta\sigma'\approx93$ kPa stays below the curve's break. Wrongly using the virgin slope $C_c\approx0.33$ would give about 18 mm, an order-of-magnitude overestimate. The footing base is at 1.5 m; the 2 m dimension on Figure 4(b) is the water table. The printed $G=2.7$ is not needed because the unit weights are given.
