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

Question 4 of 5: Consolidation Settlement of a Clay Layer under a Footing

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

Question 4: Consolidation Settlement of a Clay Layer under a Footing (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 $4\ \text{m}\times4\ \text{m}$ footing carrying $1200\ \text{kN}$, base at $1.5\ \text{m}$ depth. Profile (Figure 4b): sand $0$–$6\ \text{m}$ ($\gamma=18.3\ \text{kN/m}^3$, water table at $2\ \text{m}$), clay $6$–$9\ \text{m}$ ($\gamma=19.2\ \text{kN/m}^3$), sand & gravel below. Figure 4(a) is used directly: void ratio is read off the loading branch at the initial and final effective stresses, after checking where they sit relative to the preconsolidation break. $G=2.7$, $\gamma_w=9.81\ \text{kN/m}^3$.

Find. The consolidation settlement of the $3\ \text{m}$ clay layer beneath the footing centre.

Approach. Evaluate everything at the mid-height of the clay ($z=7.5\ \text{m}$). Compute the in-situ effective stress $\sigma_0'$, add the net footing stress increment $\Delta\sigma'$ (spread from the base by the four-quadrant $m$–$n$ influence factor, checked by the 2:1 rule), then place $\sigma_0'$ and $\sigma_0'+\Delta\sigma'$ on Figure 4(a) to get the void-ratio change and the settlement $s_c = H\,\Delta e/(1+e_0)$.

1. In-situ effective stress at mid-clay. Sand above WT $\gamma=18.3$; below WT $\gamma'=18.3-9.81=8.49$; clay $\gamma'=19.2-9.81=9.39$: $$\sigma_0' = 18.3(2)+8.49(4)+9.39(1.5)=36.6+34.0+14.1=\boxed{84.6\ \text{kPa}}.$$
2. Net footing pressure. Gross contact pressure $q=1200/(4\times4)=75\ \text{kPa}$; remove the overburden excavated at the base: $$q_{net}=75-\gamma D_f = 75-18.3(1.5)=47.6\ \text{kPa}.$$
3. Stress increment at mid-clay. Depth below base $z=7.5-1.5=6.0\ \text{m}$. Split the footing into four $2\times2$ quadrants meeting over the centre: $m=n=B/z=2/6=0.333$, giving the corner influence factor $I\approx0.045$: $$\Delta\sigma' = 4\,I\,q_{net}=4(0.045)(47.6)=8.5\ \text{kPa}.$$ Check by the 2:1 method: $\Delta\sigma'=q_{net}BL/[(B+z)(L+z)]=47.6(16)/(10)(10)=7.6\ \text{kPa}$ — consistent.
4. Where the stresses sit on Figure 4(a). $\sigma_0'=84.6$ kPa and $\sigma_1'=\sigma_0'+\Delta\sigma'=84.6+8.5=93.1$ kPa both lie on the flat, early part of the loading curve. The curve only bends into the steep virgin line at about $\sigma_c'\approx150$–$250$ kPa (Casagrande construction). Between $\sigma'=300$ and $400$ kPa the virgin slope is $C_c=(0.874-0.831)/\log(400/300)\approx0.34$, but the footing load never reaches that branch, so using $C_c$ here would be wrong.
5. Void ratios read from the curve: $$e(84.6)\approx0.961,\qquad e(93.1)\approx0.959,\qquad \Delta e\approx0.003.$$ This is the same as using the local recompression slope $C_r=\Delta e/\log(93.1/84.6)\approx0.07$.
6. Consolidation settlement. With $H=3.0$ m and $e_0=e(\sigma_0')\approx0.96$: $$s_c = \frac{H\,\Delta e}{1+e_0}=\frac{3.0(0.0027)}{1.96}.$$ $$\boxed{s_c \approx 0.004\ \text{m} \approx 4\ \text{mm}}$$ Using the printed $e_0\approx1.03$ label instead gives the same 4 mm.

The settlement is modest because the $4\times4\ \text{m}$ footing is small relative to the $6\ \text{m}$ depth to the clay, so the load has spread widely and $\Delta\sigma'$ is only $\sim9$ kPa at mid-clay. On top of that, the clay is over-consolidated for this stress range, so it responds on its stiff recompression branch. Applying the virgin $C_c\approx0.34$ would give about 22 mm, more than five times too much.