18-Env-A3 Geotechnical and Hydrogeological Engineering · Undated paper
Question 5 of 6: Consolidation Settlement Beneath a Square Footing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 18-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. Six questions, each 20 marks, equal value; all six are solved below.
Find. (a) Initial and final effective vertical stress at the midpoint of the clay (z = 1 m) below the footing centre. (b) Ultimate primary consolidation settlement. (c) Change in void ratio.
Fig. 3 — Square footing on a 2 m normally-consolidated clay layer resting on impermeable rock; Point A is the clay's midpoint directly below the footing centre.
Approach. Find the initial effective stress from the clay's own submerged unit weight, add the Boussinesq stress increase from the footing (four-corner method) to get the final effective stress, then apply the standard normally-consolidated settlement formula.
Initial effective stress. $\gamma_{sat}=(G_s+e_0)\gamma_w/(1+e_0)=(2.75+2.5)/3.5\times9.81=14.72\text{ kN/m}^3$, so $\gamma_b=14.72-9.81=4.905\text{ kN/m}^3$. At $z=1\text{ m}$: $$\sigma_0' = \gamma_b\,z = 4.905\times1 = \boxed{4.91\text{ kPa}}$$
Stress increase from the footing. Contact pressure $q_0=P/(B\times L)=8000/(2\times2)=2000\text{ kPa}$. Splitting the 2×2 m footing into four 1×1 m quadrants around the centre point, each has $m=n=(B/2)/z=1/1=1$, giving the standard Boussinesq corner influence factor $I_\sigma(1,1)=0.1752$. Summing all four quadrants: $$\Delta\sigma = 4\,I_\sigma\,q_0 = 4\times0.1752\times2000 = \boxed{1402\text{ kPa}}$$
(b) Consolidation settlement. The clay is normally consolidated, so loading proceeds entirely along the virgin compression line (only $C_c$ applies): $$S_c = \dfrac{C_c H}{1+e_0}\log_{10}\!\left(\dfrac{\sigma_f'}{\sigma_0'}\right) = \dfrac{0.40\times2}{3.5}\log_{10}\!\left(\dfrac{1406.7}{4.91}\right) = 0.2286\times2.457 = \boxed{0.562\text{ m}}$$
(c) Change in void ratio. $$\Delta e = C_c\log_{10}\!\left(\dfrac{\sigma_f'}{\sigma_0'}\right) = 0.40\times2.457 = \boxed{0.983}$$ Check: $S_c=\Delta e\,H/(1+e_0)=0.983\times2/3.5=0.562\text{ m}$ — matches (b).
Check: the recompression index $C_r=0.10$ is given but not used — since the clay is explicitly normally consolidated, its preconsolidation pressure equals $\sigma_0'$, so the entire loading path from $\sigma_0'$ to $\sigma_f'$ lies on the virgin compression line and only $C_c$ applies. $C_r$ would only enter if the clay were overconsolidated with $\sigma_f'$ exceeding a preconsolidation pressure greater than $\sigma_0'$.
