18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2019
Question 5 of 6: Primary Consolidation Beneath a Circular Footing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2019 — 18-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. All SIX questions are answered below (the exam marks only the first five as submitted; all six are solved here as a complete study resource). Marking scheme: each question 20 marks, equal value.
Given. A single, normally-consolidated clay layer with the footing and soil data below.
Given data
Quantity
Value
Clay thickness, $H$
4 m (on impermeable rock)
Footing diameter, $D$
1 m (circular)
Column load, $P$
4000 kN
Specific gravity of solids, $G_s$
2.51
Initial void ratio, $e_0$
3.9
Compression index, $C_c$
0.45
Find. (a) Initial and final effective vertical stress at the clay's midpoint below the footing centre. (b) Ultimate primary consolidation settlement. (c) The resulting change in void ratio.
Check: the footing is stated and drawn as a 1 m diameter circular footing throughout the problem stem and figure, but parts (a) and (c) refer to a "square footing" – the source figure's own labels are internally inconsistent. Solved here using the circular geometry (the only one with a stated diameter and the only one drawn), consistent with the problem's own dominant description.
Figure 3: circular footing on a single-drainage clay layer over impermeable rock; Point A is the layer's midpoint directly below the footing centre.
Approach. Get the initial effective stress at the midpoint from the clay's own buoyant unit weight, add the Boussinesq stress increase from the circular footing load at that depth, then apply the standard normally-consolidated settlement and void-ratio-change formulas to the full (single-drainage) layer thickness.
(a) Initial effective stress. The clay is saturated with the water table at the ground surface, so effective stress uses the buoyant unit weight: $$\gamma_b = \dfrac{(G_s-1)\gamma_w}{1+e_0} = \dfrac{(1.51)(9.81)}{4.9} = 3.02\text{ kN/m}^3$$ At the midpoint, $z=H/2=2$ m: $$\sigma_0' = \gamma_b z = 3.02\times2 = \boxed{6.05\text{ kPa}}$$
Stress increase from the footing (Boussinesq, circular loaded area, on the centreline). Contact pressure $q=P/A=4000/(\pi(0.5)^2)=5093$ kPa. Check: this contact pressure is extremely high for a 1 m footing (roughly 20× a typical safe bearing pressure for clay) – the load and footing size are used exactly as printed, per the exam's own "state your assumptions" instruction, rather than substituted for a more realistic pair. $$\Delta\sigma = q\left[1-\left(\dfrac{1}{1+(a/z)^2}\right)^{1.5}\right],\quad a=0.5\text{ m},\ z=2\text{ m}$$ $$\Delta\sigma = 5093\left[1-\left(\dfrac{1}{1.0625}\right)^{1.5}\right] = 5093(0.0869) = 442.7\text{ kPa}$$ $$\sigma_1' = \sigma_0'+\Delta\sigma = 6.05+442.7 = \boxed{448.8\text{ kPa}}$$
(b) Ultimate primary consolidation. Both stresses fall in the virgin-compression range for this normally-consolidated clay (so $C_r$ is not needed – there is no unload–reload branch here), and only the top of the layer drains (impermeable rock below), so the full 4 m thickness is used directly (not halved): $$S_c = \dfrac{C_cH}{1+e_0}\log_{10}\left(\dfrac{\sigma_1'}{\sigma_0'}\right) = \dfrac{0.45\times4}{4.9}\log_{10}\left(\dfrac{448.8}{6.05}\right) = 0.3673\times1.870 = \boxed{0.687\text{ m}}$$
(c) Change in void ratio. $$\Delta e = C_c\log_{10}\left(\dfrac{\sigma_1'}{\sigma_0'}\right) = 0.45\times1.870 = \boxed{0.842}$$ (check: $S_c=\Delta e\,H/(1+e_0)=0.842(4)/4.9=0.687$ m, matching step 3.)
