Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, compass and ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 are compulsory; Question 6 offers five 5-mark optional items of which the source asks for three.
Given. Profile (Fig. Q-4): sand $0$–$5$ m ($\gamma_{sat}=17\text{ kN/m}^3$, water table at 1.5 m depth) over 8 m of overconsolidated clay ($\gamma_{sat}=20\text{ kN/m}^3$) over impermeable rock. $e_0=0.72$, $C_c=0.28$, $C_r=0.054$, $c_v=2.68\times10^{-4}\text{ cm}^2/\text{s}$, $\sigma'_p=180$ kPa, $\Delta\sigma'=65.4$ kPa at mid-clay (depth 9 m).
Find. (a) Stress profiles; (b) $S_c$; (c) $t_{50}$; (d) final stresses; (e) $S_c$ at 1 yr; (f) effect of a sand layer at the clay base.
Approach. Compute $\sigma_0$, $u_0$, $\sigma'_0$ at mid-clay, compare $\sigma'_0+\Delta\sigma'$ against $\sigma'_p$ to select $C_r$ or $C_c$, then apply the settlement and Terzaghi time-factor equations with the correct drainage path length.
Fig. Q4a — total and effective vertical stress profiles before construction ($\gamma_w=9.81\text{ kN/m}^3$); mid-clay $\sigma'_0=91.4$ kPa sits well left of $\sigma'_p=180$ kPa, confirming overconsolidation.
(a) In-situ stresses at mid-clay (depth 9 m), before construction. $\sigma_0=(1.5+3.5)(17)+4(20)=85+80=\boxed{165\text{ kPa}}$. $u_0=\gamma_w(9-1.5)=9.81\times7.5=\boxed{73.6\text{ kPa}}$. $\sigma'_0=165-73.6=\boxed{91.4\text{ kPa}}$ (profiles plotted above; note $\sigma'_0<\sigma'_p=180$ kPa, confirming the clay is overconsolidated at this depth, as stated).
(b) Primary consolidation settlement. Final effective stress $\sigma'_f=\sigma'_0+\Delta\sigma'=91.4+65.4=156.8$ kPa. Since $\sigma'_f=156.8\text{ kPa}<\sigma'_p=180$ kPa, the entire stress increase stays within the recompression range, so $C_r$ (not $C_c$) governs: $S_c=\dfrac{C_rH}{1+e_0}\log_{10}\!\dfrac{\sigma'_f}{\sigma'_0}=\dfrac{0.054\times8}{1.72}\log_{10}\!\dfrac{156.8}{91.4}=\boxed{58.9\text{ mm}}$.
(c) Time for 50% settlement. Only the sand above the clay is permeable (rock below is impermeable) → single drainage, $H_{dr}=H=8\text{ m}=800\text{ cm}$. $T_{50}=0.197$ (standard). $t_{50}=\dfrac{T_{50}H_{dr}^2}{c_v}=\dfrac{0.197\times800^2}{2.68\times10^{-4}}=4.70\times10^{8}\text{ s}=\boxed{14.9\text{ years}}$.
(d) Final stresses at mid-clay. After full consolidation the excess pore pressure has dissipated back to hydrostatic ($u_f=u_0$), so $\sigma_f=\sigma_0+\Delta\sigma=165+65.4=\boxed{230.4\text{ kPa}}$, $u_f=\boxed{73.6\text{ kPa}}$, $\sigma'_f=230.4-73.6=\boxed{156.8\text{ kPa}}$ (consistent with step 2).
(e) Settlement after 1 year. $T_v=\dfrac{c_vt}{H_{dr}^2}=\dfrac{2.68\times10^{-4}\times3.156\times10^{7}}{800^2}=0.0132$. Since $T_v<0.197$: $U=100\sqrt{4T_v/\pi}=\boxed{13.0\%}$. $S_c(1\text{yr})=U\times S_c=0.130\times58.9=\boxed{7.6\text{ mm}}$.
(f) If a sand layer also underlies the clay (double drainage). Only the drainage path length changes ($H_{dr}=H/2=4\text{ m}=400\text{ cm}$, since water can now escape both up and down) — $S_c=58.9$ mm is unaffected (it depends on stress change, not drainage path). Question 3 (time for 50%): $t_{50}\propto H_{dr}^2$, so $t_{50}$ drops by a factor of 4 to $\boxed{3.7\text{ years}}$. Question 5 (settlement in 1 year): $T_v$ increases 4$\times$ to $0.0529$, giving $U=\boxed{25.9\%}$ and $S_c(1\text{yr})=0.259\times58.9=\boxed{15.3\text{ mm}}$ — consolidation proceeds noticeably faster with a second drainage face, even though the ultimate settlement is unchanged.
