24-Bld-A6 Geotechnical Materials and Analysis · December 2018
Question 4 of 7
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-BLD-A6 Geotechnical Materials and Analysis — National Examinations, December 2018. 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory; Section B directs "answer any three of Q4–Q7," but for completeness this solution answers all four.
Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.
Figure 3 (reproduced, schematic): each 2 m soil sample is fed by a reservoir at elevation 5 m and drained through a static tube to a reservoir at elevation 3 m (elevations measured above the sample base) — situation (1) feeds the top from the high reservoir (downward flow), situation (2) feeds the bottom from the high reservoir (upward flow). X-X is the mid-height plane, 1 m from either face.
Check: Figure 3's plumbing is read from the exam's own sketch — in situation (1) the tall (elevation 5 m) reservoir connects directly to the TOP of the sample and the short (elevation 3 m) reservoir connects via a static outer tube to the BOTTOM (net downward flow); situation (2) reverses this (net upward flow). Both connecting tubes are open, rigid, and carry no head loss themselves — only the soil offers flow resistance — so each tube's total head equals its own reservoir's free-surface elevation throughout.
Given. Sample length $L=2\text{ m}$; $\gamma_{sat}=20\text{ kN/m}^3$; reservoir elevations (above the sample base) 5 m and 3 m, connected as above.
Find. $\sigma'$ on plane X-X (mid-height) for both situations.
Approach. Total head is constant along each open connecting tube (equal to that reservoir's surface elevation) and varies linearly through the homogeneous soil itself; interpolate the head at X-X, recover the pore pressure, and combine with the total stress built up from the soil's own (submerged) top surface. Cross-check with the direct seepage-force form $\sigma'=\gamma'z\pm i\gamma_w z$.
Hydraulic gradient (both situations). $\Delta h=5-3=2\text{ m}$ over $L=2\text{ m}$, so $i=1.0$ in magnitude both ways — only the direction differs. The critical gradient is $i_{cr}=\gamma'/\gamma_w=(20-9.81)/9.81=1.039$, so both cases sit just below quick condition.
Situation (1) — downward. $h_{top}=5\text{ m}$, $h_{bot}=3\text{ m}$; at X-X (mid-depth), $h=4\text{ m}$, pressure head $=4-1=3\text{ m}$, $u_{XX}=9.81(3)=29.43\text{ kPa}$. The soil's own top surface sits under 3 m of standing water ($\sigma_{top}=29.4\text{ kPa}$, $\sigma'_{top}=0$); descending 1 m into the soil, $\sigma_{XX}=29.4+20(1)=49.4\text{ kPa}$, so $\sigma'_{XX}=49.4-29.4=\boxed{20.0\text{ kPa}}$ (check: $\gamma'z+iz\gamma_w=10.19+9.81=20.0$ ✓).
Situation (2) — upward. $h_{top}=3\text{ m}$, $h_{bot}=5\text{ m}$; at X-X, $h=4\text{ m}$ again, pressure head $=3\text{ m}$, $u_{XX}=29.43\text{ kPa}$. Now the top surface carries only 1 m of standing water ($\sigma_{top}=9.81\text{ kPa}$); $\sigma_{XX}=9.81+20(1)=29.8\text{ kPa}$, so $\sigma'_{XX}=29.8-29.4=\boxed{0.38\text{ kPa}}$ (check: $\gamma'z-iz\gamma_w=10.19-9.81=0.38$ ✓).
Situation (2)'s effective stress has collapsed to under 2% of the no-flow value ($\gamma'z=10.19\text{ kPa}$) because $i=1.0$ sits only 0.04 below $i_{cr}$ — a small additional head would trigger a quick (boiling) condition in that sample.
