24-Bld-A6 Geotechnical Materials and Analysis · May 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, May 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.
Check: the exam text gives no soil unit weight for Question 4. Consistent with typical values of 19–20 kN/m³, this solution adopts a saturated unit weight $\gamma_{sat}=19\text{ kN/m}^3$ for the submerged soil below the ground surface, flagged explicitly here rather than left unstated.
Find. (i) the flow net; (ii) the seepage rate $q$ under the wall per unit length; (iii) the effective stress at Point A.
Figure 3 with flow net sketched: water seeps from the upstream (4.5 m head) side, around the sheet-pile tip at 6.0 m depth, to the downstream (0.5 m head) side. The narrow 2.60 m gap beneath the tip forces most flow lines to converge there, concentrating head loss in that region.
Approach. Sketch flow lines and equipotential lines to form curvilinear "squares" satisfying $q=k\Delta H(N_f/N_d)$; because the wall penetrates deep into the stratum (ratio $d_1/D=0.70$), the flow net is heavily "squeezed" toward the narrow $d_2=D-d_1=2.60\text{ m}$ gap beneath the tip, so a fine numerically-solved flow net (finite-difference solution of Laplace's equation for head, equivalent to an infinitely-refined hand-drawn net) is used here to fix $N_f/N_d$ precisely, then the same head field gives the pore pressure at Point A directly.
Set up the flow domain. Model the two half-planes on either side of the impermeable sheet pile (embedded to $d_1=6.0$ m in the $D=8.6$ m stratum), fixed head $H_u=4.5$ m upstream and $H_d=0.5$ m downstream at the ground surface, no-flow at the impervious base and at the sheet pile itself, and solve $\nabla^2h=0$ by finite differences (5-point stencil, harmonic-mean conductance across the wall to enforce zero flux through it). A convergence check (grid halved from $0.2\times0.05$ m to $0.1\times0.025$ m) changed the flow ratio by only $\sim1.3\%$, confirming the solution is grid-independent.
Read off the effective flow-net ratio. Integrating the horizontal Darcy flux across a vertical section clear of the wall gives, for $\Delta H=4.0$ m: $$\frac{N_f}{N_d}=\frac{q/k}{\Delta H}\approx0.070$$ — equivalent to a hand-sketched net of roughly 3 flow channels crossing about 43 equipotential drops, consistent with the deep penetration ratio forcing a long, narrow seepage path under the tip.
Compute the seepage rate. $$q=k\,\Delta H\left(\frac{N_f}{N_d}\right)=1.5\times10^{-5}\times4.0\times0.0704$$ $$\boxed{q\approx4.2\times10^{-6}\ \text{m}^3/\text{s per m of wall}}$$
Extract the head at Point A from the same solution. Point A sits at the pile-tip elevation ($y=d_1=6.0$ m below ground) on the downstream side, right where the flow net compresses most. The solved total head there is $h_A\approx2.28$ m (referenced to the ground surface datum), giving a pressure head of $h_A+y_A=2.28+6.00=8.28$ m and $$u_A=\gamma_w(h_A+y_A)=9.81\times8.28\approx81.2\text{ kPa}.$$
Total and effective stress at A. The downstream soil column above A carries the 0.5 m standing water plus 6.0 m of saturated soil: $$\sigma_A=\gamma_w H_d+\gamma_{sat}d_1=9.81(0.5)+19(6.0)=4.9+114.0=118.9\text{ kPa}$$ $$\boxed{\sigma'_A=\sigma_A-u_A=118.9-81.2\approx37.7\text{ kPa}}$$ For comparison, the *static* (no-flow) effective stress at the same point would be $\sigma_A-\gamma_w(H_d+d_1)=118.9-63.8=55.1$ kPa — seepage reduces the effective stress at A by about a third, because upward-tending flow near the tip counteracts some of the downward gravitational effective stress.
