Question 5 of 6: Seepage

Question 5: Seepage (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $L=20$ m, $H=10$ m (pervious layer thickness), $h_t=10$ m (headwater above ground; downstream water sits at ground level, i.e.\ zero tailwater), $D=1$ m sheet-pile embedment, $\gamma_w=10\text{ kN/m}^3$, $k=6\times10^{-3}\text{ cm/s}$. From the flow net: $N_f=3$ flow channels, $N_d=8$ equipotential drops.

Find. (1) $q$ per metre run of dam. (2) $h_{total}$, elevation head, pressure head at A, B, C, D. (3) Pore pressure at points 1–5 and the resulting uplift force.

Approach. Seepage quantity follows directly from $q=k\,\Delta h_{total}\,(N_f/N_d)$. Total head at any point is fixed by which equipotential (drop number $n$) it lies on: $h_{total}=h_{headwater}-n(\Delta h_{total}/N_d)$; pressure head follows from $h_{total}-z$, and pore pressure from $u=\gamma_w\times(\text{pressure head})$.

[Figure not reproduced: Figure Q5 (redrawn, schematic) — flow net beneath the dam/sheet-pile cutoff. Equipotential lines (blue) are numbered by drop from the headwater; point A sits on drop 1, points 1–5 on drops 2–6 respectively (B coincides with point 2's drop, C with point 4's drop), and D on drop 7. See the official exam paper.]

1. Part (1) — seepage quantity. With $k=6\times10^{-3}\text{ cm/s}=6\times10^{-5}\text{ m/s}$ and downstream water at ground level (so the full head loss is $\Delta h_{total}=h_t=10$ m): $$q=k\,\Delta h_{total}\left(\frac{N_f}{N_d}\right)=6\times10^{-5}(10)\left(\frac{3}{8}\right)=\boxed{2.25\times10^{-4}\text{ m}^3\text{/s per m run}}$$ equivalently $\boxed{19.4\text{ m}^3/\text{day per m run}}$.
2. Part (2) — head at each equipotential drop. Head loss per drop: $\Delta h=h_t/N_d=10/8=1.25$ m. Taking the elevation datum at the impervious layer ($z=0$), the ground surface/dam base sits at $z=H=10$ m and the headwater surface at $z=H+h_t=20$ m, so $$h_{total}(n)=20-1.25n$$ For A ($n=1$), B ($n=3$, shares point 2's equipotential), C ($n=5$, shares point 4's), D ($n=7$): $$h_{total,A}=18.75\text{ m},\quad h_{total,B}=16.25\text{ m},\quad h_{total,C}=13.75\text{ m},\quad h_{total,D}=11.25\text{ m}$$
3. Part (2) — pressure head and pore pressure. Pressure head $\psi=h_{total}-z$; pore pressure $u=\gamma_w\psi$. $$\psi_A=18.75-10=\boxed{8.75\text{ m}},\ u_A=\boxed{87.5\text{ kPa}}$$ $$\psi_B=16.25-15=\boxed{1.25\text{ m}},\ u_B=\boxed{12.5\text{ kPa}}$$ $$\psi_C=13.75-6=\boxed{7.75\text{ m}},\ u_C=\boxed{77.5\text{ kPa}}$$ $$\psi_D=11.25-9=\boxed{2.25\text{ m}},\ u_D=\boxed{22.5\text{ kPa}}$$ All four are positive, as required for saturated soil beneath the dam.
4. Part (3) — pore pressure at points 1–5. All five points sit at the dam-base elevation $z=H=10$ m, on drops $n=2,3,4,5,6$ respectively. $$u(n)=\gamma_w\big[h_{total}(n)-10\big]=10\big[(20-1.25n)-10\big]=10(10-1.25n)$$ $$u_1=75\text{ kPa},\ u_2=62.5\text{ kPa},\ u_3=50\text{ kPa},\ u_4=37.5\text{ kPa},\ u_5=25\text{ kPa}$$ The pressure falls linearly (equal drops, equal 5 m spacing).
5. Part (3) — total uplift force. Trapezoidal integration of $u(x)$ over the 20 m base (equivalently, since the profile is exactly linear, the average ordinate times the base length): $$U_{plift}=\frac{u_1+u_5}{2}\times L=\frac{75+25}{2}(20)=\boxed{1000\text{ kN per m run of dam}}$$