Question 5 of 6: Seepage

Question 5: Seepage (20 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. $L=30\text{ m}$ (dam base length), $H=20\text{ m}$ (permeable layer thickness), $h_t=10\text{ m}$ (reservoir level above the downstream water level, i.e. total head loss), $D=3\text{ m}$ (depth of the dam base below the ground surface, per the $D$ dimension on Figure Q5), $\gamma_{sat}=21.3\text{ kN/m}^3$, $\gamma_w=9.81\text{ kN/m}^3$, $k=3\times10^{-3}\text{ cm/s}$; $z_A=10\text{ m}$, $z_B=15\text{ m}$, $z_C=6\text{ m}$, $z_D=9\text{ m}$ (depths below the ground surface, measured downward as drawn). The printed "points a, b, c, d and e are 7.5 m apart" refers to the base points labelled 1–5 on the figure ($4\times7.5=30\text{ m}=L$).

Find. (a) Seepage rate per unit length of dam. (b) Total stress, pore-water pressure and effective stress at points A, B, C, D. (c) Pore-water pressure diagram and total uplift force along the dam base between points 1 and 5.

Approach. Count flow channels and equipotential drops on the printed net; seepage follows from $q=k\,h_t\,N_f/N_d$. Each drop removes $h_t/N_d$ of total head, so the drop index $n$ at a point gives its total head. Take the datum at the downstream ground/tailwater level with $z$ positive downward, so elevation head $=-z$ and pressure head $=h+z$; then $u=\gamma_w(h+z)$, $\sigma$ comes from the overlying soil plus any free water above the ground, and $\sigma'=\sigma-u$.

1. Part (a) — seepage rate. $k=3\times10^{-3}\text{ cm/s}=3\times10^{-5}\text{ m/s}$. $$q=k\,h_t\,\frac{N_f}{N_d}=(3\times10^{-5})(10)\left(\frac{4}{12}\right)=\boxed{1.0\times10^{-4}\text{ m}^3\text{/s per m}}\ (=0.10\text{ L/s per m run of dam})$$
2. Total head at A–D. Each drop removes $h_t/N_d=10/12=0.833\text{ m}$; the upstream ground is at $h=10\text{ m}$ and the downstream ground at $h=0$: $$h=h_t-n\frac{h_t}{N_d}:\quad h_A=10-1(0.833)=\boxed{9.17\text{ m}},\quad h_B=10-4(0.833)=\boxed{6.67\text{ m}}$$ $$h_C=10-8(0.833)=\boxed{3.33\text{ m}},\quad h_D=10-11(0.833)=\boxed{0.83\text{ m}}$$
3. Part (b) — pore-water pressure at A–D. Pressure head $=h-(-z)=h+z$: $$u_A=9.81(9.17+10)=\boxed{188.0\text{ kPa}},\quad u_B=9.81(6.67+15)=\boxed{212.6\text{ kPa}}$$ $$u_C=9.81(3.33+6)=\boxed{91.6\text{ kPa}},\quad u_D=9.81(0.83+9)=\boxed{96.5\text{ kPa}}$$
4. Part (b) — total and effective stress at A–D. A lies upstream of the dam beneath $h_t=10\text{ m}$ of reservoir water, which loads the ground surface; D lies downstream where the tailwater is at ground level. B and C lie beneath the dam; the dam's self-weight is not given, so their total stress is taken from the soil column $\gamma_{sat}z$ alone (state this assumption on the exam — any dam load would add equally to $\sigma$ and $\sigma'$ there). $$\sigma_A=\gamma_wh_t+\gamma_{sat}z_A=9.81(10)+21.3(10)=311.1\text{ kPa},\quad \sigma_B=21.3(15)=319.5\text{ kPa}$$ $$\sigma_C=21.3(6)=127.8\text{ kPa},\quad \sigma_D=21.3(9)=191.7\text{ kPa}$$ $$\sigma'=\sigma-u:\quad \sigma'_A=\boxed{123.1\text{ kPa}},\ \sigma'_B=\boxed{107.0\text{ kPa}},\ \sigma'_C=\boxed{36.2\text{ kPa}},\ \sigma'_D=\boxed{95.2\text{ kPa}}$$ All four effective stresses are positive, so no point is near a quick condition; the lowest is at C, beneath the downstream half of the dam where the seepage is rising toward the exit.
5. Part (c) — pore pressure along the base. The base points are at depth $D=3\text{ m}$ below the ground surface, on drops 2, 4, 6, 8, 10, so $u_i=\gamma_w\left(h_t-n_i\dfrac{h_t}{N_d}+D\right)$: $$u_1=9.81(8.33+3)=\boxed{111.2\text{ kPa}},\quad u_2=9.81(6.67+3)=\boxed{94.8\text{ kPa}},\quad u_3=9.81(5.00+3)=\boxed{78.5\text{ kPa}}$$ $$u_4=9.81(3.33+3)=\boxed{62.1\text{ kPa}},\quad u_5=9.81(1.67+3)=\boxed{45.8\text{ kPa}}$$ Successive points are 2 drops apart over equal 7.5 m spacings, so the pressure falls linearly by $2(0.833)(9.81)=16.35\text{ kPa}$ per segment.
6. Part (c) — total uplift force. The diagram is a trapezoid, so the trapezoidal sum over the four 7.5 m segments equals the mean end pressure times $L$: $$U=\frac{u_1+u_5}{2}L=\frac{111.2+45.8}{2}(30)=\boxed{2354\text{ kN per m run of dam}}$$