22-Agric-A2 Soil Physics and Mechanics · December 2019
Question 3 of 6: Seepage Beneath a Concrete Dam (Flow Net)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A2 Soil Physics & Mechanics,
National Exams December 2019 — a three-hour open-book examination;
any non-communicating calculator is permitted. The cover page states that five (5)
questions constitute a complete exam paper and that only the first five as they appear
in the answer book are marked, that each question is of equal value, and that some
questions require a written answer whose clarity and organization matter for marks. All
six printed questions are worked here, because the set is a study resource rather than a
timed attempt; on exam day a candidate submits only the first five, in order.
Find. Total seepage Q (a); piezometric head at Point A (b); FS against
heave at the toe (c).
Flow net beneath the dam (schematic): 4 flow channels, 12 equipotential
drops, head loss H = 3.5 m; Point A sits on the third equipotential line from upstream,
within the bottom-most flow channel.
Approach. Read the seepage quantity, the head at any point, and the exit
gradient directly off the flow-net parameters (Nf, Nd, H), then compare
the exit gradient at the toe against the soil's critical (buoyant) gradient for the heave
check.
a) Total seepage flow rate. With k converted to m/s
($4.2\times10^{-4}\ \text{cm/s}=4.2\times10^{-6}\ \text{m/s}$), the standard flow-net
discharge per unit length of dam, scaled by the crest length L = 100 m:
$$q'=kH\frac{N_f}{N_d}=(4.2\times10^{-6})(3.5)\frac{4}{12}=4.90\times10^{-6}\ \text{m}^3/\text{s per m}$$
$$\begin{aligned}
Q&=q'L=4.90\times10^{-6}(100)=4.90\times10^{-4}\ \text{m}^3/\text{s}\\
&=\boxed{1.76\ \text{m}^3/\text{hr}}
\end{aligned}$$
b) Piezometric head at Point A. Each of the 12 equipotential drops
dissipates an equal share of the total head loss, $\Delta h=H/N_d=3.5/12=0.2917\ \text{m}$;
Point A sits on the third equipotential line counted from the upstream (high head) side, so
three drops have already occurred by the time flow reaches A. Taking the downstream
tailwater as the datum (head = 0):
$$h_A=H-3\Delta h=3.5-3(0.2917)=\boxed{2.63\ \text{m}}$$
above the downstream tailwater level.
c) Factor of safety against heave at the toe. The saturated and buoyant
unit weights of the sandy soil from Gs = 2.65 and n = 0.30
($e=n/(1-n)=0.4286$):
$$\begin{aligned}
\gamma_{sat}&=\frac{(G_s+e)}{1+e}\gamma_w=\frac{(2.65+0.4286)}{1.4286}(9.81)=21.14\ \text{kN/m}^3\\
\gamma'&=21.14-9.81=11.33\ \text{kN/m}^3
\end{aligned}$$
Terzaghi's exit-gradient method compares the critical (buoyant) gradient against the actual
exit gradient over the embedment depth d = 1.75 m using the head lost in the LAST
equipotential drop:
$$\begin{aligned}
i_{exit}&=\frac{\Delta h}{d}=\frac{0.2917}{1.75}=0.1667\\
i_{cr}&=\frac{\gamma'}{\gamma_w}=\frac{11.33}{9.81}=1.155
\end{aligned}$$
$$FS_{heave}=\frac{i_{cr}}{i_{exit}}=\frac{1.155}{0.1667}=\boxed{6.93}$$
