Question 3 of 6: Seepage & Uplift Beneath a Concrete Gravity Dam

Question 3: Seepage & Uplift Beneath a Concrete Gravity Dam (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.

Find. (a) seepage volume beneath the dam, m³/day; (b) the uplift pressure distribution along the dam base and whether it is a concern.

Approach. With a flat, uniform-thickness pervious layer and the headwater/tailwater both flooding its surface directly at the dam's toes (no unflooded gap), the flow net for this cross-section reduces to the single-fragment case: horizontal, one-dimensional Darcy flow under the dam's own base, over the full layer thickness. The same linear head loss along the base gives the uplift-pressure distribution directly.

1. Part (a) — seepage rate. The total head drop across the seepage path is the headwater elevation above grade minus the tailwater elevation (also relative to grade): $\Delta H=12-0=12$ m. Applying Darcy's law across the dam footprint (thickness $a$, path length $B$), $$q=\frac{k\,a\,\Delta H}{B}=\frac{10\times40\times12}{40}=\boxed{120\ \text{m}^3/\text{day per m}}.$$ Multiplying by the dam's crest length $L=20$ m, $$Q=q\,L=120\times20=\boxed{2400\ \text{m}^3/\text{day}}.$$
2. Part (b) — total head along the base. Because the flow is essentially horizontal under a flat base, the total head falls off LINEARLY from the upstream toe ($x=0$, $h=\Delta H=12$ m) to the downstream toe ($x=B=40$ m, $h=0$): $$h(x)=12-\frac{12}{40}x=12-0.3x\ \ \text{(m, measured above the tailwater datum)}.$$ Since the dam base is flat (elevation head constant along it), the pressure head equals the total head at every point, so the uplift pressure is $$u(x)=\gamma_w\,h(x)=9.81\,(12-0.3x)\ \text{kPa}.$$ At the heel ($x=0$): $u=9.81\times12=\boxed{117.7\ \text{kPa}}$. At the toe ($x=40$ m): $u=\boxed{0}$ — a straight-line (triangular) distribution between the two.
3. Total uplift force and comment. Integrating the triangular distribution over the base and the dam's crest length, $$U=\tfrac12\,u_{\max}\,B\,L=\tfrac12\times117.7\times40\times20=\boxed{47{,}088\ \text{kN}\approx47.1\ \text{MN}}.$$ This is a substantial uplift force acting over the full 40 m × 20 m base — it reduces the effective normal stress (and hence the frictional sliding resistance) along the base by up to 117.7 kPa at the heel, tapering to none at the toe. Because the uplift is largest exactly where the driving reservoir thrust is also largest (near the heel), it works directly against both the sliding and the overturning factors of safety; without further reduction, a dam of this footprint and seepage path should be checked against sliding/overturning WITH this uplift included, and if the margin is tight, mitigation (a cutoff wall or grout curtain near the heel, and/or a toe drainage/relief-well system to reduce the pressure before it reaches the base) is the standard remedy.