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Seepage expressions: no drain vs toe drain

Short answer

For a homogeneous dam without drain, Casagrande’s method uses a basic parabola with focus at the downstream toe: q = k y₀, y₀ = √(h²+d²) – d, and an exit correction is needed. With a toe drain, the focus is at the drain’s upstream end, giving q = k y₀ with no exit correction and the line entering the drain horizontally.

Write with neat sketches the expressions for computing seepage and the phreatic surface in earthen dams for two cases: homogeneous dam without drain and dam with toe drain.

Seepage through a homogeneous earth dam on an impervious base is found by Casagrande's method, which treats the phreatic line as a basic parabola with its focus FF at the d/s toe (or at the start of the drain), based on Dupuit's assumption. Notation: hh = reservoir depth above the base, bb = horizontal projection of the wetted u/s slope, BB = base width, α\alpha = angle of the d/s face.

Case 1: Homogeneous dam without a drain

       A'  entry corrected 0.3b
   h ~~~~\___  phreatic line (parabola)
        /     \____
       /  dam      \___ a (exit point on d/s face)
   ///////////////////F (toe) \\\
        |<----- d ------>|
  • Horizontal distance from the corrected entry to the toe: d=B−0.7bd=B-0.7b.
  • Basic parabola: y0=h2+d2−dy_0=\sqrt{h^2+d^2}-d and y=y02+2y0xy=\sqrt{y_0^2+2y_0x}, with xx from FF.
  • Seepage (per unit length): q=k y0q=k\,y_0.
  • Exit point on the d/s face (distance aa along the slope from the toe):
a+Δa=y01−cos⁡α,a=(1−C)(a+Δa)a+\Delta a=\frac{y_0}{1-\cos\alpha},\qquad a=(1-C)(a+\Delta a)

where C=Δa/(a+Δa)C=\Delta a/(a+\Delta a) is read from Casagrande's chart (0.36 at 30∘30^\circ, 0.26 at 90∘90^\circ, 0.10 at 150∘150^\circ, 0 at 180∘180^\circ).

  • For α<30∘\alpha<30^\circ, Schaffernak's formula is also used:
q=k asin⁡αtan⁡α,a=dcos⁡α−d2cos⁡2α−h2sin⁡2αq=k\,a\sin\alpha\tan\alpha,\qquad a=\frac{d}{\cos\alpha}-\sqrt{\frac{d^2}{\cos^2\alpha}-\frac{h^2}{\sin^2\alpha}}

Case 2: Dam with a toe drain

   h ~~~~\___  phreatic line
        /     \___
       /  dam      \___
   ///////////////[ toe drain ]\\\
        |<--- d --->|F (start of drain)
  • The parabola is drawn with the focus at the u/s end of the toe drain, so d=B−0.7b−ℓd=B-0.7b-\ell (ℓ\ell = drain length).
  • y0=h2+d2−dy_0=\sqrt{h^2+d^2}-d, y=y02+2y0xy=\sqrt{y_0^2+2y_0x} and q=k y0q=k\,y_0.
  • The line enters the drain nearly horizontally (α=180∘\alpha=180^\circ, C=0C=0), so no exit correction is required and the line meets the drain at a=y01−cos⁡180∘=y02a=\dfrac{y_0}{1-\cos180^\circ}=\dfrac{y_0}{2} from the focus. The d/s face stays dry above the drain, which improves the stability of the d/s slope.

✓ Answer checked by Claude Opus 5.5 (high effort)

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