Section 4: Soil and Water Conservation Engineering
Soil Mechanics
Subtopic: Permeability and seepage analysis
Question Statement:
A homogeneous earthen embankment has an upstream water level of 12 m. The top width of the embankment is 3 m, and its upstream and downstream slopes (horizontal : vertical) are 2:1 and 2.5:1 respectively. A freeboard of 2 m is provided. The coefficient of permeability of the fill material is 2.5 × 10⁻⁵ m·s⁻¹. If a 20 m long horizontal filter is placed inward from the downstream toe, the seepage discharge per unit length of the embankment is n × 10⁻⁵ m³·s⁻¹·m⁻¹. The value of n is ________. (Rounded off to two decimal places)
Official Answer Key:
5.91 to 5.95
Step-by-Step Solution & Explanation
This is solved by Casagrande’s basic-parabola method, with the focus at the upstream end of the horizontal filter.
Step 1 — embankment geometry.
Total height = water level + freeboard = 12 + 2 = 14 m
Base width = (u/s slope projection) + (top width) + (d/s slope projection)
= 2(14) + 3 + 2.5(14) = 28 + 3 + 35 = 66 m
Step 2 — locate the focus F (upstream end of the filter), measured from the upstream toe:
x_F = 66 − 20 = 46 m
Step 3 — locate the corrected entry point A of the parabola.
The water surface meets the upstream slope at horizontal distance Δ = 2 × 12 = 24 m from the u/s toe.
Casagrande’s entry correction places the starting point of the basic parabola 0.3Δ upstream of that point:
x_A = 24 − 0.3(24) = 24 − 7.2 = 16.8 m, at height h = 12 m
Step 4 — horizontal distance from A to the focus:
d = 46 − 16.8 = 29.2 m
Step 5 — seepage discharge per unit length:
q = k(√(d² + h²) − d)
= 2.5 × 10⁻⁵ × (√(29.2² + 12²) − 29.2)
= 2.5 × 10⁻⁵ × (√(852.64 + 144) − 29.2)
= 2.5 × 10⁻⁵ × (31.57 − 29.2)
= 2.5 × 10⁻⁵ × 2.37
= 5.92 × 10⁻⁵ m³·s⁻¹·m⁻¹
Hence n = 5.92.
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