Question 5 of 7: Filter Design and Drainage-Layer Capacity

Question 5: Filter Design and Drainage-Layer Capacity (8 + 12 = 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) the D₁₅ and D₈₅ sizes that a soil filter placed against this subgrade must have; (b) the steady-state discharge capacity of the drainage layer per metre of width, and the times to 50 percent and 95 percent drainage.

Approach. Part (a): read D₁₅, D₅₀ and D₈₅ off the subgrade curve, apply the three Terzaghi/Cedergren filter criteria to bound the filter's D₁₅, then fix D₈₅ by keeping the filter's gradation curve roughly parallel to the subgrade's and checking it against the pipe-bridging requirement. Part (b): Darcy's law along the cross slope for the steady capacity, and the Casagrande–Shannon time factor for the transient drawdown.

1. Part (a) — read the controlling sizes off the subgrade gradation. Two of the three fall exactly on printed sieves: $$D_{15} = 0.15\ \text{mm} \ (15 \text{ percent passing}), \qquad D_{85} = 1.18\ \text{mm} \ (85 \text{ percent passing})$$ and D₅₀ is interpolated logarithmically between 0.30 mm (35 percent) and 0.60 mm (65 percent): $$D_{50} = 0.30 \times 2^{(50-35)/(65-35)} = 0.30\sqrt{2} = 0.424\ \text{mm}$$
2. Apply the retention (anti-piping) criterion. The filter must be fine enough that subgrade particles cannot wash through its pore throats: $$\frac{D_{15}^{F}}{D_{85}^{S}} \le 5 \quad\Longrightarrow\quad D_{15}^{F} \le 5 \times 1.18 = 5.90\ \text{mm}$$
3. Apply the permeability criterion. The filter must also be coarse enough to be substantially more permeable than the soil it drains, otherwise it simply becomes part of the subgrade: $$\frac{D_{15}^{F}}{D_{15}^{S}} \ge 5 \quad\Longrightarrow\quad D_{15}^{F} \ge 5 \times 0.15 = 0.75\ \text{mm}$$ The two criteria together define the design window $$\boxed{0.75\ \text{mm} \le D_{15}^{F} \le 5.90\ \text{mm}}$$
4. Select the filter D₁₅. Taking the geometric mean of the window keeps equal proportional margin against piping and against clogging: $$\sqrt{0.75 \times 5.90} = 2.10\ \text{mm} \quad\Longrightarrow\quad \boxed{D_{15}^{F} = 2.0\ \text{mm}}$$
5. Fix the filter D₈₅. A filter is specified as a band, and the standard way to generate it is to shift the subgrade curve to the right by the ratio just chosen, keeping it roughly parallel so that the filter is itself internally stable. The subgrade has $$\frac{D_{85}^{S}}{D_{15}^{S}} = \frac{1.18}{0.15} = 7.87$$ so the parallel filter has $$D_{85}^{F} = 2.0 \times 7.87 = 15.7\ \text{mm} \quad\Longrightarrow\quad \boxed{D_{85}^{F} \approx 16\ \text{mm}}$$
6. Check the selection against the remaining criteria. The uniformity criterion is satisfied, $$D_{50}^{F} = 2.0 \times \frac{0.424}{0.15} = 5.66\ \text{mm} \le 25\,D_{50}^{S} = 25 \times 0.424 = 10.6\ \text{mm}\ \checkmark$$ and the filter is coarse enough to bridge the openings of a slotted collector pipe: for a typical 3 mm slot the requirement D₈₅^F ≥ 2 × slot width = 6 mm is met more than twice over. The resulting specification is a well-graded sand to fine gravel running roughly from 0.75 mm to 20 mm.
7. Part (b) — compute the steady-state capacity. Under steady infiltration the drainage layer flows full and parallel to the cross slope, so Darcy's law with a hydraulic gradient equal to the slope gives, per metre of width, $$q = k\,H\,S = 0.035 \times 0.200 \times 0.04 = 2.80 \times 10^{-4}\ \text{m}^{3}/\text{s per m}$$ $$\boxed{q = 0.28\ \text{L/s per metre of width} = 24.2\ \text{m}^{3}/\text{day per metre}}$$ Equivalently, expressed as the infiltration the layer can accept over its 6 m length, 24.2/6 = 4.03 m/day, which is far above any credible pavement infiltration rate — the layer is not capacity-limited.
8. Form the time constant for the transient drawdown. The Casagrande–Shannon formulation writes the time to drain as a dimensionless time factor multiplied by $$m = \frac{n_{e}\,L^{2}}{k\,H} = \frac{0.25 \times 6^{2}}{0.035 \times 0.200} = \frac{9.0}{0.0070} = 1286\ \text{s} = 0.357\ \text{h}$$ with the slope factor $$S_{1} = \frac{L\,S}{H} = \frac{6 \times 0.04}{0.200} = 1.20$$
9. Evaluate the time factors. For S₁ of order one or more the drawdown is slope-driven: the drying front travels down the layer at the kinematic speed kS/n_e, so the degree of drainage is U = t kS/(n_eL) and the time factor reduces to T = U/S₁: $$T_{50} = \frac{0.50}{1.20} = 0.417, \qquad T_{95} = \frac{0.95}{1.20} = 0.792$$ The horizontal-layer limit of the same solution, T₅₀ = 0.50, provides the upper bound used as a check below.
10. Compute the drainage times. $$t_{50} = T_{50}\,m = 0.417 \times 1286 = 536\ \text{s}, \qquad t_{95} = T_{95}\,m = 0.792 \times 1286 = 1018\ \text{s}$$ $$\boxed{t_{50} = 8.9\ \text{min} \quad\text{and}\quad t_{95} = 17.0\ \text{min}}$$ The bound confirms the order of magnitude: even treating the layer as perfectly flat gives t₅₀ = 0.50 × 1286 s = 10.7 min, so the answer is bracketed within two minutes regardless of how the slope is handled, and complete drainage takes n_eL/(kS) = 17.9 min.
11. Interpret the result for design. AASHTO rates the quality of drainage by the time taken to remove water: 2 hours is "excellent" and 1 day is "good". At under 20 minutes to 95 percent drainage this layer is comfortably in the excellent class, which justifies drainage coefficients m₂ = m₃ in the range 1.25–1.40 for the granular layers it serves — a structural saving of roughly 25 percent of the granular thickness compared with the m = 1.00 assumed in Question 3.

Final Results.