Subgrade Intrusion and Trench-Drain Capacity: 16-Civ-B7 Transportation Planning and Engineering, May 2016, Q5

Question 5: Subgrade Intrusion and Trench-Drain Capacity (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.

Quantity	Symbol	Value
Subgrade gradation	—	100 % at 2.0 mm; 85 % at 0.300 mm; 60 % at 0.150 mm; 42 % at 0.075 mm; 15 % at 0.040 mm
Subbase 15 % size	D₁₅	2.1 mm
Drain pipe diameter	D	4 in = 0.3333 ft
Manning roughness	n	0.01
Pipe slope	S	2.5 % = 0.025 ft/ft
Outlet spacing	L	300 ft

Find. (a) whether the proposed subbase satisfies the filter criterion against the given subgrade, and (b) the lateral inflow per unit length that just fills the pipe between outlets.

Subgrade grain-size distribution with D85 read at the 85 per cent passing line, compared with the proposed subbase D15 of 2.1 mm.

Approach. Part (a) reads the subgrade's 85 per cent size straight off the gradation and compares the subbase D₁₅ with it through the Terzaghi piping criterion. Part (b) computes the full-flow capacity of the drain with Manning's equation, converts it to a daily volume, and spreads that volume over the 300 ft that drains to each outlet.

Part (a) — read the subgrade D₈₅. D₈₅ is the sieve size through which 85 per cent of the subgrade passes. The table gives exactly 85 per cent passing the 300 µm sieve, so no interpolation is needed: $$D_{85(\text{subgrade})} = 0.300\ \text{mm}$$
Apply the filter (piping) criterion. A coarse layer will not allow the finer material beneath it to migrate upward into its voids provided $$\frac{D_{15(\text{subbase})}}{D_{85(\text{subgrade})}} \le 5$$ Substituting the proposed subbase, $$\frac{2.1}{0.300} = 7.0$$ $$\boxed{\frac{D_{15}}{D_{85}} = 7.0 > 5 \Rightarrow \text{criterion violated}}$$
Interpret and give the remedy. The voids in the proposed subbase are large enough for the subgrade fines to work up into them under repeated wheel loading and the pore-pressure pulses that accompany it. The consequences are progressive contamination and loss of permeability in the drainage layer, loss of support beneath the pavement, and eventually rutting and fatigue cracking. To satisfy the criterion the subbase would need $$D_{15} \le 5(0.300) = 1.5\ \text{mm}$$ so either a finer, better-graded subbase must be specified, or the existing material must be separated from the subgrade by a well-graded transition (filter) layer or, in current Canadian practice, by a geotextile separator selected on its apparent opening size. It is worth noting for completeness that the companion permeability criterion, $D_{15(\text{subbase})} \ge 5\,D_{15(\text{subgrade})}$, is easily met here — the difficulty is entirely one of retention, not of drainage.
Part (b) — geometry of the full pipe. Working in foot units with $D = 4/12 = 0.3333$ ft, a pipe flowing just full has $$\begin{aligned} A &= \frac{\pi D^2}{4} = \frac{\pi (0.3333)^2}{4} = 0.08727\ \text{ft}^2 \\ R &= \frac{D}{4} = 0.08333\ \text{ft} \end{aligned}$$
Manning's equation in U.S. customary units. The full-flow capacity is $$Q = \frac{1.486}{n}A R^{2/3} S^{1/2} = \frac{1.486}{0.01}(0.08727)(0.08333)^{2/3}(0.025)^{1/2}$$ Evaluating the terms, $(0.08333)^{2/3} = 0.19079$ and $(0.025)^{1/2} = 0.15811$, so $$\boxed{Q = 0.391\ \text{ft}^3/\text{s}}$$ The corresponding full-flow velocity is $v = Q/A = 0.391/0.08727 = 4.48$ ft/s, which is above the 3 ft/s usually wanted for self-cleansing and below the roughly 10 ft/s at which abrasion becomes a concern — the pipe is on a sensible grade.
Convert to a daily volume. Over a full day, $$Q_{\text{day}} = 0.391(86\,400) = 33\,798\ \text{ft}^3/\text{day}$$
Spread the capacity over the drained length. Each outlet serves 300 ft of trench, and the pipe is fullest immediately upstream of the outlet, so the whole of the capacity must be supplied by that 300 ft: $$q = \frac{Q_{\text{day}}}{L} = \frac{33\,798}{300}$$ $$\boxed{q_{\max} = 113\ \text{ft}^3/\text{day per ft of drain}}$$ Any greater inflow would surcharge the pipe, raise the water table into the pavement structure and defeat the purpose of the drain; closer outlet spacing, a larger pipe or a steeper grade would then be required. As a sanity check on magnitude, 113 ft³/day/ft over a 24 ft-wide pavement is about 4.7 ft³/day per square foot, far more than any realistic infiltration rate — the drain is generously sized for its purpose.

Quantity	Result
Subgrade D₈₅	0.300 mm
D_15(subbase) / D_85(subgrade)	7.0 (limit 5)
(a) Verdict	subgrade intrusion WILL occur — criterion violated
Maximum acceptable subbase D₁₅	1.5 mm
Pipe area / hydraulic radius	0.0873 ft² / 0.0833 ft
Full-flow capacity, Q	0.391 ft³/s (33 798 ft³/day)
Full-flow velocity	4.48 ft/s
(b) Maximum lateral inflow	113 ft³/day per ft

Topic. Filter criteria between pavement layers, and the hydraulic capacity of a longitudinal edge drain.

Key relations. Retention, $D_{15(\text{filter})}/D_{85(\text{soil})}\le 5$; permeability, $D_{15(\text{filter})}/D_{15(\text{soil})}\ge 5$; Manning full-flow, $Q=(1.486/n)AR^{2/3}S^{1/2}$ with $A=\pi D^2/4$ and $R=D/4$; lateral inflow $q = Q/L$ for an outlet spacing L.

Why this works. The retention criterion is a statement about pore size. The voids between the particles of a granular layer scale with its own fine fraction, and the constriction they present is roughly one-fifth of the D₁₅ size; if the soil beneath has particles smaller than that constriction — which is what a ratio above five means — the fines can be carried through by seeping water and by the pumping action of traffic. The drain calculation rests on a different idea: a pipe on a uniform grade reaches a definite capacity when it flows just full, and because inflow accumulates along the run, the section just upstream of an outlet carries everything collected since the previous one. Dividing capacity by spacing therefore converts a pipe capacity into the design quantity an engineer actually needs, an allowable inflow per unit length that can be compared against infiltration through joints and cracks.

Common pitfall. Inverting the criterion and comparing the subbase D₈₅ with the subgrade D₁₅, which reverses the conclusion. In part (b), using the SI form of Manning's equation with foot units — the coefficient is 1.486 for feet and seconds, and 1.0 only in metres — and forgetting the final division by the 300 ft outlet spacing, which leaves an answer three hundred times too large.

Source. Das, Principles of Geotechnical Engineering, 9th ed., §8.9 (filter design criteria); Garber & Hoel, 5th ed., §18.5 (pavement drainage and edge drains); Chow, Open-Channel Hydraulics, Ch. 5 (uniform flow, Manning's equation).

Question 5 of 7: Subgrade Intrusion and Trench-Drain Capacity

Question 5: Subgrade Intrusion and Trench-Drain Capacity (20 marks)