Question 2 of 6: USCS Classification & Landfill Cover Suitability

Question 2: USCS Classification & Landfill Cover Suitability (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. Grain-size curve read from Figure 1 (percent RETAINED on each sieve size), plus Atterberg limits.

Liquid limit $LL=50\%$, plastic limit $PL=30\%$ (plasticity index $PI=20$).

[Figure not reproduced: Figure 1 — grain-size distribution replotted as percent passing (100 − percent retained) from the source's percent-retained curve. See the official exam paper.]

Find. (a) USCS group symbol and group name; (b) suitability of this soil as landfill cover material.

Approach. First split the sample into gravel/sand/fines fractions from the percent-passing curve; because the fines fraction turns out to be under 5%, the Atterberg limits do not govern and the soil is classified purely on gradation ($C_u$, $C_c$) as a clean sand or gravel.

1. Part (a) — fines/sand/gravel split. Percent retained is the complement of percent passing, so percent passing $=100-\text{retained}$ (table above). At the #200 sieve (0.075 mm), only $100-98=2\%$ passes — fines $=2\%$. The curve plateaus at 20% retained beyond 4.0 mm (the #4/4.75 mm sieve, effectively), so $$\text{gravel}=100-80=20\%,\qquad \text{sand}=80-2=78\%,\qquad \text{fines}=2\%.$$ Since fines $<5\%$, this is a CLEAN coarse-grained soil: Atterberg limits are not used for classification (they matter only once fines $\ge 5\%$), and because sand (78%) exceeds gravel (20%), the base group is a SAND.
2. Grading coefficients from the passing curve. Reading $D_{10}$, $D_{30}$, $D_{60}$ by log-linear interpolation between the tabulated points ($D_{10}$ falls exactly on a plotted point): $$D_{10}=0.10\ \text{mm},\qquad D_{30}=0.29\ \text{mm},\qquad D_{60}=0.63\ \text{mm}.$$ $$C_u=\frac{D_{60}}{D_{10}}=\frac{0.626}{0.100}=6.26,\qquad C_c=\frac{D_{30}^2}{D_{10}D_{60}}=\frac{0.292^2}{0.100\times0.626}=\boxed{C_u=6.26,\ C_c=1.36}.$$
3. Apply the well-graded-sand test. USCS requires $C_u\ge6$ AND $1\le C_c\le3$ for SW; both are satisfied ($6.26\ge6$, $1\le1.36\le3$), so the soil is well-graded. With gravel at 20% ($\ge15\%$), ASTM D2487 group-naming adds "with gravel": $$\boxed{\text{Group symbol SW}\ \ -\ \ \text{Group name: Well-graded sand with gravel}}.$$

Part (b). A modern municipal solid waste landfill final cover is a layered system: a vegetative/topsoil layer, a protective/drainage layer (to shed infiltrating water laterally before it can reach the waste), and — doing the actual containment work — a low-permeability BARRIER layer (compacted clay at $k\lesssim10^{-7}$–$10^{-9}\ \text{cm/s}$, or a geomembrane/GCL) intended to minimize water infiltration into the waste mass and limit leachate generation. A clean, well-graded SAND like this one is exactly the wrong material for that barrier role: an SW soil with $D_{10}\approx0.1$ mm and only 2% fines is highly permeable ($k$ typically $10^{-2}$–$10^{-4}$ cm/s for a well-graded medium sand), so infiltrating precipitation would pass through it with little resistance, defeating the purpose of a low-permeability cap. It is UNSUITABLE, in its natural state, as the barrier component of the cover. The same properties that disqualify it as a barrier — high permeability, good compactability, free-draining behaviour, and resistance to erosion once vegetated — make it a good candidate for the OTHER layers of the cover system: the drainage layer directly above the barrier (where high $k$ is exactly what is wanted, to move infiltrating water laterally to a collection system), the protective/foundation layer that cushions the barrier from equipment traffic and root penetration, or general grading fill. In short: not suitable as the impermeable cap itself, but a reasonable and even desirable material for the drainage or protective layers of the same cover system.