Classification: 18-Geol-A6 Soil Mechanics, December 2019, Q1

Question 1: Classification (10 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. The percent-finer sieve data above for two soils, plus Atterberg limits: Soil A — $LL=30\%$, $PL=27\%$; Soil B — $LL=40\%$, $PL=30\%$.

Find. Plot both grain-size distribution curves and assign each soil its Unified Soil Classification System (USCS) group symbol.

Grain-size distribution curves for Soils A and B (semi-log), with Soil A's $D_{10}$, $D_{30}$, $D_{60}$ marked.

Approach. Read $D_{10}$, $D_{30}$, $D_{60}$ off the curve to get $C_u$ and $C_c$ for the coarse fraction, split each soil into gravel/sand/fines by the No. 4 and No. 200 sieves, then walk the USCS coarse-/fine-grained branch using the fines percentage and the plasticity chart.

Split Soil A into gravel, sand, and fines. Percent retained on No. 4 (4.76 mm) $=100-72=28\%$ gravel; percent between No. 4 and No. 200 (75 $\mu$m) $=72-8=64\%$ sand; percent passing No. 200 $=8\%$ fines. The coarse fraction (gravel+sand$=92\%$) exceeds half the sample, so Soil A is coarse-grained; within the coarse fraction, sand ($64\%$) exceeds gravel ($28\%$), so the base symbol is S (sand).
Grade Soil A's coarse fraction. Interpolating the semi-log curve (log-linear between bracketing sieve points) for the 10/30/60% finer sizes: $$D_{10}=0.106\text{ mm}, \quad D_{30}=0.336\text{ mm}, \quad D_{60}=3.26\text{ mm}$$ $$C_u=\frac{D_{60}}{D_{10}}=\frac{3.26}{0.106}=30.8 \qquad C_c=\frac{(D_{30})^2}{D_{10}D_{60}}=\frac{0.336^2}{0.106\times 3.26}=0.33$$
Apply the fines-content branch. Soil A's fines ($8\%$) fall in the $5$–$12\%$ borderline band, which requires a dual symbol (gradation symbol + plasticity symbol). Gradation: SW needs $C_u>6$ (met, $30.8$) and $1SP. Plasticity: $I_p=LL-PL=30-27=3\%<4$, so the fines plot below the $I_p=4$ cut → plasticity half is M. $$\boxed{\text{Soil A} = \text{SP--SM (poorly graded sand with silt)}}$$
Split and classify Soil B. Gravel $=100-91=9\%$; sand $=91-55=36\%$; fines $=55\%$. Fines exceed $50\%$, so Soil B is fine-grained, classified from its position on the plasticity (A-line) chart rather than its gradation. $I_p=LL-PL=40-30=10\%$. Liquid limit $LL=40\%<50\%$, so the "L" (low-to-medium plasticity) branch applies.
Locate Soil B relative to the A-line. A-line: $I_p=0.73(LL-20)=0.73(40-20)=14.6\%$. Since the measured $I_p=10\% < 14.6\%$, Soil B plots below the A-line (and is well clear of the $I_p=4$–$7$ borderline hatch), placing it in the silt (M) region with $LL<50\%$: $$\boxed{\text{Soil B} = \text{ML (inorganic silt of low plasticity, sandy)}}$$

Final results
Quantity	Soil A	Soil B
Gravel / Sand / Fines	28% / 64% / 8%	9% / 36% / 55%
$C_u$, $C_c$ (coarse fraction)	30.8, 0.33	— (fine-grained)
$I_p$	3%	10%
USCS group symbol	SP–SM	ML

Topic: Unified Soil Classification System (USCS).

Key relations: $C_u=D_{60}/D_{10}$; $C_c=(D_{30})^2/(D_{10}D_{60})$; A-line $I_p=0.73(LL-20)$; coarse/fine split at 50% passing No. 200; gravel/sand split at No. 4.

Why this works: USCS first separates soils by which half of the sample (by mass) is coarser or finer than 75 $\mu$m, because that fraction controls whether strength/permeability behaviour is governed by grain packing (coarse) or by plasticity/water content (fine). Coarse soils are then graded by their size-distribution shape ($C_u$, $C_c$); fine soils are placed on the plasticity chart because the A-line empirically separates clay-like (above) from silt-like (below) engineering behaviour at a given liquid limit.

Common pitfall: forgetting the 5–12% fines borderline band requires a dual symbol (most students jump straight to a single SW/SP or SM/SC symbol), and swapping the A-line's $LL-20$ for $LL$ itself.

Source: Das, Principles of Geotechnical Engineering, Ch. 4 (soil classification); ASTM D2487 USCS criteria (reproduced on the exam's own classification sheet, p. 15).

Question 1 of 6: Classification

Question 1: Classification (10 marks)