Classification: 18-Geol-A6 Soil Mechanics, May 2016, Q1

Question 1: Classification (15 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. Percent-passing grain-size data for Soils A and B (Table Q1, below); Soil A is non-plastic; Soil B has $LL=70\%$, $PL=25\%$.

Find. USCS group symbol and group name for Soil A and Soil B.

Approach. Plot both gradations on a semi-log grain-size chart, read $D_{10}$, $D_{30}$, $D_{60}$ for Soil A to get $C_u$, $C_c$, and locate Soil B's $LL$/$PI$ on the plasticity chart relative to the A-line.

Sieve	Size (mm)	Soil A, % passing	Soil B, % passing
3 in	75	100	100
2 in	50	100	100
1 in	25	95	100
0.75 in	19	90	100
0.375 in	9.5	75	100
No. 4	4.76	70	100
No. 8	2.38	55	100
No. 20	0.84	35	97
No. 40	0.42	25	92
No. 100	0.15	15	82
No. 200	0.075	7	75

Fig. Q1 — grain-size distribution, Soils A and B, with $D_{10}$/$D_{30}$/$D_{60}$ construction for Soil A.

Soil A — read $D_{10}$, $D_{30}$, $D_{60}$ (log-linear interpolation between bracketing sieve points). $D_{60}$ (60% passing, between 4.76 mm/70% and 2.38 mm/55%): $D_{60}=\boxed{3.00\text{ mm}}$. $D_{30}$ (30% passing, between 0.84 mm/35% and 0.42 mm/25%): $D_{30}=\boxed{0.594\text{ mm}}$. $D_{10}$ (10% passing, between 0.15 mm/15% and 0.075 mm/7%): $D_{10}=\boxed{0.0973\text{ mm}}$.
Soil A — gradation coefficients. $C_u=\dfrac{D_{60}}{D_{10}}=\dfrac{3.00}{0.0973}=\boxed{30.8}$; $\quad C_c=\dfrac{D_{30}^2}{D_{60}D_{10}}=\dfrac{0.594^2}{3.00\times0.0973}=\boxed{1.21}$.
Soil A — group symbol. Passing No. 200 $=7\%<50\%$ → coarse-grained. Of the coarse fraction (93%), sand-size (No.4–No.200) $=70-7=63\%$ exceeds gravel-size (retained No.4) $=30\%$ → a SAND. Fines $=7\%$ falls in the 5–12% dual-symbol band (neither "clean" nor "with fines"), so both the grading symbol and the fines symbol are reported. $C_u=30.8\ge6$ and $1\le C_c=1.21\le3$ → well graded (W); fines stated non-plastic → silt (M). Gravel $=30\%\ge15\%$ of the total sample → the group name carries a "with gravel" modifier. $\Rightarrow \boxed{\text{SW-SM, well-graded sand with silt and gravel}}$.
Soil B — group symbol. Passing No. 200 $=75\%\ge50\%$ → fine-grained. $PI=LL-PL=70-25=\boxed{45}$. A-line: $PI_{A}=0.73(LL-20)=0.73(70-20)=36.5$. Since $PI=45>PI_A=36.5$, the point plots above the A-line → clay (C); $LL=70\ge50$ → high plasticity (H). (Check: U-line $PI_U=0.9(LL-8)=55.8>45$, so the point is a valid, physically possible plot.) All of Soil B passes No. 4 (0% gravel); the sand-size fraction $=100-75=25\%\ge15\%$ of total → "with sand" modifier. $\Rightarrow \boxed{\text{CH, fat clay with sand}}$.

Quantity	Soil A	Soil B
$D_{10}$, $D_{30}$, $D_{60}$ (mm)	0.0973, 0.594, 3.00	—
$C_u$, $C_c$	30.8, 1.21	—
$PI$ (%)	NP	45
USCS group	SW-SM	CH
Group name	Well-graded sand with silt and gravel	Fat clay with sand

Concept & Theory

Topic. Unified Soil Classification System (USCS) for coarse- and fine-grained soils.

Key relations. $C_u=D_{60}/D_{10}$; $C_c=D_{30}^2/(D_{60}D_{10})$; A-line $PI=0.73(LL-20)$; U-line $PI=0.9(LL-8)$.

Why this works. $C_u$ and $C_c$ quantify how continuously the particle sizes are distributed (well- vs. poorly-graded); the A-line separates soils whose plasticity is dominated by clay minerals (above, C) from those dominated by silt/organics (below, M/O) at the same liquid limit.

Common pitfall. Classifying by the 5%/12% fines cutoffs without checking the dual-symbol band in between, and forgetting the "with sand"/"with gravel" name modifiers once the secondary fraction reaches 15% of the total sample.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 4 (USCS classification).

Question 1 of 6: Classification

Question 1: Classification (15 marks)