Grain-Size Gradation & USCS Classification: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2016, Q2

Question 2: Grain-Size Gradation & USCS Classification (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 distribution curves for soils A, B, C (inorganic) and D (organic), read from the printed semi-log chart; for soil D, $LL=52\%$, $PL=22\%$.

Given data — $D_{10}$, $D_{30}$, $D_{60}$ digitized from the chart (mm)
Soil	$D_{10}$	$D_{30}$	$D_{60}$
A	0.25	1.55	23.7
B	0.11	0.135	0.22
C	0.0094	0.050	0.61

Figure 1 — grain-size distribution, read from the printed chart. $D_{10}$/$D_{30}$/$D_{60}$ markers are shown for soils A and C, the two extremes needed in part (a).

Find. (a) which inorganic soil (A, B or C) has the highest $C_u$, and which has the lowest $C_c$; (b) USCS group symbol and group name for organic soil D.

Approach. Read $D_{10}$, $D_{30}$, $D_{60}$ for each curve and compute $C_u=D_{60}/D_{10}$ and $C_c=D_{30}^2/(D_{10}D_{60})$; for soil D, compute the plasticity index and compare it against the A-line on the plasticity chart, using the fact that it is stated to be organic.

Part (a) — gradation coefficients for A, B, C. With the digitized diameters above, $$\begin{aligned} C_{u,A}&=\frac{23.7}{0.25}=\boxed{94.8}, \\ C_{u,B}&=\frac{0.22}{0.11}=\boxed{2.0}, \\ C_{u,C}&=\frac{0.61}{0.0094}=\boxed{64.9}. \end{aligned}$$ $$\begin{aligned} C_{c,A}&=\frac{1.55^2}{0.25\times23.7}=\boxed{0.405}, \\ C_{c,B}&=\frac{0.135^2}{0.11\times0.22}=\boxed{0.753}, \\ C_{c,C}&=\frac{0.050^2}{0.0094\times0.61}=\boxed{0.436}. \end{aligned}$$ Soil A spans by far the widest range of particle sizes on the chart (its curve is the most gently sloped of the three, running from fine sand to gravel), so it has both the largest $D_{60}/D_{10}$ ratio and — because for a curve that is close to log-linear over its 10–60% range $C_c=10^{-10/b}$ where $b$ is the %-per-decade slope — the smallest $C_c$ as well: $$\boxed{\text{Soil A has the highest } C_u\ (\approx95)\ \textbf{and}\ \text{the lowest } C_c\ (\approx0.41).}$$
Part (b) — plasticity index and A-line for soil D. $$PI=LL-PL=52-22=\boxed{30\%}.$$ The A-line value at this liquid limit is $$PI_{A\text{-line}}=0.73(LL-20)=0.73\times32=\boxed{23.4\%}.$$ Since $PI=30\%>23.4\%$, soil D plots ABOVE the A-line (and below the U-line, $PI_{U}=0.9(LL-8)=39.6\%$, so the point is a valid one).
Classify. $LL=52\%\ge50\%$ places soil D in the high-plasticity ("H") group. An above-A-line, high-plasticity, ORGANIC fine soil (given) is classified $$\boxed{\text{Group symbol OH}\ -\ \text{Group name: Organic clay (high plasticity)}}.$$ (An inorganic soil at the same point would be CH; the "O" prefix and the fact that its position is above the A-line — the CLAY side — is what fixes the group name as organic CLAY rather than organic silt.)

Check: $D_{10}$/$D_{30}$/$D_{60}$ for A, B, C were read from the printed chart (two independent readings per curve) — treat to about ±15% engineering tolerance for chart reading; the classification bands (Cu highest/Cc lowest = soil A) are robust to that tolerance since the gap to the next soil is large on both metrics. Soil D's classification assumes the source's statement that it is organic is itself the basis for the "O" prefix (per ASTM D2487, an organic soil is normally confirmed by an oven-dried/not-dried liquid-limit ratio <0.75, which is not printed on this exam but is given as a stated fact of the problem).

Quantity	Value
$C_u$ (A, B, C)	94.8, 2.0, 64.9
$C_c$ (A, B, C)	0.405, 0.753, 0.436
(a) Highest $C_u$ / lowest $C_c$	Soil A (both)
Soil D: $PI$ / A-line $PI$	30% / 23.4% (above A-line)
(b) USCS classification, soil D	OH — Organic clay (high plasticity)

Concept & Theory

Topic. Gradation coefficients from a grain-size distribution curve, and USCS classification of a fine-grained organic soil via the plasticity (Casagrande) chart.

Key relations. $C_u=D_{60}/D_{10}$; $C_c=D_{30}^2/(D_{10}D_{60})$; A-line $PI=0.73(LL-20)$; U-line $PI=0.9(LL-8)$; $LL\ge50\%\Rightarrow$ high-plasticity ("H") group; organic + above A-line + $LL\ge50\Rightarrow$ OH (organic clay).

Why this works. $C_u$ measures how spread out the particle sizes are (a steep, near-vertical curve like B has $C_u$ close to 1 — nearly one particle size); $C_c$ measures the curve's shape/skew between $D_{10}$ and $D_{60}$ and is largely set by the curve's slope on a log-linear plot, independent of where the curve sits — so the same soil (A) can simultaneously have the highest $C_u$ and the lowest $C_c$. The A-line separates clays (above) from silts (below) at the same plasticity; organic soils get an "O" prefix instead of "C"/"M" because organic matter depresses the liquid limit relative to an inorganic soil of the same plasticity index.

Common pitfall. Assuming the steepest (most "uniform-looking") curve must be an extreme on both $C_u$ and $C_c$ — in fact a uniform soil's $C_c$ is usually close to 1 (neither the highest nor the lowest), not the minimum.

Source. Das, Principles of Geotechnical Engineering, ch. 2 (grain-size analysis) and ch. 5 (USCS classification, plasticity chart).

Question 2 of 6: Grain-Size Gradation & USCS Classification

Question 2: Grain-Size Gradation & USCS Classification (20 marks)