Particle Size Distribution and Gradation: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2017, Q3

Question 3: Particle Size Distribution and Gradation (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. Mass retained on each sieve (tabulated above); total sample mass $=\sum m_i=2387.8$ g (sum of all retained masses plus the pan).

Find. (a) the gradation curve and the gravel/sand/fines split; (b) $C_u$ and $C_c$.

Approach. Accumulate retained mass down the stack to get cumulative percent retained, subtract from 100 for percent passing, plot against $\log$(grain size), then read $D_{10}$, $D_{30}$, $D_{60}$ off the curve by log-linear interpolation.

Cumulative retained & percent passing. With total mass 2387.8 g: $$\%\text{passing}=100-\frac{\sum_{\text{coarser}} m_i}{2387.8}\times100.$$

Percent passing by sieve
Sieve	Size (mm)	Retained (g)	Cum. retained (g)	% passing
#4	4.750	198.8	198.8	91.67
#10	2.000	386.4	585.2	75.49
#20	0.850	426.5	1011.7	57.63
#40	0.425	363.7	1375.4	42.40
#60	0.250	287.9	1663.3	30.34
#100	0.150	293.0	1956.3	18.07
#140	0.106	97.6	2053.9	13.98
#200	0.075	116.4	2170.3	9.11
pan	<0.075	217.5	2387.8	0

Fig. Q3 — particle size distribution curve (semi-log), with $D_{10}$, $D_{30}$, $D_{60}$ marked.

Part (a) — gravel/sand/fines split. Percent coarser than 4.75 mm (retained on #4) is the gravel fraction; percent finer than 0.075 mm (passing #200, i.e. the pan) is the fines fraction; sand is everything between: $$\%\text{Gravel}=8.33\%,\qquad \%\text{Fines}=9.11\%,\qquad \%\text{Sand}=100-8.33-9.11=\boxed{82.57\%}.$$
Part (b) — $D_{10}$, $D_{30}$, $D_{60}$ and gradation coefficients. Log-linear interpolation on the percent-passing curve gives $D_{10}=0.080$ mm, $D_{30}=0.246$ mm, $D_{60}=0.952$ mm, so $$C_u=\frac{D_{60}}{D_{10}}=\frac{0.952}{0.080}=\boxed{11.9},\qquad C_c=\frac{D_{30}^2}{D_{10}D_{60}}=\frac{0.246^2}{0.080\times0.952}=\boxed{0.80}.$$

Check: with $C_u=11.9\ (>6)$ but $C_c=0.80$ falling just outside the well-graded window $1\le C_c\le3$, this sand classifies as poorly graded (SP) rather than well-graded (SW) under USCS — a wide size spread alone does not guarantee a smooth, well-graded curve.

Quantity	Value
% Gravel (>4.75 mm)	8.33%
% Sand	82.57%
% Fines (<0.075 mm)	9.11%
$D_{10}$	0.080 mm
$D_{30}$	0.246 mm
$D_{60}$	0.952 mm
Coefficient of uniformity, $C_u$	11.9
Coefficient of curvature, $C_c$	0.80

Concept & Theory

Topic. Mechanical sieve analysis and gradation classification.

Key relations. $\%\text{passing}=100-\dfrac{\sum_{\text{coarser}}m_i}{m_{total}}\times100$; $C_u=D_{60}/D_{10}$; $C_c=D_{30}^2/(D_{10}D_{60})$.

Why this works. Sieving physically separates particles by size, so the cumulative retained mass IS the cumulative count of everything coarser than that opening; converting to percent passing and plotting against log(size) turns a lognormal-ish natural particle-size distribution into an approximately straight, easily-interpolated curve.

Common pitfall. Computing percent RETAINED (not passing) against the wrong total (forgetting the pan), and confusing well-graded ($C_u>4$–6 AND $1\le C_c\le3$, both conditions) with merely well-spread ($C_u$ large alone, which this sample is).

Source. Das, Principles of Geotechnical Engineering (9th ed.), Ch. 2 — Grain-Size Analysis and Soil Classification.

Question 3 of 6: Particle Size Distribution and Gradation

Question 3: Particle Size Distribution and Gradation (20 marks)