Soil classification from a grading curve

Abstract

Classifying a soil under the Unified Soil Classification System needs five numbers from the laboratory: the percentage passing the No. 200 and No. 4 sieves, the particle sizes D10, D30 and D60, and the liquid limit and plasticity index. This article works one sample through the whole sequence with the arithmetic shown, including the logarithmic interpolation that D10 and D30 require, and shows why a soil with a coefficient of uniformity of nearly 20 is still classified as poorly graded.

Classification is the first thing done to a soil sample and the last thing anyone checks. It is also entirely mechanical: given five numbers from the laboratory, the group symbol follows from a table with no judgement required. This article works one sample all the way through, showing every step of the arithmetic.

Every figure below is recomputed from the input data by a script that runs on each build of this site. If a number here is wrong, the build fails.

What the classification needsPermalink to “What the classification needs”

The Unified Soil Classification System, ASTM D2487, sorts soil first by how much of it is fine. Two sieves do that work:

SieveApertureSeparates
No. 44.75 mmgravel from sand
No. 2000.075 mmsand from fines (silt and clay)

A soil with more than 50% retained on the No. 200 sieve is coarse-grained; 50% or more passing it is fine-grained. Within a coarse-grained soil, more than 50% of the coarse fraction retained on the No. 4 sieve makes it a gravel, and 50% or more passing makes it a sand.

Note the phrase coarse fraction. The No. 4 split is applied to the material coarser than the No. 200 sieve, not to the whole sample. Applying it to the whole sample is a common and silent error, and it changes the answer whenever the fines content is appreciable.

The samplePermalink to “The sample”

A sieve analysis, expressed as percentage passing:

Sieve aperture (mm)19.09.54.752.361.180.6000.3000.1500.075
Passing (%)100887460473522126

Take the fractions straight off the table:

fines=6%,sand=746=68%,gravel=10074=26%\text{fines} = 6\%, \qquad \text{sand} = 74 - 6 = 68\%, \qquad \text{gravel} = 100 - 74 = 26\%

Fines are 6%, well under 50%, so the soil is coarse-grained. The coarse fraction is 1006=94%100 - 6 = 94\% of the sample, and the part of it passing the No. 4 sieve is

6894×100=72.3%\frac{68}{94} \times 100 = 72.3\%

which is 50% or more, so this is a sand.

D10, D30 and D60 — the logarithmic stepPermalink to “D10, D30 and D60 — the logarithmic step”

The three characteristic sizes are defined by the percentage passing: D60D_{60} is the aperture through which 60% of the sample passes, and so on. Only D60D_{60} falls exactly on a sieve here — the 2.36 mm sieve passes exactly 60%, so

D60=2.360 mmD_{60} = 2.360\ \text{mm}

D30D_{30} and D10D_{10} have to be interpolated, and the interpolation is logarithmic, because a grading curve is plotted against log(aperture)\log(\text{aperture}). Between two sieves with apertures a1>a2a_1 > a_2 passing p1>p2p_1 > p_2, the size at a target percentage pp is

D=exp ⁣[lna2+pp2p1p2(lna1lna2)]D = \exp\!\left[\ln a_2 + \frac{p - p_2}{p_1 - p_2}\left(\ln a_1 - \ln a_2\right)\right]

For D30D_{30}, between the 0.600 mm sieve (35%) and the 0.300 mm sieve (22%):

30223522=813=0.6154,D30=0.300×(0.6000.300)0.6154=0.300×20.6154=0.4596 mm\frac{30 - 22}{35 - 22} = \frac{8}{13} = 0.6154, \qquad D_{30} = 0.300 \times \left(\frac{0.600}{0.300}\right)^{0.6154} = 0.300 \times 2^{0.6154} = 0.4596\ \text{mm}

For D10D_{10}, between 0.150 mm (12%) and 0.075 mm (6%):

106126=23,D10=0.075×22/3=0.1191 mm\frac{10 - 6}{12 - 6} = \frac{2}{3}, \qquad D_{10} = 0.075 \times 2^{2/3} = 0.1191\ \text{mm}

What linear interpolation costsPermalink to “What linear interpolation costs”

Interpolating linearly on aperture instead — reading the curve as though the axis were linear — gives D30=0.4846D_{30} = 0.4846 mm and D10=0.1250D_{10} = 0.1250 mm. Carried through:

D10D_{10} (mm)D30D_{30} (mm)CuC_uCcC_c
Logarithmic (correct)0.11910.459619.820.75
Linear (wrong)0.12500.484618.880.80

The error in CcC_c is 6%. Here it changes nothing, because both values are far from the threshold. Near a threshold it decides the group symbol, and nothing in the output would tell you which interpolation produced it.

Uniformity and curvaturePermalink to “Uniformity and curvature”

Two ratios summarise the shape of the grading curve. ERDC/CRREL TR-15-4 gives them as

Cu=D60D10,Cc=(D30)2D60×D10C_u = \frac{D_{60}}{D_{10}}, \qquad C_c = \frac{\left(D_{30}\right)^2}{D_{60} \times D_{10}}

For this sample:

Cu=2.3600.1191=19.82,Cc=0.459622.360×0.1191=0.75C_u = \frac{2.360}{0.1191} = 19.82, \qquad C_c = \frac{0.4596^2}{2.360 \times 0.1191} = 0.75

The criteria for a clean sand are Cu6C_u \ge 6 and 1Cc31 \le C_c \le 3. This sample passes the first comfortably and fails the second.

So it is poorly graded, SP, despite Cu20C_u \approx 20. That is the point of having two ratios. A large CuC_u says only that the sample spans a wide range of sizes; it says nothing about whether the range is filled. CcC_c below 1 means the intermediate sizes are under-represented — a gap-graded soil, which spans a wide range and still packs badly. One number cannot express both, and quoting CuC_u alone is how gap-graded material gets described as well graded.

Fines between 5% and 12%: a dual symbolPermalink to “Fines between 5% and 12%: a dual symbol”

The fines content decides how much further the classification has to go:

  • Below 5% — a clean sand or gravel. The group symbol is settled by CuC_u and CcC_c.
  • Above 12% — the fines govern, and the symbol becomes SM, SC, GM or GC depending on where the fines plot on the plasticity chart.
  • Between 5% and 12% — both matter, and ASTM requires a dual symbol: SW-SM, SW-SC, SP-SM or SP-SC for sands, and the equivalent GW-GM, GW-GC, GP-GM, GP-GC for gravels.

This sample has 6% fines, so it lands in the middle case. It is SP from the grading, and the second half of the symbol depends on whether those fines are silty or clayey.

The plasticity chartPermalink to “The plasticity chart”

Fines are classified by Atterberg limits, plotted as plasticity index against liquid limit. Two straight lines govern the chart:

A-line:PI=0.73(LL20)U-line:PI=0.9(LL8)\text{A-line:}\quad PI = 0.73\left(LL - 20\right) \qquad\qquad \text{U-line:}\quad PI = 0.9\left(LL - 8\right)
The plasticity chart, with the A-line and U-line Plasticity index plotted against liquid limit. The A-line, PI equals 0.73 times LL minus 20, separates clays above it from silts below it. The U-line, PI equals 0.9 times LL minus 8, lies above the A-line and is an upper bound on results from natural soils. A vertical line at liquid limit 50 separates low from high plasticity. The worked sample, liquid limit 42 and plasticity index 22, plots above the A-line and below the U-line, in the CL region. 0 20 40 50 60 80 100 0 10 20 30 40 50 60 Liquid limit (%) Plasticity index U-line A-line CL ML CH MH
The plasticity chart. Line positions are computed from the two equations; the plotted point is the worked sample.

The A-line is the one that classifies. Above it a fine-grained soil is a clay, below it a silt. The U-line classifies nothing — it is an upper bound on what natural soils have been observed to do. A result plotting above the U-line is not an unusual soil; it is almost certainly a test that needs repeating.

That is a genuinely useful thing to know at the bench, and it is why the U-line belongs on the chart even though no group symbol depends on it.

The sample’s finesPermalink to “The sample’s fines”

The fines from this sample have LL=42LL = 42 and PI=22PI = 22. Evaluate both lines at LL=42LL = 42:

PIA=0.73(4220)=16.06,PIU=0.9(428)=30.60PI_{\text{A}} = 0.73\left(42 - 20\right) = 16.06, \qquad PI_{\text{U}} = 0.9\left(42 - 8\right) = 30.60

The measured PI=22PI = 22 is above 16.06 and below 30.60: above the A-line, so the fines are clayey; below the U-line, so the result is plausible. With LL=42<50LL = 42 < 50 the fines are CL, a lean clay.

The answerPermalink to “The answer”

Coarse-grained; sand, because 72.3% of the coarse fraction passes the No. 4 sieve. Poorly graded, because Cc=0.75C_c = 0.75 falls outside 1 to 3. Fines at 6%, so a dual symbol, and the fines are clayey:

SP-SC\boxed{\textbf{SP-SC}}

Poorly graded sand with clay.

Every step above is arithmetic on measured numbers. Which is exactly why it is worth automating — the calculation is short enough that nobody checks it, and long enough that a transposed digit survives. The same argument applies to the Mohr circle, where the standing error is a factor of two rather than a logarithm.

Classification is where the chain starts: a constitutive model in KATAI 2D is only as good as the parameters, and the parameters are only as good as the group symbol that decided which correlations were allowed to apply.

ReferencesPermalink to “References”

  • ASTM D2487, Standard Practice for Classification of Soils for Engineering Purposes (Unified Soil Classification System). ASTM International. The governing standard; the criteria applied above are its criteria.
  • ASTM D6913, Standard Test Methods for Particle-Size Distribution (Gradation) of Soils Using Sieve Analysis.
  • ASTM D4318, Standard Test Methods for Liquid Limit, Plastic Limit, and Plasticity Index of Soils.
  • García-Gaines, R.A. and Frankenstein, S. (2015). USCS and the USDA Soil Classification System: Development of a Mapping Scheme. ERDC/CRREL TR-15-4, U.S. Army Engineer Research and Development Center. Approved for public release. Its Table 4 and Figure 3 state the classification criteria, the definitions of CuC_u and CcC_c, and the A-line and U-line equations used above.
  • Casagrande, A. (1948). Classification and identification of soils. Transactions of the American Society of Civil Engineers, 113(1), 901–930. The origin of the plasticity chart.

The sieve data and Atterberg limits above are a worked illustration, not measurements from a real site. Everything derived from them is computed, not quoted.

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