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Types of Soil Classification

Common soil classification systems used in geotechnical work are the Textural Classification or USDA system, the AASHTO system, and the Unified Soil Classification System (USCS).

  1. USDA: based on corrected percentages of sand, silt, and clay.
  2. AASHTO: commonly used for highway subgrade materials.
  3. USCS: commonly used in geotechnical engineering and foundation work.

USDA Corrected Percentages

If a soil contains gravel or particles larger than 2 mm, correct the percentages of sand, silt, and clay based on the portion finer than 2 mm.

$$\% \text{ corrected component} = \frac{\% \text{ component}}{\% \text{ sand} + \% \text{ silt} + \% \text{ clay}}(100)$$

The corrected sand, silt, and clay percentages should add to 100 percent before using the USDA textural triangle.

AASHTO Classification Criteria

In AASHTO classification, gravel is the fraction passing the 25 mm sieve and retained on No. 10 sieve, sand is the fraction passing No. 10 and retained on No. 200, and silt-clay is the fraction passing No. 200.

$$F_{200} \le 35\% \Rightarrow \text{granular material}$$
$$F_{200} > 35\% \Rightarrow \text{silt-clay material}$$
$$PI \le 10 \Rightarrow \text{silty fine fraction}$$
$$PI \ge 11 \Rightarrow \text{clayey fine fraction}$$

AASHTO Group Index

The group index evaluates the quality of a soil as highway subgrade material. Negative values are reported as zero and final values are rounded to the nearest whole number.

$$GI = (F_{200}-35)[0.2+0.005(LL-40)] + 0.01(F_{200}-15)(PI-10)$$
$$GI_{LL} = (F_{200}-35)[0.2+0.005(LL-40)]$$
$$GI_{PI} = 0.01(F_{200}-15)(PI-10)$$

For A-1-a, A-1-b, A-2-4, A-2-5, and A-3 soils, the group index is always zero. For A-2-6 and A-2-7 soils, use the plasticity-index part only.

USCS Gradation Coefficients

The particle-size distribution curve is used to obtain effective size and gradation coefficients.

$$D_{10} = \text{effective size}$$
$$C_u = \frac{D_{60}}{D_{10}}$$
$$C_c = \frac{D_{30}^2}{D_{10}D_{60}}$$
$$S_o = \sqrt{\frac{D_{75}}{D_{25}}}$$

USCS Classification Procedure

Start with No. 200 sieve passing. If more than 50 percent passes No. 200, the soil is fine-grained. If less than 50 percent passes No. 200, the soil is coarse-grained.

$$F_{200} > 50\% \Rightarrow \text{fine-grained soil}$$
$$F_{200} < 50\% \Rightarrow \text{coarse-grained soil}$$
$$R_4 > 50\% \text{ of coarse fraction} \Rightarrow \text{gravel}$$
$$R_4 < 50\% \text{ of coarse fraction} \Rightarrow \text{sand}$$

For clean coarse-grained soils, use GW, GP, SW, or SP. With more than 12 percent fines, use GM, GC, SM, or SC. Between 5 and 12 percent fines, use a dual symbol.

Well-Graded Criteria and Plasticity Chart

Well-graded gravels and sands must satisfy both coefficient requirements. Fine-grained soils use liquid limit and the plasticity chart.

$$C_u > 4 \text{ for gravel}$$
$$C_u > 6 \text{ for sand}$$
$$1 < C_c < 3$$
$$PI = 0.73(LL - 20) \quad \text{A-line}$$
$$LL > 50\% \Rightarrow \text{high plasticity}$$
$$LL < 50\% \Rightarrow \text{low plasticity}$$

Problem: AASHTO Percent Gravel, Sand, and Fines

From a sieve analysis of a soil sample, 54 percent passes the No. 10 sieve and 4 percent passes the No. 200 sieve. Find the percentages of gravel, sand, and silt-clay under AASHTO.

$$\% \text{ gravel} = 100 - 54 = 46\%$$
$$\% \text{ sand} = 54 - 4 = 50\%$$
$$\% \text{ silt-clay} = F_{200} = 4\%$$

Answer: Gravel = 46 percent, sand = 50 percent, and fines = 4 percent.

Problem: Effective Size, Uniformity, and USCS

A soil sample has 4 percent passing No. 200 and is predominantly sand. From the grain-size curve, $D_{10}$ is about 0.15 mm and $D_{60}$ is about 2.2 mm. Classify the soil using USCS if $C_c = 1.28$.

$$D_{10} \approx 0.15 \text{ mm}$$
$$C_u = \frac{D_{60}}{D_{10}} = \frac{2.2}{0.15} = 14.67 \approx 15$$
$$1 < C_c = 1.28 < 3$$

Since $F_{200} = 4\%$, the soil is a clean coarse-grained soil. Since it is sand, $C_u > 6$, and $1 < C_c < 3$, it is well-graded sand.

Answer: SW, well-graded sand.

Problem: Coefficient of Gradation

For a soil sample, determine the coefficient of gradation from $D_{10}$, $D_{30}$, and $D_{60}$ obtained from the grain-size distribution curve.

$$C_c = \frac{D_{30}^2}{D_{10}D_{60}}$$

Read $D_{10}$, $D_{30}$, and $D_{60}$ at 10 percent, 30 percent, and 60 percent finer from the semi-log grain-size curve, then substitute directly into the formula.

Problem: Percentages from Sieve Curve

From a sieve curve: for soil A, 98 percent passes No. 4; for soil B, 95 percent passes No. 4 and 34 percent passes No. 200; for soil C, 12 percent passes No. 200. Determine gravel for A, sand for B, and silt-clay for C.

$$\% \text{ gravel for A} = 100 - 98 = 2\%$$
$$\% \text{ sand for B} = 95 - 34 = 61\%$$
$$\% \text{ silt-clay for C} = 12\%$$

Problem: AASHTO Soil A

Soil A has $F_{200} = 17\%$, liquid limit = 35 percent, and plastic limit = 20 percent. Classify it using AASHTO.

$$F_{200}=17\% < 35\% \Rightarrow \text{granular material}$$
$$PI = LL - PL = 35 - 20 = 15$$
$$GI = 0.01(17 - 15)(15 - 10) = 0.10 \approx 0$$

Answer: A-2-6 (0).

Problem: AASHTO Soil B

Soil B has $F_{200} = 33\%$, liquid limit = 46 percent, and plastic limit = 29 percent. Classify it using AASHTO.

$$F_{200}=33\% < 35\% \Rightarrow \text{granular material}$$
$$PI = 46 - 29 = 17$$
$$GI = 0.01(33 - 15)(17 - 10) = 1.26 \approx 1$$

Answer: A-2-7 (1).

Problem: AASHTO Soil C

Soil C has $F_{200} = 63\%$, liquid limit = 47 percent, and plastic limit = 24 percent. Classify it using AASHTO.

$$F_{200}=63\% > 35\% \Rightarrow \text{silt-clay material}$$
$$PI = 47 - 24 = 23$$
$$LL - 30 = 47 - 30 = 17$$
$$PI > LL - 30 \Rightarrow A\text{-}7\text{-}6$$
$$GI = (63 - 35)[0.2 + 0.005(47 - 40)] + 0.01(63 - 15)(23 - 10)$$
$$GI = 12.82 \approx 13$$

Answer: A-7-6 (13).

Concept: AASHTO Soil Fractions

What type of soil is the fraction passing No. 10 and retained on No. 200 under AASHTO? What type passes No. 200?

$$\text{Passing No. 10 and retained on No. 200} = \text{sand}$$
$$\text{Passing No. 200} = \text{silt and clay}$$

Concept: AASHTO Plasticity

In AASHTO, fine fractions with $PI \ge 11$ are clayey, while fine fractions with $PI \le 10$ are silty. Sizes larger than 75 mm are boulders and cobbles.

$$PI \ge 11 \Rightarrow \text{clayey}$$
$$PI \le 10 \Rightarrow \text{silty}$$

Concept: USCS Coarse and Fine Soils

In USCS, gravelly and sandy soils with less than 50 percent passing No. 200 are coarse-grained soils. Soils with 50 percent or more passing No. 200 are fine-grained soils. The boundary between low and high plasticity is $LL = 50$.

$$F_{200} < 50\% \Rightarrow \text{coarse-grained}$$
$$F_{200} \ge 50\% \Rightarrow \text{fine-grained}$$
$$LL = 50 \Rightarrow \text{low/high plasticity boundary}$$

Concept: USCS Gravel and Sand Fractions

Under USCS, the fraction passing the 76.2 mm sieve and retained on No. 4 is gravel. The fraction passing No. 4 and retained on No. 200 is sand.

$$\text{Passing 76.2 mm and retained on No. 4} = \text{gravel}$$
$$\text{Passing No. 4 and retained on No. 200} = \text{sand}$$

Problem: USCS Soil A

Soil A has 0 percent passing No. 200, 98 percent passing No. 4, $D_{10}=0.5$, $D_{30}=0.9$, and $D_{60}=1.4$. Classify the soil using USCS.

$$F_{200}=0\% < 50\% \Rightarrow \text{coarse-grained}$$
$$\% \text{ passing No. 4}=98\% > 50\% \Rightarrow \text{sand}$$
$$C_u = \frac{1.4}{0.5}=2.8$$
$$C_c = \frac{0.9^2}{0.5(1.4)}=1.16$$

The sand does not satisfy $C_u > 6$, so it is poorly graded.

Answer: SP.

Problem: USCS Soil B

Soil B has 38 percent passing No. 200, 100 percent passing No. 4, liquid limit = 46, and plastic limit = 29. Classify the soil using USCS.

$$F_{200}=38\% < 50\% \Rightarrow \text{coarse-grained}$$
$$\text{Coarse fraction}=100-38=62\%$$
$$\frac{1}{2}(62)=31\%$$
$$\% \text{ passing No. 4}=100\% > 31\% \Rightarrow \text{sand}$$
$$PI=46-29=17>7$$

With more than 12 percent fines and $PI>7$, the soil is clayey sand.

Answer: SC.

Problem: USCS Soil C

Soil C has 68 percent passing No. 200 and 12 percent finer than 0.002 mm. From the distribution, sand is 32 percent, silt is 56 percent, and clay is 12 percent. Classify the soil using USCS.

$$F_{200}=68\% > 50\% \Rightarrow \text{fine-grained}$$
$$\% \text{ silt}=68-12=56\%$$

The dominant fine fraction is silt.

Answer: M, silt.

Problem: USCS Percent Gravel, Sand, and Fines

Using USCS, a sample has 90 percent passing No. 4 and 4 percent passing No. 200. Determine the percentage of gravel, sand, and fines.

$$\% \text{ gravel}=100-90=10\%$$
$$\% \text{ sand}=90-4=86\%$$
$$\% \text{ fines}=4\%$$

Check: $10+86+4=100\%$.

Problem: Corrected USDA Percentages

A soil has sand = 12 percent, silt = 24 percent, and clay = 44 percent. If it contains 20 percent particles larger than 2 mm, the actual sum of sand, silt, and clay is 80 percent. Compute the corrected sand, silt, and clay percentages.

$$\% \text{ corrected sand}=\frac{12}{80}(100)=15\%$$
$$\% \text{ corrected silt}=\frac{24}{80}(100)=30\%$$
$$\% \text{ corrected clay}=\frac{44}{80}(100)=55\%$$

Answer: The soil plots as clay in the USDA system.

Concept: USCS Plasticity Chart and the A-Line

For fine-grained soils in the USCS, classification depends on both the liquid limit and the plasticity index. The A-line separates clays from silts and organic soils.

$$\text{A-line: } PI = 0.73(LL-20)$$

Soils that plot above the A-line with $PI > 7$ are classified as clays (C). Soils below the A-line or with $PI < 4$ are silts or organic soils (M or O). The boundary between low-plasticity (L) and high-plasticity (H) groups is at $LL = 50$ percent. A soil with $LL < 50$ and plotting above the A-line with $PI > 7$ is CL. The same location but with $LL > 50$ is CH. Soils near the boundary band between 4 and 7 plasticity index are dual-symbol, such as CL-ML.

Problem: USCS Classification of Gravel with Gradation Data

A soil has 8 percent passing the No. 200 sieve, 15 percent passing the No. 4 sieve, and gradation coefficients $C_u = 8$ and $C_c = 2.1$. Classify this soil using USCS.

$$F_{200}=8\% < 50\% \Rightarrow \text{coarse-grained}$$
$$\%\text{ retained on No. 4}=100-15=85\% > 50\% \Rightarrow \text{more than half coarse fraction is gravel}$$

Because $F_{200} < 5\%$, use gradation coefficients to subclassify the gravel:

$$C_u=8\ge4 \quad \text{and} \quad C_c=2.1 \in [1,3] \Rightarrow \text{well-graded gravel GW}$$

Answer: GW, well-graded gravel.

Problem: USCS Classification Using the A-Line

A soil has 72 percent passing the No. 200 sieve, a liquid limit of 42 percent, and a plastic limit of 22 percent. Classify the soil using USCS.

$$F_{200}=72\% > 50\% \Rightarrow \text{fine-grained}$$
$$PI=42-22=20$$
$$\text{A-line: } PI=0.73(LL-20)=0.73(42-20)=16.06$$

Since $PI=20 > 16.06$, the soil plots above the A-line and is a clay. Since $LL=42 < 50$, it is low plasticity.

Answer: CL, lean clay.

Problem: AASHTO Group Index Calculation

A subgrade soil has 62 percent passing the No. 200 sieve, a liquid limit of 48 percent, and a plasticity index of 22. Compute the group index and comment on the soil quality as a subgrade material.

$$GI = (F_{200}-35)[0.2+0.005(LL-40)]+0.01(F_{200}-15)(PI-10)$$
$$GI_{LL}=(62-35)[0.2+0.005(48-40)]=27[0.2+0.04]=27(0.24)=6.48$$
$$GI_{PI}=0.01(62-15)(22-10)=0.01(47)(12)=5.64$$
$$GI=6.48+5.64=12.12\approx12$$

A group index of 12 indicates a poor subgrade material. Typical classification is A-7-6 (highly plastic clay). Higher group index means poorer performance as a subgrade.

Answer: $GI = 12$; the soil is a poor subgrade material.

Problem: USCS Classification of Sandy Soil with Fines and Atterberg Data

A soil has 55 percent passing the No. 4 sieve, 20 percent passing the No. 200 sieve, a liquid limit of 35 percent, and a plasticity index of 12. Classify this soil using USCS.

$$F_{200}=20\% < 50\% \Rightarrow \text{coarse-grained}$$
$$\%\text{ passing No. 4}=55\% > 50\% \Rightarrow \text{sand (more than half coarse fraction passes No. 4)}$$

With $F_{200}=20\% > 12\%$, classification is based on Atterberg limits of the fines.

$$\text{A-line check: } PI=0.73(35-20)=10.95 < 12$$

The soil plots above the A-line with $PI = 12 > 7$, so the fines are clayey.

Answer: SC, clayey sand.

Problem: AASHTO Classification of Silty Clay Subgrade

A highway subgrade soil has 80 percent passing No. 200, a liquid limit of 55 percent, and a plastic limit of 30 percent. Determine whether the soil is granular or silt-clay material, compute the plasticity index, and find the group index.

$$F_{200}=80\% > 35\% \Rightarrow \text{silt-clay material}$$
$$PI=55-30=25>11 \Rightarrow \text{clayey fine fraction, group A-7}$$
$$GI_{LL}=(80-35)[0.2+0.005(55-40)]=45[0.2+0.075]=45(0.275)=12.375$$
$$GI_{PI}=0.01(80-15)(25-10)=0.01(65)(15)=9.75$$
$$GI=12.375+9.75=22.125\approx22$$

Since $LL-30=55-30=25=PI$, this is A-7-5. A group index of 22 represents very poor subgrade.

Answer: Silt-clay material, $PI=25$, $GI=22$; classified as A-7-5, very poor subgrade.