The two most widely used classification systems are the American Association of State Highway and Transportation Officials (AASHTO) and the Unified Soil Classification System (USCS). Our discussion here will involve only the USCS system.
The Unified Soil Classification System (ASTM 2004: Designation D-2487) clas-sifies soils based on their grain-size distribution curves and their Atterberg limits.
As shown in Table 1.2, a soil is called coarse-grained if it has less than 50% pass-ing sieve No. 200. Soils in this group can be sandy soils (S) or gravelly soils (G). It follows that a soil is called fine-grained if it has more than 50% passing sieve No.
200. Soils in this group include inorganic silts (M), inorganic clays (C), or organic silts and clays (O). The system uses the symbol W for well-graded soils, P for poorly graded soils, L for low-plasticity soils, and H for high-plasticity soils. The combined symbol GW, for example, means well-graded gravel, SP means poorly graded sand, and so on. Again, to determine the exact designation of a soil using the Unified Soil Classification System, you will need to have the grain-size distri-bution curve and the Atterberg limits of that soil. Then you can use Table 1.2 to get the soil symbol.
Example 1.2 Using the Unified Soil Classification System, classify a soil that has 95% passing a No. 10 sieve; 65% passing No. 40; and 30% passing No. 200. The soil has a liquid limit of 25 and a plastic limit of 15.
SOLUTION: The soil has 30% passing a No. 200 sieve, therefore, it is a coarse-grained soil according to the first column in Table 1.2. The soil has 95% passing a No. 10 sieve, so it must have at least 95% passing a No. 4 sieve (No. 4 has a larger opening size than No. 10). This means that the soil has less than 5% gravel (see Figure 1.7). According to the second column in Table 1.2, the soil is classified as sand. But since it has 30% fines, it is a sandy soil with fines according to the third column in Table 1.2.
Criteria Symbol Coarse-grained soils: less
than 50% passing No.
200 sieve
Gravel: more than 50%
of coarse fraction retained on No. 4 sieve
Clean gravels: less than 5% fines
Cu≥ 4 and 1 ≤ Cc≤ 3 GW
Cu< 4 and/or 1 > Cc> 3 GP
Gravels with fines: more than 12% fines
PI< 4 or plots below A line (Fig. 1.6) GM PI> 7 and plots on or above A line
(Fig. 1.6)
GC Sands: 50% or more of
coarse fraction passes No. 4 sieve
Clean sands: less than 5%
fines
Cu≥ 6 and 1 ≤ Cc≤ 3 SW
Cu< 6 and/or 1 > Cc> 3 SP
Sands with fines:
more than 12% fines
PI< 4 or plots below A line (Fig. 1.6) SM PI> 7 and plots on or above A line
(Fig. 1.6)
SC
Fine-grained soils: 50%
or more passing No.
200 sieve
Silts and clays: LL< 50 Inorganic PI> 7 and plots on or above A line
(Fig. 1.6)
CL
PI< 4 or plots below A line (Fig. 1.6) ML
Organic LL(oven dried)
LL(not dried) < 0.75 OL
Silts and clays: LL≥ 50 Inorganic PI plots on or above A line (Fig. 1.6) CH
PI plots below A line (Fig. 1.6) MH
Organic LL(oven dried)
LL(not dried) < 0.75 OH
Highly organic soils Primarily organic matter, dark in color, and organic odor Pt
15
No. 4
FIGURE 1.7 Particle-size distribution for Example 1.2.
The plasticity index of the soil is PI= LL − PL = 25 − 15 = 10 > 7. Also, the point with LL= 25 and PI = 10 plots above the A line in Figure 1.6. Therefore, the soil is classified as SC= clayey sand according to the fourth and fifth columns in Table 1.2.
1.7 COMPACTION
Compaction involves applying mechanical energy to partially saturated soils for densification purposes. The densification process brings soil particles closer to each other, thus decreasing the size of the voids by replacing air pockets with soil solids. Theoretically, we can achieve 100% saturation by replacing all air pockets by soil solids if we apply enough mechanical energy (compaction), but that is practically impossible. With proper compaction, the soil becomes stronger (higher shear strength), less compressible when subjected to external loads (i.e., less future settlement), and less permeable, making the soil a good construction material for highway embankments, ramps, earth dams, dikes, backfill for retaining walls and bridge abutments, and many other applications.
Soils are compacted in layers (called lifts) with each layer being compacted to develop a final elevation and/or shape. Compaction machines such as smooth rollers, pneumatic rollers, and sheepsfoot rollers are generally used for this purpose.
The compaction energy generated by a compactor is proportional to the pressure applied by the compactor, its speed of rolling, and the number of times it is rolled (number of passes). Usually, a few passes are needed to achieve the proper dry unit weight, provided that the proper moisture content is used for a particular soil. The required field dry density is 90 to 95% of the maximum dry density that can be achieved in a laboratory compaction test (standard proctor test or modified proctor test: ASTM 2004: Designation D-698 and D-1557) carried out on the same soil.
The standard proctor test is a laboratory test used to determine the maximum dry unit weight and the corresponding optimum moisture content for a given com-paction energy and a given soil. The soil specimen is obtained from the borrow site, which is usually an earthcut that is close to the construction site. The soil is first dried and crushed and then mixed with a small amount of water in a uniform manner. The resulting moisture content should be well below the natural moisture content of the soil. The Proctor test involves placing the moist soil in three equal layers inside an extended mold (removable extension). The inside volume of the mold (without the extension) is exactly 1000 cm3. Each soil layer is compacted using 25 blows from a 2.5-kg hammer. Each blow is applied by raising the ham-mer 305 mm and releasing it (free fall). The 25 blows are distributed uniformly to cover the entire surface of each layer. After compacting the third layer, the mold extension is removed and the soil is carefully leveled and weighed. Knowing the weightW of the moist soil and its volume V , we can calculate the unit weight as γ = W/V . A small sample is taken from the compacted soil and dried to measure the moisture contentω. Now we can calculate the dry unit weight of the compacted soil asγd = γ/(1 + ω). Once this is done, the soil sample is crushed and added to the remainder of the soil in the mixing pan. The moisture content of the soil is increased (1 to 2%) by adding more water. The test is repeated in the same manner as described above. We need to repeat the test at least four to five times to establish a compaction curve such as the one shown in Figure 1.8.
The compaction curve shown in Figure 1.8 provides the relationship between the dry unit weight and the moisture content for a given soil subjected to a specific compaction effort. It is noted from the figure that the dry unit weight increases
γd-max
ωopt
γd (kN/m3)
10 11 12 13 14 15 16 17 18 19 20
16 17 18 19 20
0.95γd-max
13.75% 17.3%
ω (%)
FIGURE 1.8 Compaction curve.
as the moisture content increases until the maximum dry unit weight is reached.
The moisture content associated with the maximum dry unit weight is called the optimum water content. As shown in the figure, when the moisture content is increased beyond the optimum water content, the dry unit weight decreases. This is caused by the water that is now occupying many of the voids and making it more difficult for the soil to compact further. Note that the degree of saturation corresponding to the optimum moisture content is 75 to 80% for most soils (i.e., 75 to 80% of the voids are filled with water).
The compaction curve provides valuable information: the maximum dry unit weight and the optimum water content that can be conveyed to the compaction contractor by specifying the required relative compaction, RC, defined as
RC= γd−field
γd−max × 100% (1.25)
in whichγd−fieldis the required dry unit weight of the compacted soil andγd−max
is the maximum dry unit weight obtained from the laboratory compaction test.
Usually, the required relative compaction is 90 to 95%. This is because it is very difficult (and costly) to achieve a field dry unit weight that is equal to the maximum dry unit weight obtained from the laboratory compaction test.
It is not enough to specify the relative compaction RC alone. We need to specify the corresponding moisture content that must be used in the field to achieve a specific RC. This is due to the nature of the bell-shaped compaction curve that can have two different moisture contents for the same dry unit weight. Figure 1.8 shows that we can use either ω = 13.75% or ω = 17.3% to achieve a dry unit weightγd−field= 18.5 kN/m3, which corresponds to RC= 95%.
In general, granular soils can be compacted in thicker layers than silt and clay.
Granular soils are usually compacted using kneading, tamping, or vibratory com-paction techniques. Cohesive soils usually need kneading, tamping, or impact. It is to be noted that soils vary in their compaction characteristics. Soils such as GW, GP, GM, GC, SW, SP, and SM (the Unified Soil Classification System, Table 1.2) have good compaction characteristics. Other soils, such as SC, CL, and ML, are characterized as good to poor. Cohesive soils with high plasticity or organic con-tents are characterized as fair to poor. At any rate, the quality of field compaction needs to be assured by measuring the in situ dry unit weight of the compacted soil at random locations. Several test methods can be used for this purpose:
1. The sand cone method (ASTM 2004: Designation D-1556) requires that a small hole be excavated in a newly compacted soil layer. The soil removed is weighed (W ) and its moisture content (ω) is determined. The volume (V ) of the hole is determined by filling it with Ottawa sand that has a known unit weight. The field dry unit weight can be calculated asγd−field= γ/(1 + ω), in whichγ is calculated as γ = W/V .
2. There is a method similar to the sand cone method that determines the volume of the hole by filling it with oil (instead of sand) after sealing the surface
of the hole with a thin rubber membrane. This method is called the rubber balloon method (ASTM 2004: Designation D-2167).
3. The nuclear density method uses a low-level radioactive source that is inserted, via a probe, into the center of a newly compacted soil layer. The source emits rays through the compacted soil that are captured by a sensor at the bottom surface of the nuclear density device. The intensity of the captured radioac-tivity is inversely proportional to soil density. The apparatus is calibrated using the sand cone method for various soils, and it usually provides reliable estimates of moisture content and dry unit weight. The method provides fast results, allowing the user to perform a large number of tests in a short time.
PROBLEMS
1.1 Refer to Figure 1.3. For soil B: (a) determine the percent finer than sieves No. 4, 10, 100, and 200; (b) determine d10, d30, and d60; (c) calculate the uniformity coefficient; and (d) calculate the coefficient of gradation.
1.2 Refer to the phase diagram shown in Figure 1.9. In this phase diagram it is assumed that the total volume of the soil specimen is 1 unit. Show that (a) γd = Gsγw(1− n), and (b) γ = Gsγw(1− n)(1 + ω).
Solids Air
Vs Vv Ww
Ws 0
W V= 1
Water
FIGURE 1.9
1.3 For a moist soil specimen, the following are given:V = 0.5 m3,W = 9.5 kN, ω = 7.3%, and Gs = 2.7. Determine the bulk unit weight γ, the dry unit weightγd, the void ratioe, the porosity n, and the degree of saturation S.
1.4 The field unit weight of a compacted soil is γ = 17.5 kN/m3, and its mois-ture content isω = 7%. Calculate the relative density of the compacted soil knowing that itsemax= 0.9, emin= 0.5, and Gs = 2.7.
1.5 A moist soil hasGs = 2.65, γ = 20 kN/m3, andω = 15.2%. Calculate its dry unit weightγd, void ratioe, porosity n, and degree of saturation S.
1.6 A 1.3-m3 soil specimen weighs 25.7 kN and has a moisture content of 11%.
The specific gravity of the soil solids is 2.7. Using the fundamental equations, (1.1) to (1.10), calculate its bulk unit weightγ, dry unit weight γd, void ratio e, porosity n, volume of water Vw, and degree of saturationS.
1.7 Four standard Proctor tests were performed on a clayey soil with the following results:
Bulk unit weight (kN/m3) Moisture content (%)
20.6 13.3
21.96 14.4
22.5 16
21.7 18.5
Plot the compaction curve and obtain the maximum dry unit weight and the corresponding optimum moisture content.
1.8 The same soil as that in Figure 1.8 is used to construct an embankment. It was compacted using a moisture content below the optimum moisture content and a relative compaction of 95%. What is the compacted unit weight of the soil in the field? If the specific gravity of the soil solids is 2.68, what are the soil’s in situ porosity and degree of saturation?