Applied Soil Mechanics (Hardcover)

Chapter One

1.1 SOIL FORMATION

Soil is a three-phase material consisting of solid particles, water, and air. Its mechanical behavior is largely dependent on the size of its solid particles and voids. The solid particles are formed from physical and chemical weathering of rocks. Therefore, it is important to have some understanding of the nature of rocks and their formation.

A rock is made up of one or more minerals. The characteristics of a particular rock depend on the minerals it contains. This raises the question: What is a mineral? By definition, a mineral is a naturally occurring inorganic element or compound in a solid state. More than 4000 different minerals have been discovered but only 10 elements make up 99% of Earth's crust (the outer layer of Earth): oxygen (O), silicon (Si), aluminum (Al), iron (Fe), calcium (Ca), sodium (Na), potassium (K), magnesium (Mg), titanium (Ti), and hydrogen (H). Most of the minerals (74%) in Earth's crust contain oxygen and silicon. The silicate minerals, containing oxygen and silicon, comprise 90% of all rock-forming minerals. One of the interesting minerals in soil mechanics is the clay mineral montmorillonite (an expansive clay), which can expand up to 15 times its original volume if water is present. When expanding, it can produce pressures high enough to damage building foundations and other structures.

Since its formation, Earth has been subjected to continuous changes caused by seismic, volcanic, and climatic activities. Moving from the surface to the center of Earth, a distance of approximately 6370 km, we encounter three different layers. The top (outer) layer, the crust, has an average thickness of 15 km and an average density of 3000 kg/[m.sup.3]. By comparison, the density of water is 1000 kg/[m.sup.3] and that of iron is 7900 kg/[m.sup.3]. The second layer, the mantle, has an average thickness of 3000 km and an average density of 5000 kg/[m.sup.3]. The third, the core, contains primarily nickel and iron and has an average density of 11,000 kg/[m.sup.3].

Within the crust, there are three major groups of rocks:

1. Igneous rocks, which are formed by the cooling of magma. Fast cooling occurs above the surface, producing igneous rocks such as basalt, whereas slow cooling occurs below the surface, producing other types of igneous rocks, such as granite and dolerite. These rocks are the ancestors of sedimentary and metamorphic rocks.

2. Sedimentary rocks, which are made up of particles and fragments derived from disintegrated rocks that are subjected to pressure and cementation caused by calcite and silica. Limestone (chalk) is a familiar example of a sedimentary rock.

3. Metamorphic rocks, which are the product of existing rocks subjected to changes in pressure and temperature, causing changes in mineral composition of the original rocks. Marble, slate, and schist are examples of metamorphic rocks.

Note that about 95% of the outer 10 km of Earth's crust is made up of igneous and metamorphic rocks, and only 5% is sedimentary. But the exposed surface of the crust contains at least 75% sedimentary rocks.

Soils Soils are the product of physical and chemical weathering of rocks. Physical weathering includes climatic effects such as freeze-thaw cycles and erosion by wind, water, and ice. Chemical weathering includes chemical reaction with rainwater. The particle size and the distribution of various particle sizes of a soil depend on the weathering agent and the transportation agent.

Soils are categorized as gravel, sand, silt, or clay, depending on the predominant particle size involved. Gravels are small pieces of rocks. Sands are small particles of quartz and feldspar. Silts are microscopic soil fractions consisting of very fine quartz grains. Clays are flake-shaped microscopic particles of mica, clay minerals, and other minerals. The average size (diameter) of solid particles ranges from 4.75 to 76.2 mm for gravels and from 0.075 to 4.75 mm for sands. Soils with an average particle size of less than 0.075 mm are either silt or clay or a combination of the two.

Soils can also be described based on the way they were deposited. If a soil is deposited in the vicinity of the original rocks due to gravity alone, it is called a residual soil. If a soil is deposited elsewhere away from the original rocks due to a transportation agent (such as wind, ice, or water), it is called a transported soil.

Soils can be divided into two major categories: cohesionless and cohesive. Cohesionless soils, such as gravelly, sandy, and silty soils, have particles that do not adhere (stick) together even with the presence of water. On the other hand, cohesive soils (clays) are characterized by their very small flakelike particles, which can attract water and form plastic matter by adhering (sticking) to each other. Note that whereas you can make shapes out of wet clay (but not too wet) because of its cohesive characteristics, it is not possible to do so with a cohesionless soil such as sand.

1.2 PHYSICAL PARAMETERS OF SOILS

Soils contain three components: solid, liquid, and gas. The solid components of soils are the product of weathered rocks. The liquid component is usually water, and the gas component is usually air. The gaps between the solid particles are called voids. As shown in Figure 1.1a, the voids may contain air, water, or both. Let us discuss the soil specimen shown in Figure 1.1a. The total volume (V) and the total weight (W) of the specimen can be measured in the laboratory. Next, let us separate the three components of the soil as shown in Figure 1.1b. The solid particles are gathered in one region such that there are no voids in between, as shown in the figure (this can only be done theoretically). The volume of this component is [V.sub.s] and its weight is [W.sub.s]. The second component is water, whose volume is [V.sub.w] and whose weight is [W.sub.w]. The third component is the air, which has a volume [V.sub.a] and a very small weight that can be assumed to be zero. Note that the volume of voids ([V.sub.v]) is the sum of [V.sub.a] and [V.sub.w]. Therefore, the total volume is V = [V.sub.v] + [V.sub.s] = [V.sub.a] + [V.sub.w] + [V.sub.s]. Also, the total weight W = [W.sub.w] + [W.sub.s].

In the following we present definitions of several basic soil parameters that hold important physical meanings. These basic parameters will be used to obtain relationships that are useful in soil mechanics.

The void ratio e is the proportion of the volume of voids with respect to the volume of solids:

e = [V.sub.v] / [V.sub.s] (1.1)

The porosity n is given as

n = [V.sub.v]/ V (1.2)

Note that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.3)

or

n = e / 1 + e (1.4)

The degree of saturation is defined as

S = [V.sub.w] / [V.sub.v] (1.5)

Note that when the soil is fully saturated, all the voids are filled with water (no air). In that case we have [V.sub.v] = [V.sub.w]. Substituting this into (1.5) yields S = 1 (or 100% saturation). On the other hand, if the soil is totally dry, we have [V.sub.w] = 0; therefore, S = 0 (or 0% saturation).

The moisture content (or water content) is the proportion of the weight of water with respect to the weight of solids:

[omega] = [W.sub.w] / [W.sub.s] (1.6)

The water content of a soil specimen is easily measured in the laboratory by weighing the soil specimen first to get its total weight, W. Then the specimen is dried in an oven and weighed to get [W.sub.s]. The weight of water is then calculated as [W.sub.w] = W - [W.sub.s]. Simply divide [W.sub.w] by [W.sub.s] to get the moisture content, (1.6).

Another useful parameter is the specific gravity [G.sub.s], defined as

[G.sub.s] = [[gamma].sub.s] / [[gamma].sub.w] = [W.sub.s]/[V.sub.s] / [[gamma].sub.w] (1.7)

where [[gamma].sub.s] is the unit weight of the soil solids (not the soil itself) and [[gamma].sub.w] is the unit weight of water ([[gamma].sub.w] = 9.81 kN/[m.sup.3]). Note that the specific gravity represents the relative unit weight of solid particles with respect to water. Typical values of [G.sub.s] range from 2.65 for sands to 2.75 for clays.

The unit weight of soil (the bulk unit weight) is defined as

[gamma] = W/V (1.8)

and the dry unit weight of soil is given as

[[gamma].sub.d] = [W.sub.s]/V (1.9)

Substituting (1.6) and (1.9) into (1.8), we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

or

[[gamma].sub.d] = [gamma] / 1 + [omega] (1.10)

Let us assume that the volume of solids [V.sub.s] in Figure 1.1b is equal to 1 unit (e.g., 1 [m.sup.3]). Substitute [V.sub.s] = 1 into (1.1) to get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.11)

Thus,

V = [V.sub.s] + [V.sub.v] = 1 + e (1.12)

Substituting [V.sub.s] = 1 into (1.7) we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.13)

Substitute (1.13) into (1.6) to get

[W.sub.w] = [[omega][W.sub.s] = [[omega][[gamma].sub.w][G.sub.s] (1.14)

Finally, substitute (1.12), (1.13), and (1.14) into (1.8) and (1.9) to get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.15)

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.16)

Another interesting relationship can be obtained from (1.5):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.17)

Equation (1.17) is useful for estimating the void ratio of saturated soils based on their moisture content. For a saturated soil S = 1 and the value of [G.sub.s] can be assumed (2.65 for sands and 2.75 for clays). The moisture content can be obtained from a simple laboratory test (described earlier) performed on a soil specimen taken from the field. An approximate in situ void ratio is calculated as e = [[omega][G.sub.s] [approximately equal to] (2.65 - 2.75)[omega].

For a fully saturated soil, we have e = [omega][G.sub.s] -> [G.sub.s] = e/[omega]. Substituting this into (1.15), we can obtain the following expression for the saturated unit weight:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1.18)

Example 1.1 A 0.9-[m.sup.3] soil specimen weighs 17 kN and has a moisture content of 9%. The specific gravity of the soil solids is 2.7. Using the fundamental equations (1.1) to (1.10), calculate (a) [gamma], (b) [[gamma].sub.d], (c) e, (d) n, (e) [V.sub.w], and (f) S.

SOLUTION: Given: V = 0.9 [m.sup.3], W = 17 kN, [omega] = 9%, and [G.sub.s] = 2.7.

(a) From the definition of unit weight, (1.8):

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(b) From (1.10):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(c) From (1.9):

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From the phase diagram (Figure 1.1b), we have

[W.sub.w] = W - [W.sub.s] = 17 kN - 15.6 kN = 1.4 kN

From (1.7):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Also, from the phase diagram (Figure 1.1b), we have

[V.sub.v] = V - [V.sub.s] = 0.9 [m.sup.3] - 0.5886 [m.sup.3] = 0.311 [m.sup.3]

From (1.1) we get

e = [V.sub.v]/[V.sub.s] = 0.311 [m.sup.3]/0.5886 [m.sup.3] = 0.528

(d) Equation (1.2) yields

n = [V.sub.v]/V = 0.311 [m.sup.3]/0.9 [m.sup.3] = 0.346

(e) From the definition of the unit weight of water,

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(f) Finally, from (1.5):

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1.2.1 Relative Density

The compressibility and strength of a granular soil are related to its relative density [D.sub.r], which is a measure of the compactness of the soil grains (their closeness to each other). Consider a uniform sand layer that has an in situ void ratio e. It is possible to tell how dense this sand is if we compare its in situ void ratio with the maximum and minimum possible void ratios of the same sand. To do so, we can obtain a sand sample from the sand layer and perform two laboratory tests (ASTM 2004: Test Designation D-4253). The first laboratory test is carried out to estimate the maximum possible dry unit weight [[gamma].sub.d]-max (which corresponds to the minimum possible void ratio [e.sub.min]) by placing a dry sand specimen in a container with a known volume and subjecting the specimen to a surcharge pressure accompanied with vibration. The second laboratory test is performed to estimate the minimum possible dry unit weight [[gamma].sub.d]-min (which corresponds to the maximum possible void ratio [e.sub.max]) by pouring a dry sand specimen very loosely in a container with a known volume. Now, let us define the relative density as

[D.sub.r] = [e.sub.max] - e / [e.sub.max] - [e.sub.min] (1.19)

This equation allows us to compare the in situ void ratio directly with the maximum and minimum void ratios of the same granular soil. When the in situ void ratio e of this granular soil is equal to [e.sub.min], the soil is at its densest possible condition and [D.sub.r] is equal to 1 (or [D.sub.r] = 100%). When e is equal to [e.sub.max], the soil is at its loosest possible condition, and its [D.sub.r] is equal to 0 (or [D.sub.r] = 0%). Note that the dry unit weight is related to the void ratio through the equation

[[gamma].sub.d] = [G.sub.s][[gamma].sub.w] / 1 + e (1.20)

It follows that

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1.3 MECHANICAL PROPERTIES OF SOIL

Soil engineers usually classify soils to determine whether they are suitable for particular applications. Let us consider three borrow sites from which we need to select a soil that has the best compaction characteristics for a nearby highway embankment construction project. For that we would need to get details about the grain-size distribution and the consistency of each soil. Then we can use available charts and tables that will give us the exact type of each soil. From experience and/or from available charts and tables we can determine which of these soils has the best compaction characteristics based on its classification.

Most soil classification systems are based on the grain-size distribution curve and the Atterberg limits for a given soil. The grain-size analysis is done using sieve analysis on the coarse portion of the soil (> 0.075 mm in diameter), and using hydrometer analysis on the fine portion of the soil (< 0.075 mm in diameter). The consistency of soil is characterized by its Atterberg limits as described below.

1.3.1 Sieve Analysis

A set of standardized sieves is used for the analysis. Each sieve is 200 mm in diameter and 50 mm in height. The opening size of the sieves ranges from 0.075 mm for sieve No. 200 to 4.75 mm for sieve No. 4. Table 1.1 lists the designation of each sieve and the corresponding opening size. As shown in Figure 1.2, a set of sieves stacked in descending order (the sieve with the largest opening size is on top) is secured on top of a standardized shake table. A dry soil specimen is then shaken through the sieves for 10 minutes. As shown in Figure 1.3, the percent by weight of soil passing each sieve is plotted as a function of the grain diameter (corresponding to a sieve number as shown in Table 1.1). It is customary to use a logarithmic horizontal scale on this plot.

(Continues...)

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Excerpted from Applied Soil Mechanics with ABAQUS Applicationsby Sam Helwany Copyright © 2007 by John Wiley & Sons, Ltd. Excerpted by permission.
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