Download
1 / 120
1.22k likes | 1.27k Views
Prepared From The Coduto’s Text Book by Instr. Nurullah AKBULUT. 4) Soil Composition.
E N D
Prepared From The Coduto’s Text Book by Instr. Nurullah AKBULUT 4) Soil Composition .in engineering practice difficulties with soils are almost exclusively do not to the soils themselves but to the water contained in their voids. On a planet without any water there would be NO need for soil mechanics. Karl Terzaghi, 1939
Once the soil and rock samples have been brought to the laboratory, we need to conduct appropriate tests to develop data for our analyses. Some of these tests measure familiar engineering properties, such as shear strength, while others focus on the sample's composition and structure. The composition of soil and rock is quite different from that of other civil engineering materials, such as steel, concrete, or wood. These differences include: • Soil and rock are natural materials, not manufactured products • Soil is a particulate materialthat consists of individual particles • Soil can contain all three phases of matter(solid, liquid, and gas) simultaneously, and these three phases can be present in varying proportions. Rock also can contain all three phases, although the liquid and gas phases may be confined to the fissures .
The most Important Properties of Soils • Elastoplastic material • Particulate structure • Heterogenity • Non-Isotropi • Three phases of matter This chapter discusses the methods we use to assess the composition of soils and the parameters we use to describe this composition
Most civil engineering materials consist of a continuous mass held together with molecular bonds, and the mechanical properties of such materials depend on their chemical makeup and on the nature of these bonds. For example, the shear strength of steel depends on the strength of the molecular bonds, and shear failure requires breaking them. In contrast, soil is a particulate material that consists of individual particles assembled together as shown in Figure 4.1. Its engineering properties depend largely on the interaction between these particles, and only secondarily on their internal properties. This is especially true in gravels, sands, and silts. For example, when soils fail in shear, they do so because the particles begin to roll and slide past each other, not because the particles break internally. Breakage of individual particles is typically minimal. Thus, the shear strength depends on factors such as the coefficient of friction between the particles, the tightness of packing, and so on, rather than the chemical bonds inside the particles. Clays also have a particulate structure, but the nature of the particles is quite different, as discussed later in this chapter. In clays, there is much more interaction between the particles and the pore water, so their behavior is more complex than that of other soils.
THE THREE PHASES • Soil also is different from most civil engineering materials in that it can simultaneously contain solid, liquid, and gas phases. • The liquid and gas phases are contained in the voidsor pores between the solid particles. • The three phases often interact, and these interactions have important effects on the soil's behavior.
The Mineral Skeleton-Three Phase Solid Particles Volume Voids
The solid phase: is always present in soil, and usually consists of particles derived from rocks. It also can include organic material. • The liquid phase: is usually present, and most often consists of water. However, it also can include other materials, such as:gasoline,leachate, seawater, petroleum seeps. • The gas phase: If the liquid phase does not completely fill the voids, then the remaining space is occupied by the gas phase. It is usually air, but can include other gasses, such as: methane, carbon diokside, hydrogen sulfide, e.t.c.
Clearly, components other than water and air are very important. However, they generally represent only a small portion of the soil weight and volume. Therefore, for purposes of this chapter, we will simply refer to the liquid and gas phases as "pore water" or “water” and "pore air“ or “air”.
WEIGHT-VOLUME RELATIONSHIPS It is helpful to identify the relative proportions of solids, water, and air in a soil, because these proportions have a significant effect on its behavior. Therefore, geotechnical engineers have developed quantitative methods of assessing these components.
Three Phase System Air Wa=0 Va Vv Water Ww Vw V W Solid Vs Ws Volume Weight, or Mass
Definitions of Weight-Volume Parameters Moisture Content or Water Content A small w indicates a dry soil, while a large w indicates a wet one. Values in the field are usually between 3 and 70%.But values greater than 100% are sometimes found in soft soils below the groundwater table, which simply means such soils have more water than solids.
Degree of Saturation • This is similar to moisture content in that both are equal to zero when there is no water. However, S has a maximum value of 100%, which occurs when all of the voids are filled with water. We use the term saturated to describe this condition. • Soils below the groundwater table are generally saturated. • Values of S above the groundwater table are usually between 5 and 100%, although values approaching zero can be found in very arid areas. • Capillary effectscan draw water upward from the groundwater table, often producing soils with S =100% well above the groundwater table. Therefore, do not use saturation computations to determine the location of the groundwater table. Instead, use observation wells.
In all calculations, w = 9.81 kN/ m3= 62.4 Ib/ft3, rw= 1000 kg/m3 for fresh water.
The unit weight of undisturbed soil samples can easily be determined in the laboratory by measuring their physical dimensions and weighing them. This method produces reliable assessments of g for many soils. • However, it is affected by sample disturbance, especially in sandy and gravelly soils. • Sometimes unit weight measurements are made on supposedly "undisturbed" samples that in reality have significant disturbance. • Such measurements are very misleading, so it is best to not even attempt unit weight measurements on poor quality samples • For most geotechnical computations, unit weight is more useful than density because we use it to compute stresses due to the weight of the soil.
Specific Gravity of Solids • The specific gravity of any material is the ratio of its density to that of water. In the case of soils, we compute it for the solid phase only, and express the results as the specific-gravity of solids; • This is quite different from the specific gravity of the entire soil mass, which would include solid, water, and air. Therefore, do not make the common mistake of computing Gs as g / gw. • Nearly all real soils have 2.60 <GS< 2.80, which is a very narrow range. The additional precision obtained by performing a test is generally not worth the expense.
For most practical problems, it is sufficient to estimate ,from the following list: • Clean, light colored sand of quartz and feldspar2.65 • Dark colored sand2.72 • Sand-silt-clay mixtures 2.72 • Clay2.65 Nevertheless, some unusual soils have olivene values well outside these limits. For example, the olivene sands in Hawaii have Gsvalues as high as 4.50, while organic soils have low Gs values, sometimes less than 2.0.
Void Ratio • The relative volumes of voids and solids may be expressed as the void ratio • Densely packed soils have a low void ratio. • Typical values in the field range from 0.1 to 2.5.
Porosity • It is a similar parameter and related with e • It typically is between 9 and 70% • Sometimes geotechnical engineers divide the porosity into two parts: the water porosity, nw, and the air porosity, na:
Relative Density • It is a special weight-volume parameter used in sandy andgravelly soils. • The values of emaxandeminrepresent the soil in very dense and very loose conditions, respectively. Thus, loose soils have low values of Drwhile dense soils have high values. • In theory, the lowest possible value of Dr is 0% and the highest possible value Dr is 100%. • Thus, Dris often more useful than e because we can easily compare the field value to the lowest and highest possible values.
These are not intended to be used in lieu of laboratory or in-situ tests, butcould be used to check test results or for preliminary analyses.
Derived Equationswith Weight Volume Relationships by Phase Diagrams
For all of these equations, parameters normally expressed as a percentage must be inserted in decimal form. • For example, a degree of saturation of 45% in Equation would be expressed as 0.45 not 45
From the definition of void ratio, if the volume of solids is 1 unit then the volume of voids ise units. • The mass of solids is then Gsw and, from the definition of water content, the mass of water is Gsw. • The volume of water is thus wGs. These volumes and masses are represented in figure. • Then a number of relationships can also be derived by using the phase diagram.
Solving Weight-Volume Problems We often encounter problems where one or more of the weight-volume parameters is known and others need to be determined. For example, some parameters, such as moisture content, can be measured in the laboratory, while others, such as void ratio, cannot. Therefore, we need to have a means of computing them. Often we can perform these computations using the derived equations presented last. However, if this method does not work, we go back to fundamentals and solve the problem using a phase diagram as follows: • Draw a phase diagram and annotate all of the dimensions presented in the problem statement. Remember to set Wa = 0. If the soil is saturated, we also can set Va = 0. • Sometimes no weights or volumes are given in the problem statement, or the only dimensions given are equal to zero (i.e., no air or no water). If this is the case, then the given data is applicable to the entire soil strata, regardless of the sample size. However, the phase diagram analysis method requires that a certain quantity of soil be specified, and will not work unless we do so. Therefore, we must assume a quantity. Any one dimension may be assumed (but only one!). Usually we assume V= 1 m3or V= 1 ft3. • Using basic equations and obvious addition and subtraction from the phase diagram (i.e., Va= V - Vs - Vw), determine all of the remaining dimensions. 4. Compute the required parameters using basic equations from these dimensions in phase diagram
Example 1 A 27.50 Ib soil sample has a volume of 0.220 ft3, a moisture content of 10.2%, and a specific gravity of solids of 2.65. Compute the unit weight, dry unit weight, degree of saturation, void ratio, and porosity.
Solution using a phase diagram • Although the fundamental and derived equations were sufficient to solve this problem, and would be the easiest method, we also will illustrate a solution using a phase diagram. • IN EXAMINATIONS and QUIZES, YOU MUST SOLVE RELEVANT PROBLEMS BY USING ONLY THIS METHOD
Step 1: Draw and annotate a phase diagram Step 2: Assume dimension Not applicable to this problem, because weights and volumes have been given
Example2 A certain soil has the following properties: Gs= 2.71 n= 41.9% w = 21.3% Find the degree of saturation, S and the unit weight, by using phase relations?
Solution Step 1: Draw and annotate phase diagram Step 2: Assume dimension. No weights or volumes were stated, so all of the given data applies to the entire soil strata. Therefore, we will develop a phase diagram for an assumed volume of 1 m3
Example 3 The standard method of measuring the specific gravity of solids (ASTM D854) uses a calibrated glass flask known as a pycnometer, as shown in Figure 4.6. The pycnometer is first filled with water and set on a balance to find its mass. Then, it is refilled with a known mass of dry soil plus water so the total volume is the same as before. Again, its mass is determined. From this data, we can compute GsUsing this technique on a certain soil sample, we have obtained the following data: • Mass of soil = 81.8 g • Moisture content of soil = 11.2% • Mass of pycnometer+water = 327.12 g • Mass of pycnometer+soil+water = 373.18 g • Volume of pycnometer = 250.00 ml Compute Gs for this soil.
