CAPITAL HUMANO
B) Capital humano adquirido
B.3) Experiencia
1. Indicadores basados en la educación formal recibida
Interactive Concept Learning
Access Chapter 3, Section 3.3 to view an animation of soil phases and to interactively learn about phase relationships. Take Quiz 3.3 to test your understanding of the concepts. Develop problem-solving skills by problem-solving problems interactively using the Problem Solver in the left sidebar.
Soil is composed of solids, liquids, and gases (Fig. 3.2a). The solid phase may be mineral, organic matter, or both. As mentioned before, we will not deal with organic matter in this textbook. The spaces between the solids (soil particles) are called voids. Water is often the predominant liquid and air is the predominant gas. We will use the terms water and air instead of liquids and gases. The soil water is commonly called porewater and it plays a very important role in the behavior of soils under load. If all the voids are filled by water, the soil is saturated. Otherwise, the soil is unsaturated. If all the voids are filled with air, the soil is said to be dry.
We can idealize the three phases of soil as shown in Fig. 3.2b. The physical properties of soils are influenced by the relative proportions of each of these phases. The total volume of the soil is the sum of the volume of solids ðVsÞ, volume of water ðVwÞ, and volume of air ðVaÞ: that is,
V¼ Vsþ Vwþ Va¼ Vsþ Vv (3.1)
where
Vv¼ Vwþ Va
is the volume of voids. The weight of the soil is the sum of the weight of solidsðWsÞ and the weight of water ðWwÞ. The weight of air is negligible. Thus,
W¼ Wsþ Ww (3.2)
The following definitions have been established to describe the proportion of each constituent in soil.
Each equation can be presented with different variables. The most popular and convenient forms are given. You should try to memorize these relationships. When you work on these relationships, think about a bread dough in which you have to reconstruct the amount of the constituent ingredients, for example, the amount of flour or water. If you add too much water to a bread dough it becomes softer and more malleable. The same phenomenon occurs in fine-grained soils.
Air
Water
Solids Solids
Void
(a) Soil (b) Idealized soil
Idealization
V W
Va
Vw
Vs
Ww
Ws Wa = 0
FIGURE 3.2 Soil phases.
1. Water content (w) is the ratio, often expressed as a percentage, of the weight of water to the weight of solids:
w¼Ww
Ws
100% (3.3)
The water content of a soil is found by weighing a sample of the soil and then placing it in an oven at 110 5C until the weight of the sample remains constant; that is, all the absorbed water is driven out.
For most soils, a constant weight is achieved in about 24 hours. The soil is removed from the oven, cooled, and then weighed. Example 3.2 illustrates the measurements and calculations required to determine the water content. It is a common mistake to use the total weight in the denominator.
Remember it is the weight (or mass) of the solids.
2. Void ratio (e) is the ratio of the volume of void space to the volume of solids. Void ratio is usually expressed as a decimal quantity.
e¼Vv
Vs (3.4)
3. Specific volumeðV0Þ is the volume of soil per unit volume of solids:
V0¼ V Vs
¼ 1 þ e (3.5)
This equation is useful in relating volumes as shown in Example 3.3 and in the calculation of volumetric strains (Chapter 5).
4. Porosity (n) is the ratio of the volume of voids to the total volume. Porosity is usually expressed as a percentage.
n¼Vv
V (3.6)
Porosity and void ratio are related by the expression
n¼ e
1þ e (3.7)
Let us prove Eq. (3.7). We will start with the basic definition [Eq. (3.6)] and then algebraically manipulate it to get Eq. (3.7). The total volume will be decomposed into the volume of solids and the volume of voids and then both the numerator and denominator will be divided by the volume of solids;
that is,
n¼Vv
V ¼ Vv
Vsþ Vv
¼ Vv=Vs
Vs=Vsþ Vv=Vs
¼ e
1þ e
The porosity of soils can vary widely. If the particles of coarse-grained soils were spheres, the maximum and minimum porosities would be 48% and 26% respectively. This is equivalent to maximum and minimum void ratios of 0.91 and 0.35, respectively. The void ratios of real coarse-grained soils vary between 1 and 0.3. Clay soils can have void ratios greater than 1.
3.3 PHASE RELATIONSHIPS 35
5. Specific gravityðGsÞ is the ratio of the weight of the soil solids to the weight of water of equal volume:
Gs¼ Ws
Vsgw (3.8)
where gw¼ 9:81 kN/m3 is the unit weight of water. We will use gw¼ 9:8 kN/m3 in this book. The specific gravity of soils ranges from approximately 2.6 to 2.8. For most problems, Gs can be assumed, with little error, to be equal to 2.7.
Two types of container are used to determine the specific gravity. One is a pycnometer, which is used for coarse-grained soils. The other is a 50 mL density bottle, which is used for fine-grained soils.
The container is weighed and a small quantity of dry soil is placed in it. The mass of the container and the dry soil is determined. De-aired water is added to the soil in the container. The container is then agitated to remove air bubbles. When all air bubbles have been removed, the container is filled with de-aired water. The mass of container, soil, and water is determined. The contents of the container are discarded and the container is thoroughly cleaned. De-aired water is added to fill the container and the mass of the container and water is determined.
Let m1be the mass of the container; m2be the mass of the container and dry soil; m3be the mass of the container, soil, and water; and m4be the mass of the container and water. The mass of dry soil is ms¼ m2 m1, the mass of water displaced by the soil particles is m5¼ m4 m3þ ms, and Gs¼ ms=m5. 6. Degree of saturation (S) is the ratio, often expressed as a percentage, of the volume of water to the volume of voids:
S¼Vw
Vv
¼wGs
e or Se¼ wGs (3.9)
If S¼ 1 or 100%, the soil is saturated. If S ¼ 0, the soil is bone dry. It is practically impossible to obtain a soil with S¼ 0.
7. Unit weight is the weight of a soil per unit volume. We will use the term bulk unit weight, g, to denote unit weight:
(a) Saturated unit weightðS ¼ 1Þ:
gsat¼ Gsþ e
(c) Effective or buoyant unit weight is the weight of a saturated soil, surrounded by water, per unit volume of soil:
g0¼ gsat gw¼ Gs 1 1þ e
gw (3.13)
Typical values of unit weight of soils are given in Table 3.1.
8. Relative densityðDrÞ is an index that quantifies the degree of packing between the loosest and densest possible state of coarse-grained soils as determined by experiments:
Dr¼ emax e
emax emin (3.14)
where emax is the maximum void ratio (loosest condition), emin is the minimum void ratio (densest condition), and e is the current void ratio.
The relative density can also be written as
Dr¼ gd ðgdÞmin ðgdÞmax ðgdÞmin
ðgdÞmax gd
(3.15)
The maximum void ratio is found by pouring dry sand, for example, into a mold of volume (V) 2830 cm3using a funnel. The sand that fills the mold is weighed. If the weight of the sand is W, then by combining Eqs. (3.10) and (3.12) we get emax¼ GsgwðV=WÞ 1. The minimum void ratio is determined by vibrating the sand with a weight imposing a vertical stress of 13.8 kPa on top of the sand. Vibration occurs for 8 minutes at a frequency of 3600 Hz and amplitude of 0.064 mm. From the weight of the sandðW1Þ and the volume ðV1Þ occupied by it after vibration, we can calculate the minimum void ratio using emin¼ GsgwðV1=W1Þ 1.
The relative density correlates very well with the strength of coarse-grained soils—denser soils being stronger than looser soils. A description of sand based on relative density is given in Table 3.2.
TABLE 3.1 Typical Values of Unit Weight for Soils
Soil type gsat(kN/m3) gd(kN/m3)
Gravel 20^22 15^17
Sand 18^20 13^16
Silt 18^20 14^18
Clay 16^22 14^21
TABLE 3.2 Description Based on Relative Density Dr(%) Description
0^20 Very loose
20^40 Loose
40^70 Medium dense or firm
70^85 Dense
85^100 Very dense
3.3 PHASE RELATIONSHIPS 37
What’s next. . .Four examples will be used to illustrate how to solve a variety of problems involving the consti-tuents of soils. In the first example, we will derive some of the equations describing relationships among the soil constituents.
Interactive Problem Solving
Access Chapter 3 on the CD, click on Problem Solver on the sidebar to interactively solve problems on phase relationships. Each time you rerun a problem, the values in the problem change.
EXAMPLE 3.1 Deriving Soil Constituent Relationships Prove the following relationships:
(a) S¼wGs
e (b) gd¼ g
1þ w (c) g¼ Gsþ Se
1þ e
gw¼Gsgwð1 þ wÞ 1þ e
Strategy The proofs of these equations are algebraic manipulations. Start with the basic definition and then manipulate the basic equation algebraically to get the desired form.
Solution 3.1
(a) For this relationship, we proceed as follows:
Step 1: Write down the basic equation,
S¼Vw
Vv
Step 2: Manipulate the basic equation to get the desired equation.
You want to get e in the denominator and you have Vv. You know that Vv¼ eVsand Vwis the weight of water divided by the unit weight of water. From the definition of water content, the weight of water is wWs. Here is the algebra:
Vv¼ eVs; Vw¼Ww
gw ¼wWs
gw S¼ wWs
egwVs
¼Gsw e (b) For this relationship, we proceed as follows:
Step 1: Write down the basic equation,
gd¼Ws
V
Step 2: Manipulate the basic equation to get the new form of the equation.
gd¼Ws
V ¼W Ww
V ¼W
VwWs
V ¼ g wgd
; gdþ wgd¼ g gd¼ g
1þ w
(c) For this relationship, we proceed as follows:
Step 1: Start with the basic equation,
g¼W V
Step 2: Manipulate the basic equation to get the new form of the equation.
g¼W
V ¼Wsþ Ww
Vsþ Vv
¼Wsþ wWs
Vsþ Vv
Substituting w¼ Se=Gs and Vv¼ eVs, we obtain g¼Wsð1 þ Se=GsÞ
Vsð1 þ eÞ
¼Gsgwð1 þ Se=GsÞ
1þ e ¼Gsgwð1 þ wÞ 1þ e or
g¼ Gsþ Se 1þ e
gw &
EXAMPLE 3.2 Calculating Soil Constituents
A sample of saturated clay was placed in a container and weighed. The weight was 6 N. The clay in its container was placed in an oven for 24 hours at 105C. The weight reduced to a constant value of 5 N.
The weight of the container is 1 N. If Gs¼ 2:7, determine the (a) water content, (b) void ratio, (c) bulk unit weight, (d) dry unit weight, and (e) effective unit weight.
Strategy Write down what is given and then use the appropriate equations to find the unknowns.
You are given the weight of the natural soil, sometimes called the wet weight, and the dry weight of the soil. The difference between these will give the weight of water and you can find the water content by using Eq. (3.3). You are also given a saturated soil, which means that S¼ 1.
Solution 3.2
Step 1: Write down what is given.
Weight of sampleþ container ¼ 6 N Weight of dry sampleþ container ¼ 5 N Step 2: Determine the weight of water and the weight of dry soil:
Weight of water: Ww¼ 6 5 ¼ 1 N Weight of dry soil: Ws¼ 5 1 ¼ 4 N Step 3: Determine the water content.
w¼Ww
Ws
100 ¼1
4 100 ¼ 25%
Note: The denominator is the weight of solids, not the total weight.
3.3 PHASE RELATIONSHIPS 39
Step 4: Determine the void ratio.
e¼wGs
S ¼0:25 2:7 1 ¼ 0:675 Step 5: Determine the bulk unit weight.
g¼W
V ¼Gsgwð1 þ wÞ
1þ e ðsee Example 3:1Þ
g¼2:7 9:8ð1 þ 0:25Þ
1þ 0:675 ¼ 19:7 kN/m3
In this case the soil is saturated, so that the bulk unit weight is equal to the saturated unit weight.
Step 6: Determine the dry unit weight.
gd¼Ws
V ¼ Gs
1þ e
gw¼ 2:7
1þ 0:675 9:8 ¼ 15:8 kN/m3 or
gd¼ g
ð1 þ wÞ¼ 19:7
1þ 0:25¼ 15:8 kN/m3 Step 7: Determine the effective unit weight.
g0¼ Gs 1 1þ e
gw¼ 2:7 1 1þ 0:675
9:8 ¼ 9:9 kN/m3 or
g0¼ gsat gw¼ 19:7 9:8 ¼ 9:9 kN/m3 &
EXAMPLE 3.3 Application of Soil Constituent Relationships to a Practical Problem An embankment for a highway is to be constructed from a soil compacted to a dry unit weight of 18 kN/m3. The clay has to be trucked to the site from a borrow pit. The bulk unit weight of the soil in the borrow pit is 17 kN/m3and its natural water content is 5%. Calculate the volume of clay from the borrow pit required for 1 cubic meter of embankment. Assume Gs¼ 2:7.
Strategy This problem can be solved in many ways. We will use two of these ways. One way is direct;
the other a bit longer. In the first way, we are going to use the ratio of the dry unit weights of the compacted soil to the borrow pit soil to determine the volume. In the second way, we will use the specific volume. In this case, we need to find the void ratio for the borrow pit clay and the desired void ratio for the embankment. We can then relate the specific volumes of the embankment and the borrow pit clay.
Solution 3.3
Step 1: Find the dry unit weight of the borrow pit soil.
gd¼ g
1þ w¼ 17
1þ 0:05¼ 16:2 kN/m3
Step 2: Find the volume of borrow pit soil required.
Volume of borrow pit soil required per m3
¼ðgdÞcompacted soil
ðgdÞborrow pit soil
¼ 18
16:2¼ 1:11 m3 Alternatively:
Step 1: Define parameters for the borrow pit and embankment. Let
V01; e1¼ specific volume and void ratio; respectively; of borrow pit clay V02; e2¼ volume and void ratio; respectively; of compacted clay Step 2: Determine e1 and e2.
gd¼ g
1þ w¼ 17
1þ 0:05¼ 16:2 kN/m3 But
gd¼ Gs
1þ e1
gw and therefore
e1¼ Gs
gw
gd 1 ¼ 2:7 9:8 16:2
1 ¼ 0:633
Similarly,
e2¼ Gs
gw
gd 1 ¼ 2:7 9:8 18
1 ¼ 0:47
Step 3: Determine the volume of borrow pit material.
V10
V20¼1þ e1
1þ e2
Therefore
V10 ¼ V201þ e1 1þ e2
¼ 1 1þ 0:633 1þ 0:47
¼ 1:11 m3 &
EXAMPLE 3.4 Application of Soil Constituent Relationships to a Practical Problem If the borrow soil in Example 3.3 were to be compacted to attain a dry unit weight of 18 kN/m3at a water content of 7%, determine the amount of water required per cubic meter of embankment, assuming no loss of water during transportation.
Strategy Since water content is related to the weight of solids and not the total weight, we need to use the data given to find the weight of solids.
3.3 PHASE RELATIONSHIPS 41
Solution 3.4
Step 1: Determine the weight of solids per unit volume of borrow pit soil.
Ws¼ g
1þ w¼ 17
1þ 0:05¼ 16:2 kN/m3 Step 2: Determine the amount of water required.
Additional water¼ 7 5 ¼ 2%
Weight of water¼ Ww¼ wWs¼ 0:02 16:2 ¼ 0:32 kN Vw¼Ww
gw ¼0:32
9:8 ¼ 0:033 m3¼ 33 liters &
What’s next. . .In most soils, there is a distribution of particles with various sizes.The distribution of particle size influences the response of soils to loads and to flow of water. We will describe methods used in the laboratory to find particle sizes in soils.