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Introduction In natural occurrence, soils are three-phase systems consisting of soil solids, water, and air. This chapter discusses the weight–volume relationships of soil aggregates. Weight–Volume Relationships

• Figure 3.1a shows an element of soil of volume � and weight �as it would exist in a natural state. • To develop the weight–volume relationships, we separate the three phases; that is, solid, water, and air, as shown in Figure 3.1b. • The total volume of a given soil sample can be expressed as � = �� + �� � = �� + �� + �� (3.1) Where �� = Volume of soil solids �� = Volume of voids �� = Volume of water in the voids �� = Volume of air the voids • Assuming the weight of the air to be negligible, we can give the total weight of the sample as: � = �� + �� (3.2) Where � = Total weight �� = Weight of soil solids �� = Weight of water

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Figure 3.1 (a) Soil element in natural state; (b) three phases of the soil element Volume Relationships The volume relationships commonly used for the three phases in a soil element are void ratio, porosity, and degree of saturation. 1 Void ratio (e) The ratio of the volume of voids to the volume of solids � = ���� Eq.(3.3) 2 Porosity (n) The ratio of the volume of voids to the total volume � = ��� Eq. (3.4) 3 Degree of saturation (S) The ratio of the volume of water to the volume of voids. � = ���� Eq. (3.5)

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The relationship between void ratio and porosity can be derived as follows: � = ���� = ��� − �� = �����1 − ����� = �1 − � (3.6) Also, we have � = �1 + � (3.7) Weight Relationships The common weight relationships are moisture content and unit weight. 1 Moisture content (w) (It is also referred to as water content) and it is defined as the ratio of the weight of water to the weight of solids in a given volume of soil. � = ���� Eq.(3.8) 2 Unit weight (�) It is the weight of soil per unit volume. � = �� Eq. (3.9) The unit weight can also be expressed in terms of weight of soil solids, moisture content, and total volume. From: � = �� = �� + ��� = �� �1 + �������� = ��(1 + �)� or, � = ��(1 + �)� (3.10) Soils engineers sometimes refer to the unit weight defined by Eq. (3.9) as the moist unit weight. It is sometimes necessary to know the weight per unit volume of soil excluding water. This is referred to as the dry unit weight, �� . Thus,

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�� = ��� (3.11) From Eqs. (3.10) and (3.11), we can give the relationship among unit weight, dry unit weight, and moisture content as: �� = �1 + � (3.12) Unit weight is expressed in kilonewtons per cubic meter (kN/m3). Since the newton is a derived unit, it may sometimes be convenient to work with densities � of soil. The SI unit of density is kilograms per cubic meter (kg/m3). We can write the density equations [similar to Eqs. (3.9) and (3.11)] as � = �� (3.13) �� = ��� (3.14) where � = Density of soil (kg/m3) �� = Dry density of soil (kg/m3) � = Total mass of the soil sample (kg) �� = Mass of soil solids in the sample (kg) The unit of total volume, �, is ��. The unit weights of soil in �/�3 can be obtained from densities in ��/�� as � = �.� = 9.81� (3.15) and �� = �� .� = 9.81�� (3.15) Where � = acceleration due to gravity = 9.81 m/sec2

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Relationships among Unit Weight, Void Ratio, Moisture Content, and Specific Gravity To obtain a relationship among unit weight (or density), void ratio, and moisture content, consider a volume of soil in which the volume of the soil solids is 1, as shown in Figure 3.2. If the volume of the soil solids is 1, then the volume of voids is numerically equal to the void ratio, e [from Eq. (3.3)]. The weights of soil solids and water can be given as �� = ���� �� = ��� = ����� where �� = specific gravity of soil solids � = moisture content ��= unit weight of water

Figure 3.2 Three separate phases of a soil element with volume of soil solids equal to 1

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The unit weight of water is 9.81 kN/m3. Now, using the definitions of unit weight and dry unit weight [Eqs. (3.9) and (3.11)], we can write � = �� = �� + ��� = ���� + �����1 + � = (1 + �)����1 + � (3.17) and �� = ��� = ����1 + � (3.18) Since the weight of water in the soil element under consideration is �����, the volume occupied by it is �� = ���� = ������� = ��� Hence, from the definition of degree of saturation [Eq.(3.5)], we have � = ���� = ���� or �� = ��� (3.19) This is a very useful equation for solving problems involving three-phase relationships. If the soil sample is saturated—that is, the void spaces are completely filled with water (Figure 3.3)—the relationship for saturated unit weight can be derived in a similar manner: ���� = �� = �� + ��� = ���� + ���1 + � = (�� + �)��1 + � (3.20) Where ���� saturated unit weight of soil.

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As mentioned before, because it is convenient to work with densities, the following equations [similar to the unit-weight relationships given in Eqs. (3.17), (3.18), and (3.20)] are useful: Density � = (���)������� (3.21) Dry density �� = ������� (3.22) Saturated ���� = (����)����� (3.23) where ��= density of water = 1000 kg/m3.

Figure 3.3 Saturated soil element with volume of soil solids equal to 1 Some typical values of void ratio, moisture content in a saturated condition, and dry unit weight for soils in a natural state are given in Table 3.1.

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Table 3.1 Void Ratio, Moisture Content, and Dry Unit Weight for Some Typical Soils in a Natural State

Relationships among Unit Weight, Porosity, and Moisture Content The relationships among unit weight, porosity, and moisture content can be developed in a manner similar to that presented in the preceding section. Consider a soil that has a total volume equal to one, as shown in Figure 3.4. From Eq. (3.4), � = ���

Figure 3.4 Soil element with total volume equal to 1

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If � is equal to 1, then �� is equal to �, so �� = 1 − �. The weight of soil solids (��) and the weight of water (��) can then be expressed as follows: �� = ����(1 − �) (3.24) �� = ��� = �����(1 − �) (3.25) So, the dry unit weight equals �� = ��� = ����(1 − �)1 = ����(1 − �) (3.26) The moist unit weight equals � = �� + ��� = ����(1 − �)(1 + �) (3.27) Figures 3.5 shows a soil sample that is saturated and has � = 1. According to this figure. ���� = �� + ��� = (1 − �)���� + ���1 = [(1− �)�� + �]�� (3.28) The moisture content of a saturated soil sample can be expressed as � = ���� = ���(1− �)���� = �(1 − �)�� (3.29)

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Figure 3.5 Saturated soil element with total volume equal to 1

Example 3.1 For a saturated soil, show that ���� = ���� �1 + �1 + �� �� Solution From Eqs.(3.19) and (3.20) ���� = (����)����� (a) and � = ��� or �� = �� (b)

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Combining Eqs.(a) and (b) gives ���� = (�� + �)��1 + � = ���� �1 + �1 + �� ��

Example 3.2 A moist soil has these values: � = 7.08 × 10���� , � = 13.95 �� , � = 9.8% and �� = 2.66. Determine the following: a. � b. �� c. � d. � e. � (%) f. Volume occupied by water Solution a. From Eq. (3.13). � = �� = 13.957.08 × 10�� = 1970.3 ��/�� b. From Eq.(3.12) �� = �1 + � = 1970.31 + 9.8100 = 1794.4 ��/�� c. From Eq.(3.22) � = ������ − 1 � = (2.66)(1000)1794.4 − 1 = 0.48 d. From Eq.(3.7) � = �1 + � = 0.481 + 0.48 = 0.324 e. From Eq.(3.19) �(%) = ����� � (100)

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→ �(%) = (0.098)(2.66)0.48 (100) = 54.3% f. Mass of soil solids is �� = �1 + � = 13.951 + 0.098 = 12.7 �� Thus, mass of water is �� = � −�� = 13.95 − 12.7 = 1.25 �� Volume of water is �� = ���� = 1.251000 = 0.00125 �� Example 3.3 In the natural state, a moist soil has a volume of 0.0093 m3 and weighs 177.6 N. The oven dry weight of the soil is 153.6 N. If �� = 2.71, calculate the moisture content, moist unit weight, dry unit weight, void ratio, porosity, and degree of saturation. Solution Refer to Figure 3.6. The moisture content [Eq. (3.8)] is � = ���� = � −���� = 177.6 − 153.6153.6 = 15.6% The moist unit weight (Eq.(3.9)) is � = �� = 177.60.0093 = 19096 �/�� ≈ 19.1��/�� For dry unit weight (Eq.(3.11)), we have �� = ��� = 153.60.0093 = 16516 �/�� ≈ 16.52 ��/��

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The void ratio (Eq.(3.3) is found as follows: � = ���� �� = ���� �� = 0.15362.71 × 9.81 = 0.0058 �� �� = � − �� = 0.0093− 0.0058 = 0.0035 ��

Fig. 3-6 Illustration of Example 3-3 so � = 0.00350.0058 ≈ 0.60 For porosity [Eq.(3.7)], we have � = �1 + � = 0.601 + 0.60 = 0.375 We find the degree of saturation [Eq.(3.5)] as follows: � = ����

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�� = ���� = 0.0249.81 = 0.00245 �� so � = 0.002450.0035 × 100 = 70% Example 3.4 The dry density of a sand with a porosity of 0.387 is 1600 kg/m3. Find the void ratio of the soil and the specific gravity of the soil solids. Solution Void ratio From � = 0.387 and Eq.(3.6) � = �1 − � = 0.3871 − 0.387 = 0.631 Specific gravity of soil solids from Eq.(3.22) �� = ����1 + � where �� = dry density of soil �� = density of water 1000 kg/m3 Thus, 1600 = ��(1000)1 + 0.631 �� = 2.61

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Example 3.5 For a saturated soil, given � = 40% and �� = 2.71, determine the saturated and dry unit weights. Solution For saturated soil, from Eq. (3.19) � = ��� = (0.4)(2.71) = 1.084 From Eq.(3.20), ���� = (�� + �)��1 + � = (2.71 + 1.084)9.811 + 1.084 = 17.86 ��/�� From Eq.(3.18), �� = ����1 + � = (2.71)(9.81)1 + 1.084 = 12.76 ��/�� Example 3.6 The mass of a moist soil sample collected from the field is 465 g, and its oven dry mass is 405.76 g. The specific gravity of the soil solids was determined in the laboratory to be 2.68. If the void ratio of the soil in the natural state is 0.83, find the following: a) The moist unit weight of the soil in the field (kN/m3) b) The dry unit weight of the soil in the field (kN/m3) c) The weight of water (in kN) to be added per cubic meter of soil in the field for saturation Solution Part a. From Eq. (3.8),

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� = ���� = 465 − 405.76405.76 = 14.6% From Eq.(3.17), � = (1 + �)����1 + � = (1 + 0.146)(2.68)(9.81)1 + 0.83 = 16.46 ��/�� Part b. From Eq.(3.18), �� = ����1 + � = (2.68)(9.81)1 + 0.83 = 14.37 ��/�� Part c. From Eq.(3.20), ���� = (�� + �)��1 + � = (2.68 + 0.83)(9.81)1 + 0.83 = 18.82 ��/�� So, the weight of water to be added is ���� − � = 18.82− 16.46 = 2.36 ��/��

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Relative Density The term relative density is commonly used to indicate the in situ denseness or looseness of granular soil. It is defined as �� = ���� − ����� − ���� where ��= Relative density, usually given as a percentage � = In-situ void ratio of the soil ����= Void ratio of the soil in the loosest condition ����= Void ratio of the soil in the densest condition The values of �� may vary from a minimum of 0 for very loose soil, to a maximum of 1 for very dense soil. Soils engineers qualitatively describe the granular soil deposits according to their relative densities, as shown in Table 3.2. Table 3.2 Qualitative description of granular soil deposits

By using the definition of dry unit weight given in Eq. (3.18), we can also express relative density in terms of maximum and minimum possible dry unit weights. Thus, �� = � �� − ��(���)��(���) − ��(���)� ���(���)�� �

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Example 3.7 For a given sandy soil, ���� = 0.75, ���� = 0.46, and �� = 2.68. What is the moist unit weight of compaction (kN/m3) in the field if �� = 78% and � = 9%? Solution From Eq. (3.30), �� = ���� − ����� − ���� → 0.78 = 0.75 − �0.75 − 0.46 → � = 0.524 Again, from Eq.(3.17), � = (1 + �)����1 + � = (1 + 0.09)(2.68)(9.81)1 + 0.524 = 18.8 ��/��