CHAPTER 2 Basic Terminology and Interrelations

2.1 PHASES IN If any volume of mass is considered, it comprises in general of three phases, viz., (mineral ), air (gas) and liquid (water), as shown in Fig. 2.1.

Fully Completely Partially saturated dry saturated

Ma VW Vv Va Vv VW

Fig. 2.1 Multiphase soil model.

A soil mass, if fully saturated, will be a two-phase system of soil grains and pore water. If completely dry, it will also be a two-phased system comprising solid soil grains and pore air. A partially saturated soil will, however, be three-phase system, comprising solid soil grains, water and air. Typical two- and three-phase diagrams are shown in Fig. 2.2.

Fully Completely Partially saturated dry saturated Mass Weight M = 0 V Vv V Vv a Wa = 0 W a V M M W W W Mww× r

V M WM= g V V V M S S S S WMss= g

(a) (b) (c)

Fig. 2.2 Two-phase diagrams (a) and (b) for fully saturated and completely dry soils (c) three-phase diagram for partially saturated soil. 22

h e e = _____ also h = _____ 1 – h 1 + e

____WW (MC)w = WS

W Ws(1 + w) and Unit Weight g = __ = ______V V

WS g dry unit weight g = ___ = _____ d V 1 + w

Air Va

Vv Total Total Water weight volume Ww Vw V = W V gw

Solid Ws Vs Gsgw

Fig. 2.3

2.3 UNIT WT., , MC AND SP. GRAVITY RELATIONSHIP

Wt. of soil solids, WS = Gsgw where GS = Sp. gravity of soil solids

Wt. of water = w WS = wGsgw w = Moisture content

gw = unit wt. of water

Fig. 2.4 24 Geotechnical Engineering

ILLUSTRATIVE EXAMPLES Example 2.1 The bulk of a soil sample is 1.90 gm/cc. The moisture content is 15%. The specific gravity of the soil particles is 2.61. Calculate the dry unit weight, dry density, void ratio, and degree of saturation. Total mass M Solution: Bulk density r = ______= __ = 1.9 gm/cc Total volume V

Ww + Ws Ww Bulk unit weight g = ______; w = ___ × 100 V Ws ____w Ws 1 + ______( 100 ) ______V × g \ g = or Ws = w V 1 + ____ ( 100 )

Ws V × g g r × g 1.9 × 9.8 Dry unit weight, g = ___ = ______= ______= ______= ______= 16.2 kN/m3 d V ____w ____w ____w 15 1 + V 1 + 1 + 1 + ____ ( 100 ) 100 100 100

______V × r Dry density rd = Ms/V; M/V = 1.9, M = 1.9 V, Ms = w 1 + ____ V ( 100 ) ______1.9 V ____1.9 = w = = 1.65 gm/cc 1 + ____ V 1.15 ( 100 )

Consider 1 cc of soil (V = 1), Vs = 1.65/Gs = 1.65/2.61 = 0.63 cc

Vv = 1 – 0.63 = 0.37 cc

c = Vv /Vs = 0.37/0.63 = 0.59 V n = ___ v = 0.37/1.00 = 0.37 V Mass of water in 1 cc of soil = 1.90 – 1.65 = 0.25 gm 0.25 Degree of saturation = V /V = 100 × ____ = 67.6%. w v 0.37 Example 2.2 If a soil has a void ratio of 0.7 and a specific gravity of 2.72, calculate the following: (a) Dry unit weight (b) Saturated unit weight (c) Submerged unit weight Also compute the unit weight and at a degree of saturation of 75%. Basic Terminology and Interrelations 25

Solution: e = 0.7; Gs = 2.72

Gsgw 2.72 × 9.8 (a) Dry unit weight g = _____ = ______= 15.68 kN/m3 d 1 + e 1 + 0.7

Gs + e (b) Saturated unit weight g = ______g ; S = 1 sat 1 + e w 2.72 + 0.7 g = ______× 9.8 = 19.7 kN/m3 sat 1 + 0.7

Gs – 1 (c) Submerged (or buoyant) unit weight g = ______× g 1 + e w 2.72 – 1 g = ______× 9.8 = 9.91 kN/m3 1 + 0.7

Gs + Sr × e g = ______g 1 + e w

2.72 + (0.75 × 0.70) 2.72 + 0.525 At S = 0.75, g = ______× 9.8 = ______× 9.8 r 1 + 0.70 1.70 3.245 × 9.8 = ______= 18.7 kN/m3 1.70 Since w Gs = Sre 0.75 × 0.70 w × 2.72 = 0.75 × 0.70 \ w = ______= 19.3% 2.72

Example 2.3 A cylindrical soil specimen having a volume of 86.15 cm3 weighs 168.0 gm in its natural condition. When dried completely in an oven, the specimen weighs 130.5 gm. The value of Gs is 2.73. What is the degree of saturation of the specimen?

Solution: W = 168.0 gm, Ws = 130.5 gm

Water content w = Ww /Ws = (168.0 – 130.5)/130.5 = 37.5/130.5 = 0.287 or 28.7% 3 Gs = Ws /Vs i.e., 2.73 = 130.5/Vs or Vs = 47.8 cm

Vv = V – Vs 3 Vv = 86.1 – 47.8 = 38.3 cm

Void ratio e = Vv /Vs = 38.3/47.8 = 0.80

Now Sr e = wGs

Sr × 0.8 = 0.287 × 2.73 or Sr = 0.979 i.e., degree of saturation of the soil sample = 97.9% 26 Geotechnical Engineering

Example 2.4 An attempt was made to determine the water content of a given moist soil of known specific gravity, using a pycnometer. The usual laboratory procedure for specific gravity determination of dry soil is done on the wet soil. Following are the observations:

Mass of pycnometer (M1) = 545 g

Mass of pycnometer with moist soil (M2) = 790 g

Mass of pycnometer with soil and water (M3) = 1540 g

Mass of pycnometer and water (M4) = 1415 g Specific gravity of soil grains = 2.67 Determine the water content of the soil from the first principles Solution: Consider the two-phase diagrams shown in Fig. 2.5 representing the observations.

Water Water

Water Mw

Ms Soil solids Water VVw = s

(a) Mass M1 (b) Mass M4

Fig. 2.5

Thus,

M3 – M4 = Ms – (mass of an equal volume of water)

____Ms ___Ms = Ms – rw Vs = = Vw G ( rs ) ( rw ) G – 1 = M _____ s ( G ) G \ M = (M – M ) _____ s 3 4 ( G – 1 ) (M – M ) – M and w = ______2 1 s × 100 Ms Substituting the respective values 2.67 M = (1540 – 1415) ______= 199.85 g s (2.67 – 1) Basic Terminology and Interrelations 27

(790 – 545) – 199.85 and w = ______× 100 = 22.6% 199.85 Example 2.5 The bulk unit weight of a soil is 19.10 kN/m3, the water content is 12.5% and the specific gravity of soil solids 2.67. Determine the dry unit weight, void ratio, porosity and degree of saturation. g 19.1 Solution: g = _____ = ______= 16.98 kN/m3 d 1 + w 12.5 1 + ____ 100

G gw g = _____ d 1 + e

G gw 2.67 × 9.81 \ e = ____ – 1 = ______– 1 = 0.54 gd 16.98 e 0.54 \ n = _____ × 100 = ______× 100 = 35.07% 1 + e 1 + 0.54 wG 12.5 2.67 and S = ___ × 100 = ____ × ____ × 100 = 61.8% r e 100 0.54 Example 2.6 How many cubic metres of fill can be constructed at a void ratio of 0.65 from 2,21,000 m3 of borrow material that has a void ratio of 1.25?

Solution: Let eb and ef be the void ratios of the borrow material and the fill, respectively. Also let Vvb and Vvf be the volume of voids in the borrow and the fill, respectively. As the same number of soil grains obtained from the borrow are used in the fill, the volume of soil solids is same in both the cases. From Fig. 2.6

Voids Vvb

Voids Vvf

Vb

Vf

Soil solids Vs Soil solids Vs

(a) Borrow (b) Fill

Fig. 2.6 V ___Vvb ___vf eb = and ef = Vs Vs

\ Vvb = eb Vs and Vvf = ef Vs 28 Geotechnical Engineering

Total volume of soil in the borrow = Vb = Vvb + Vs i.e., Vb = eb Vs + Vs = (1 + eb) Vs

_____Vb Vs = 1 + eb

Total volume of soil in the fill = Vf = (1 + ef) Vs

Substituting for Vs V _____b 1______+ 0.65 3 Vf = (1 + ef) = × 221000 = 162066.7 m 1 + eb 1 + 1.25 Example 2.7 For a stable packing of regular spheres at the minimum density, find the void ratio and the dry unit weight. Unit weight of soil solids is 25 kN/m3. Solution: Let D be the diameter of each sphere. D3 Volume of each sphere = ____ p 6 For the arrangement in Fig. 2.7 the density will be minimum Total volume = 2D × 2D × D = 4 D3 Fig. 2.7

4D3 – 4 × p D3/6 1 – p/6 6 – p \ e = ______= ______= _____ = 0.91 4 × p D3/6 p/6 p 3 ___Ms ____Vs gs 4______× p D /6 × gs ___p gs p______× 25 Also gd = g = = = = V V 4D3 6 6 3 or gd = 13.09 kN/m

Example 2.8 In order to determine the in-place density of a highway , the bottle method was adopted. The mass of soil extracted from a hole at the surface was 4.87 kg. The hole was then filled with sand from the sand bottle and found to have a mass of 3.86 kg. While calibrating the sand bottle, to fill a container of volume 0.0048 m3 a mass of 6.82 kg of sand was needed. In moisture content determination, 28.26 g of the moist soil weighed 22.2 g after oven drying. If the specific gravity of the soil was 2.67, determine the bulk and dry of the soil and the degree of saturation of the soil.

6.82 Solution: Density of sand in sand bottle = ______= 1420.8 kg/m3 0.0048

= 1.42 Mg/m3

3.86 Volume of the hole = ______= 0.00272 m3 1420.8 Basic Terminology and Interrelations 29

4.87 Bulk density = ______= 1790.4 kg/m3 = 1.79 mg/m3 0.00272 28.26 – 22.2 Water content = ______× 100 = 27.3% 22.2 1790.4 Dry density = ______= 1406.4 kg/m3 = 1.41 mg/m3 27.3 1 + ____ 100

Grw 2.67 × 1000 e = ____ –1 = ______– 1 = 0.899 rd 1406.4 wG 27.3 2.67 S = ___ × 100 = ____ × _____ × 100 = 81.08% r e 100 0.899 Example 2.9 A fully saturated soil sample was extracted during an oil drilling. The wet mass of the sample was 3.15 kg and the volume of the sampling tube was 0.001664 m3. After analysis the soil sample was found to contain 28.2% of the liquid as kerosene and the dry mass as 2.67 kg. The specific gravity of soil grains was 2.68. Determine the bulk density, void ratio, and water content of the sample. 3.15 Solution: Bulk density = ______= 1893 kg/m3 = 1.89 mg/m3 0.001664 2.67 Volume of soil grains = ______= 0.000996 m3 2.68 × 1000 Volume of voids = 0.001664 – 0.000996 = 0.000668 m3 0.000668 \ e = ______= 0.67 0.000996 As the soil was fully saturated, Volume of liquid = volume of voids = 0.000668 m3 Volume of water = (1 – 0.282) × 0.000668 = 0.00048 m3 Mass of water = 0.48 kg 0.48 Water content = ____ × 100 = 17.98% 2.67

Exercise

1. A soil sample whose water content is 20% has a bulk density of 2.16 gm/cc. The sample undergoes air drying with significant change in void ratio. What is the water content of this sample when its bulk density is reduced to 2.0 gm/cc. (Ans. = 11.11%)