T ERMINOLOGY AND D EFINITIONS
2.4 ILLUSTRATIVE EXAMPLES
COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 21
Water content, w = weight of water weight of solids = S e
G ww . .
. γ
γ = S.e/G
or w.G = S.e
γ = total weight total volume = +
+ = +
+
( . )
( ) . ( )
( )
G S e e
G w
w w e
1
1 γ γ1 γd = weight of solids
total volume = + G
e . m
( )
γ 1 and so on.
The reader may, in a similar manner, prove the other relationships also.
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22 GEOTECHNICAL ENGINEERING
Water content, w = ( . . )
. 1668 1400
140− ×100% = 19.14%
Total volume of soil sample, V = π
4 × (3.81)2 × 7.62 cm3 = 86.87 cm3
Bulk unit weight, γ = W/V = 1668 86 87
.
. = 0.0192 N/cm3 = 18.84 kN/m3
Dry unit weight, γd = γ
( )
.
( . )
1
18 84 1 0 1914
+ =
w + kN/m3 = 15.81 kN/m3 Specific gravity of solids, G = 2.70
γd = G e . w
( )
γ
1 + γw = 9.81 kN/m3 15.81 = 2 7 9 81
1
. .
( )
×
+e (1 + e) = 2 7 9 81 15 81
. .
.
× = 1.675
∴ Void ratio, e = 0.675
Degree of saturation, S = wG
e =0 1914 2 70× 0 675
. .
. = 0.7656 = 76.56%.
Example 2.3: A soil has bulk density of 20.1 kN/m3 and water content of 15%. Calculate the water content if the soil partially dries to a density of 19.4 kN/m3 and the void ratio remains
unchanged. (S.V.U.—B.E. (R.R.)—Dec., 1971)
Bulk unit weight, γ = 20.1 kN/m3 Water content, w = 15%
Dry unit weight, γd = γ
( )
.
( . )
1
20 1 1 0 15
+ =
+
w kN/m3 = 17.5 kN/m3
But γd = G
e . w
( )
γ 1 + ;
if the void ratio remains unchanged while drying takes place, the dry unit weight also remains unchanged since G and γw do not change.
New value of γ = 19.4 kN/m3
γd = γ (1 + w)
∴ γ = γd(1 + w)
or 19.4 = 17.5 (1 + w)
(1 + w) = 19 4 17 5 .
. = 1.1086 w = 0.1086
Hence the water content after partial drying = 10.86%.
Example 2.4: The porosity of a soil sample is 35% and the specific gravity of its particles is 2.7.
Calculate its void ratio, dry density, saturated density and submerged density.
(S.V.U.—B.E. (R.R.)—May, 1971)
COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 23 Porosity, n = 35%
Void ratio, e = n/(1 – n) = 0.35/0.65 = 0.54 Specific gravity of soil particles = 2.7
Dry unit weight, γd = G e . w
( )
γ 1 + = 2 7 9 81
154
. .
.
× kN/m3 = 17.20 kN/m3
Saturated unit weight, γsat = ( ) (G e) .
e w + 1+ γ = ( . . )
. 2 70 0 54
154
+ × 9.81 kN/m3 = 20.64 kN/m3
Submerged unit weight, γ ′ = γsat – γw
= (20.64 – 9.81) kN/m3 = 10.83 kN/m3.
Example 2.5: (i) A dry soil has a void ratio of 0.65 and its grain specific gravity is = 2.80. What is its unit weight ?
(ii) Water is added to the sample so that its degree of saturation is 60% without any change in void ratio. Determine the water content and unit weight.
(iii) The sample is next placed below water. Determine the true unit weight (not consid-ering buoyancy) if the degree of saturation is 95% and 100% respectively.
(S.V.U.—B.E.(R.R.)—Feb, 1976) (i) Dry Soil
Void ratio, e = 0.65 Grain specific gravity, G = 2.80 Unit weight, γd = G
e . w
( )
. .
. γ
1
2 80 9 8 165
+ = × kN/m3 = 16.65 kN/m3. (ii) Partial Saturation of the Soil
Degree of saturation, S = 60%
Since the void ratio remained unchanged, e = 0.65 Water content, w = S e
G
. . .
=0 60 0 65.×
2 80 = 0.1393 = 13.93%
Unit weight = ( )
( ) . ( . . . )
. .
G Se e w +
+ = + ×
1
2 80 0 60 0 65
165 9 81
γ kN/m3
= 18.97 kN/m3. (iii) Sample below Water
High degree of saturation S = 95%
Unit weight = ( )
( ) . ( . . . )
. .
G Se e w +
+ = + ×
1
2 80 0 95 0 65
165 9 81
γ kN/m3
= 20.32 kN/m3
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24 GEOTECHNICAL ENGINEERING
Full saturation, S = 100%
Unit weight = ( )
( ) . ( . . )
. .
G e e w +
+ = +
1
2 80 0 65 165 9 81
γ kN/m3
= 20.51 kN/m3.
Example 2.6: A sample of saturated soil has a water content of 35%. The specific gravity of solids is 2.65. Determine its void ratio, porosity, saturated unit weight and dry unit weight.
(S.V.U.—B.E.(R.R.)—Dec., 1970) Saturated soil
Water content, w = 35%
specific gravity of solids, G = 2.65
Void ratio, e = wG, in this case.
∴ e = 0.35 × 2.65 = 0.93
Porosity, n = e
e 1
0 93 193 + = .
. = 0.482 = 48.20%
Saturated unit weight, γSat = ( ) (G e) .
e w + +
1 γ
= ( . . )
( . ) .
2 65 0 93 1 0 93+ 9 81
+ ×
= 18.15 kN/m3 Dry unit weight, γd = G
e . w
( )
γ 1 + = 2 65 9 81
193
. .
.
× = 13.44 kN/m3.
Example 2.7: A saturated clay has a water content of 39.3% and a bulk specific gravity of 1.84.
Determine the void ratio and specific gravity of particles.
(S.V.U.—B.E.(R.R.)—May, 1969) Saturated clay
Water content, w = 39.3%
Bulk specific gravity, Gm = 1.84 Bulk unit weight, γ = Gm .γw
= 1.84 × 9.81 = 18.05 kN/m3 In this case, γsat = 18.05 kN/m3
γsat = ( ) (G e).
e w + +
1 γ
For a saturated soil,
e = wG
or e = 0.393 G
∴ 18.05 = ( . )
(G . G) . ( . ) G
+ +
0 393
1 0 393 9 81
COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 25 whence G = 2.74
Specific gravity of soil particles = 2.74 Void ratio = 0.393 × 2.74 = 1.08.
Example 2.8: The mass specific gravity of a fully saturated specimen of clay having a water content of 30.5% is 1.96. On oven drying, the mass specific gravity drops to 1.60. Calculate the
specific gravity of clay. (S.V.U.—B.E.(R.R.)—Nov. 1972)
Saturated clay
Water content, w = 30.5%
Mass specific gravity, Gm = 1.96
∴ γsat = Gm .γw = 1.96 γw
On oven-drying, Gm = 1.60
∴ γd = Gm.γw = 1.60γw
γsat = 1.96.γw = ( )
( )
G e e + w
+ γ
1 ...(i)
γd = 1.60.γw = G e . w
( )
γ
1 + ...(ii)
For a saturated soil, e = wG
∴ e = 0.305G
From (i),
1.96 = ( . )
( . )
.
( . )
G G
G
G G +
+ =
+ 0 305
1 0 305
1305 1 0 305
⇒ 1.96 + 0.598G = 1.305G
⇒ G = 1960
0 707 .
. = 2.77 From (ii),
1.60 = G/(1 + e)
⇒ G = (1 + 0.305G) 1.6
⇒ G = 1.6 + 0.485G
⇒ 0.512G = 1.6
⇒ G = 1.6/0.512 = 3.123
The latter part should not have been given (additional and inconsistent data).
Example 2.9: A sample of clay taken from a natural stratum was found to be partially satu-rated and when tested in the laboratory gave the following results. Compute the degree of saturation. Specific gravity of soil particles = 2.6 ; wet weight of sample = 2.50 N; dry weight of sample = 210 N ; and volume of sample = 150 cm3. (S.V.U.—B.E.(R.R.)—Nov., 1974)
Specific gravity of soil particles, G = 2.60
Wet weight, W = 2.50 N;
Volume, V = 150 cm3
Dry weight, Wd = 2.10 N
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Water content, w = ( ) ( . . )
. W W
Wd d
− ×100= 2 5 2 1− ×
2 1 100%
= 0 40
2 10. 100%
. × = 19.05%
Bulk unit weight, γ = W/V = 2.50/150 = 0.0167 N/cm3
= 16.38 kN/m3 Dry unit weight, γd = γ
( )
.
( . )
1
16 38 1 0 1905
+ =
+
w kN/m3
= 13.76 kN/m3
Also, γd Wd N/cm3 kN/m3
= V = = =
L NM O
QP
2 10 150 0 014. / . 13 734.
But γd = G
e . w
( )
γ 1 + 13.76 = 2 6 9 81
1
. .
( )
× +e (1 + e) = 2 6 9 81
13 76
. .
.
× = 1.854 e = 0.854
Degree of saturation, S = wG
e = 0 1905 2 6× 0 854
. .
. = 0.58
= 58%
Aliter. From the phase-diagram (Fig. 2.6) V = 150 cc W = 2.50 N Wd = Ws = 2.10 N
Water
Solids V = 150 cm3
V = 69.23 cmv
3
W = 0.40 Nw Air
Vw= 40 cm
3
V = 80.77 cms
3 W = 2.10 Ns
W = 2.50 N
Fig. 2.6 Phase diagram (Example 2.9) Ww = (2.50 – 2.10) N
= 0.40 N Vw = Ww
γw = 0 40 0 01
.
. = 40 cm3,
COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 27
Vs = W W G
s s
s
γ = γw = . ×
.
. .
2 10
2 6 0 01 = 80.77 cm3 Vv = (V – Vs) = (150 – 80.77) = 69.23 cm3 Degree of saturation, S = V
V
w v
S = 40/69.23 = 0.578
∴ S = 40/69.23 = 0.578
∴ Degree of saturation = 57.8%
Thus, it may be observed that it may sometimes be simpler to solve numerical problems by the use of the soil-phase diagram.
Note. All the illustrative examples may be solved with the aid of the soil-phase diagram or the unit-phase diagram also ; however, this may not always be simple.
SUMMARY OF MAIN POINTS
1. Soil is a complex physical system ; generally speaking, it is a three-phase system, mineral grains of soil, pore water and pore air, constituting the three phases. If one of the phases such as pore water or pore air is absent, it is said to be dry or saturated in that order ; the system then reduces to a two-phase one.
2. Phase-diagram is a convenient representation of the soil which facilitates the derivation of use-ful quantitative relationships involving volumes and weights.
Void ratio, which is the ratio of the volume of voids to that of the soil solids, is a useful concept in the field of geotechnical engineering in view of its relatively invariant nature.
3. Submerged unit weight is the difference between saturated unit weight and the unit weight of water.
4. Specific gravity of soil solids or grain specific gravity occurs in many relationships and is one of the most important values for a soil.
REFERENCES
1. Alam Singh & B.C. Punmia : Soil Mechanics and Foundations, Standard Book House, Delhi-6, 1970.
2. A.R. Jumikis : Soil Mechanics, D. Van Nostrand Co., Princeton, NJ, USA, 1962.
3. T.W. Lambe and R.V. Whitman : Soil Mechanics, John Wiley & Sons, Inc., NY, 1969.
4. D.F. McCarthy : Essentials of Soil Mechanics and Foundations, Reston Publishing Company, Reston, Va, USA, 1977.
5. V.N.S. Murthy : Soil Mechanics and Foundation Engineering, Dhanpat Rai and Sons, Delhi-6, 2nd ed., 1977.
6. S.B. Sehgal : A Text Book of Soil Mechanics, Metropolitan Book Co., Ltd., Delhi, 1967.
7. G.N. Smith : Essentials of Soil Mechanics for Civil and Mining Engineers, Third Edition, Metric, Crosby Lockwood Staple, London, 1974.
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8. M.G. Spangler : Soil Engineering, International Textbook Company, Scranton, USA, 1951.
9. D.W. Taylor : Fundamentals of Soil Mechanics, John Wiely & Sons, Inc., New York, 1948.