ch.2 Soil Aggregate Basic Relationships (composition of soil terminology and definitions) 2.1Composition of soil Soil is a complex physical system. A mass of soil includes accumulated solid particles or soil grains and the void spaces that exist between the particles. The void spaces may be partially or completely filled with water or some other liquid. Void spaces not occupied by water or any other liquid are filled with air or some other gas. Phase means any homogeneous part of the system different from other parts of the system,if not, so the material will be consisting of more than one phase is said to be heterogeneous Since the volume occupied by a soil mass may generally be expected to include material in all the three states of matter solid, liquid and gas, soil is, in general, referred to as a three-phase system. When the soil voids are completely filled with water, the gaseous phase being absent, it is said to be fully saturated or merely saturated. When there is no water at all in the voids, the voids will be full of air, the liquid phase being absent ; the soil is said to be dry. (It may be noted that the dry condition is rare in nature and may be achieved in the laboratory through oven-drying). In both 6
these cases, the soil system reduces to a two-phase one as shown in Fig. 2.2 (a) and (b). These are merely special cases of the three-phase system. 2.2 Basic terminology A number of quantities or ratios are defined below, which constitute the basic terminology in soil mechanics, these quantities are use in predicting the engineering behavior of soil 7
The weight-volume relationships commonly used for the three phases in a soil element Total weight WW s +W w Total volume V V s +V w +V a Water (Moisture) Content Water content or Moisture content of a soil mass is defined as the ratio of the weight of water to the weight of solids (dry weight) of the soil mass. It is denoted by the letter symbol w and is commonly expressed as a percentage : water content of fine-grained soils > water content of coarse- grained soils Void Ratio Void ratio of a soil mass is defined as the ratio of the volume of voids to the volume of solids in the soil mass. It is denoted by the letter symbol e and is generally expressed as a decimal fraction : Porosity Porosity of a soil mass is the ratio of the volume of voids to the total volume of the soil mass. It is denoted by the letter symbol n and is commonly expressed as a percentage: 8
s +V v n vv n ss + vv +,divided by V s ee 1+ee, e nn 1 nn Degree of Saturation Degree of saturation of a soil mass is defined as the ratio of the volume of water in the voids to the volume of voids. It is designated by the letter symbol S and is commonly expressed as a percentage : The S value changes from 0% for completely dry soil conditions to 100% for fully saturated soil. The soils with 0 < S < 100% are called partially saturated soils Percent Air Voids Percent air voids of a soil mass is defined as the ratio of the volume of air voids to the total volume of the soil mass. It is denoted by the letter symbol n a and is commonly expressed as a percentage : n a aa x 100 Air Content Air content of a soil mass is defined as the ratio of the volume of air voids to the total volume of voids. It is designated by the letter symbol a c and is commonly expressed as a percentage : 9
Specific Gravity of Solids The specific gravity of soil solids is defined as the ratio of the unit weight of solids (absolute unit weight of soil) to the unit weight of water at the standard temperature (4 C). This is denoted by the letter symbol G s and is given by : G s MM ss ss ρρ ww WW ss ss γγ ww, but γγ ss WWWW/ G s γγ ss γγ ww ρ MM MMMM - Bulk density, γ, where g 9.81 Unit weight γ WW ( is one of the most important physical properties of the soil ) Units : KN/m 3, gm/cm 3, t/m 3 the unit weight must be expressed with due regard to the state of soil. γ f ( unit weight of solid constituents, n and S) Bulk Unit weight γγ bb WW wwwwww ( for a partially saturated soil, general case) γγ bb (WW ww + WW ss ) / ( ww + ss + aa ) Dry unit weight γγ dd WW ss / ( for dry soils WW ww, ww 00 ) γ d (W s ) / ( V s + V a ) γ d (W s ) / ( V) (W- W w )/ V * ww w / W s, γ d W s /V γ d W/V - w. W s /V γ d γ b - w γ d γγ dd γγ bb /(11 + ww) Saturated unit weight of soil γγ ssssss (W w + W s ) / ( V w + V s ) 10
Submerged (Buoyant) Unit Weight The Submerged unit weight or Buoyant unit weight of a soil is its unit weight in the submerged condition. In other words, it is the submerged weight of soil solids (W s ) sub per unit of total volume, V of the soil. It is denoted by the letter symbol γ sub : γ sub (W s ) sub / V (W s ) sub is equal to the weight of solids in air minus the weight of water displaced by the solids. This leads to : (W s ) sub W s V s. γ w Since the soil is submerged, the voids must be full of water ; the total volume V, then, must be equal to (V s + V w ),(W s ) sub may now be written as : (W s ) sub W W w V s. γ w W V w. γ w V s γ w W γ w (V w + V s ) W V. γ w Dividing throughout by V, the total volume, (W s)sub V 2.3 Certain important relationships (W/V) γ w or γ sub γ sat γ w 11
URelationship among: Gs, S, e, w ww w / W s, S V w / V v, e V v /V s S.e V w /V s ww w / W s ww.γγ ww w S.e / G s SS.γγ ss ww.γγ ww SS.GG ss.γγ ww V w /V s.g s w.g s S.e ------------------------------------------------------------------------------------------------ URelationship among: e, V, V s s +V v V s + v V V s + e V s V s (1+e) V s 1+ee ------------------------------------------------------------------------------------- URelationship among: S, n, n a a c ( V a+v w -V w ) / V v, but V v V a +V w a c (V v -V w )/V v, a c (V v /V v ) - (V w /V v ), a c 1-S, a c 0 (for sat.soil), a c 1 ( for dry soil ) n. a c. n a n a n(1-s) 12
2.5 Illustrative Examples Ex.1: A sample of soil obtained from a test pit is 15x15x15 cm in volume and 6.4 kg in weight, it is oven dried and the dry weight of sample is 5.7 kg. calculate the water content, wet unit weight and dry unit weight. sol: Weight of water W w W- W s 6.4-5.7 0.7 kg Wet unit weight γ wet WW 6.4 1000gg 15 15 15cccc 3 1.9 g/cm3 Dry unit weight γ dry WWWW 5.7 1000gg 15 15 15cccc 3 1.69 g/cm3 Water content w W w /W s 0.7/5.7 0.12312.3% ------------------------------------------------------------------------------------------------ Ex.2: Determine the wet density, dry unit weight, void ratio, water content, and degree of saturation for a sample of moist soil which has a mass of 18.18 kg and occupies a total volume of 0.009 m 3. when dried in an oven, the dry mass is 16.13 kg. The specific gravity of the soil solids is 2.70. sol: wet density, ρ MM 18.18 kkkk 0.009cccc 3 2020 kg/m3 Dry unit weight γ dry WW ss 16.13 kkkk 0.009cccc 3 1792 kg/m3 1.8 gm/cm 3 w WW ww 18.18 16.13 kkkk x 100% xx100% 12.7% WW ss 16.13kkkk e Vv Vs 0.0031 0.0059 0.53 V s MMMM GGGG.ρρ ww 16.13 kkkk 2.7 1 gggg /cccc 3 16.13 kkkk 2700 0.0059 m 3 V v V- V s 0.009-0.0059 0.0031 m 3 S ww.gggg ee 0.127 2.70 0.53 *100% 64.7% 13
Ex.3:A 150 cm 3 sample of well soil its weight 250 g, it is oven-dried and found that its weight is 162 g. calculate the dry unit weight, water content, void ratio and G s. sol: γ dry WWWW 162gg 150cm 3 1.08 g/ cm3 w WW ss WWWW 250 162 gg 162 gg 0.54354.3% V w WW ss 250 162gg 88 GG ww.γγ ww 1 1 gg/cccc 3 cm3 V v ( fully saturated state ) V s V-V v 150-88 62cm 3 e 88cccc 3 62cccc 3 1.42 G s WW ss 162 2.61 ss.γγ ww 62cccc 3 1gg/cccc 3 ------------------------------------------------------------------------------------------------ Ex.4:Labrotory test data on a sample of saturated soil show that the void ratio is 0.45 and the specific of soil solids is 2.65. For these conditions, determine the wet unit weight and water content of this sample of soil. sol: e V v / V s 0.45 ( V v?, V s? ) UBy using specific volume approach V s 1 V V s + ev s 1+0.451.45 W s V s.g s.γ w 1*2.65*1t/m 3 2.65t W w V w.g w.γ w 0.45*1*1t/m 3 0.45t W W s + W w 2.65+0.45 3.1t γ wet W/V 3.1/1.422.14 t/m 3 w W w /W s 0.45/2.650.1717% 14
Ex.5: Undisturbed (30*30*30) cm 3 of soil obtained from a test pit found to have a wet weight if 35 kg and its dry weight is 26kg. what would be the unit weight of such soil if it was below and above the water table, G s 2.7 sol : γ sub (W s ) sub / V (W s ) sub W V. γ w, but W35kg, V 27000 cm 3 (W s ) sub 35000-(27000) 8000 g γ sub 8000/27000 0.296 g/cm 3 γ sat (W w + W s ) / ( V w + V s ) 35000/27000 1.296 g/cm 3 H.W : determine void ratio and porosity ------------------------------------------------------------------------------------------------ Ex.6:For a soil in natural state, given : e 0.8, w 24% and G s 2.68 a) Determine the moist unit weight and dry unit weight and degree of saturation b) If this soil is completely saturated by adding water, what would be its moisture at that time?, also find the saturated unit weight. (γ w 9.81 KN/m 3 ) sol : UBy using unit volume approach V1 a) γγ bbbbbbbb G s (1-n)γ w. (1+w), but n e/(1+e) 4 9 γγ bbbbbbbb 2.68*(1-4 9 )*9.81*(1+0.24) 18.111 KN/m3 γγ dd G s (1-n)γ w 2.68*(1-4 9 )*9.81 14.61 KN/m3 S ww GGGG ee *100[ (0.24*2.68)/0.8 ]*10080.4% b) e w.g s ( sat.state ), w ee GGGG 0.8 *100 100 29.85% 2.68 γ sat G s (1-n)γ w + n γ w [2.68*(1-4 )*9.81]+[ 4 *9.81] 18.97 KN/m3 9 9 15
Ex.7:For a saturated soil, prove that γ sat [ 1+ww 1+wwwwww ]G s.γ w sol : γ sat WW WWWW +WWWW ww.wwww+wwww (1 + ww) WWWW but, w WWWW WWWW, W s G s.v s.γ w γ sat (1 + ww) GGGG..γw e by adding one, 1+e +1 + 1+e but e wg s γ sat (1 + ww) GGGG.γw 1+ee γ sat [ 1+ww 1+wwwwww ]G s.γ w 16