Basic Component of soil Weight-Volume Relationships Prepared by: [email protected] unimas. my 1 Introduction • Soil is a three-phase material consisting of a skeleton of solid particles. • The solid particles encompassing voids filled with water & air. • It is necessary that the constitution of the solidswater-air mixture can be expressed quantitatively in terms of some standard physical properties. • Soil water is commonly known as pore water • If all voids are filled with water = soil is saturated, otherwise will be known as soil is unsaturated.

not all voids are filled with water) • If all voids are filled with air = soil is dry Prepared by:[email protected] unimas. my 2 At the end of lecture, students should be able to: • Determine the proportions of the main constituents in a soil • Determine particle size distribution in a soil mass • Classify soils • Determine index properties of soils Prepared by:[email protected] unimas. my 3 Definition of Key Terms Water content (w) is the ratio of the weight of water to the weight of solids • Void ratio (e) is the ratio of the volume of voids to the volume of solids • Porosity (n) is the ratio of the volume of voids to the total volume of soil • Degree of saturation(Sr) is the ratio of the volume of water to the volume of voids • Bulk unit weight (? bulk) is the weight density, the weight of a soil per unit volume • Saturated unit weight (? sat) is the weight of a saturated soil per unit volume • Dry unit weight (? dry) is the weight of a dry soil per unit volume • Effective unit weight (? ) is the weight of soil solids in a submerged soil per unit volume Prepared by:[email protected]

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unimas. my 4 Weight-volume relationships • Soil can be idealized in 3 phases as shown in Fig. 1. • The physical properties of soils are influenced by the relative proportions of each of these phases. Solids Va Void Air Wa V Vw Water Ww W Vs Solids Ws Fig. 1 idealization Prepared by:[email protected] unimas. my 5 • Referring to Fig. 1, the total volume of the soil is the sum of the volume of solids (Vs), volume of water (Vw), and volume of air (Va): V ? Vs ? Vw ? Va ? Vs ?

Vv Where VV ? VW ? Va = volume of voids As the weight of air, Wa = 0, The weight of the soil is: W ? WS ? WW Note: *the following equations have been established based on the unit solid volume idealized phases Prepared by:[email protected] unimas. my 6 Basic quantities of soil components derived from phase diagrams Water content (w) or Moisture Content (m) ratio of weight of water to the weight of solids in soil, often expressed as percentage w? Ww x 100% Ws Void Ratio (e) volume not occupied by solid = volume of voids expressed as a decimal quantity (no dimension) olume of voids Vv Va ? Vw Void ratio, e ? ? ? volume of solids Vs Vs Specific Volume (v) volume of soil per unit volume of solids v? V ? 1? e Vs Prepared by:[email protected] unimas. my 7 Porosity (n) ratio of volume of voids to the total volume expressed as percentage n? volume of voids Vv ? total volume V The void ratio & porosity are inter-related as follows: e? n e , ? n ? 1? n 1? e 5 It can be proven by: n? Vv VV ? ? V Vs ? Vv VS Vv VS Vs V ? v ? Vs e 1? e Notes: Regularly, coarse-grained soils void ratios vary from 1 to 0. 3. Clay soils can have void ratios greater than 1

Prepared by:[email protected] unimas. my 8 Specific Gravity (Gs) the ratio of the weight of soil solids to the weight of water of equal volume W W Gs ? s ? s Ww Vs? w Vs ? Vw ?w = unit weight of water = 9. 81kN/m3 Mass of water equal volume; ?w ? Ww ? Ww ? ? wVw Vw Prepared by:[email protected] unimas. my 9 Phase Relationships –basic quantities are used to define other quantities in unit solid volume basis Volumes Va = e (1-Sr) Specific Volume , v = 1 + e e weight Air Wa = 0 Vw = wGs = Sre Water Ww = wGs? w Vs = 1 Solids Figure 2 Ws = Gs? w

Referring to Figure 2 above: Degree of Saturation (Sr) the quantity of water in the soil often expressed in percentage. defined as the ratio of the volume of water to the volume of voids Notes: If Sr = 1 or 100% (soil is saturated) If Sr = 0 (soil is perfectly dry) Sr ? volume of water Vw wGs ? ? volume of voids Vv e ? S r e ? wGs 10 Prepared by:[email protected] unimas. my Air-voids content (Av) air-voids volume of a soil is a part of voids volume which not occupied by water Air-voids volume = vol. of voids – vol. of water Refer to Fig 2 Va ? Vv ? Vw ? e ? Sr e ? e(1 ? Sr )

Av = ratio of the air-voids volume to the specific volume of soil Va e(1 ? S r ) Av ? ? ? n(1 ? S r ) v 1? e since so, wGs Sr ? e e ? wGs Av ? 1? e Prepared by:[email protected] unimas. my 11 Unit Weight is the weight of a soil per unit volume. *Unit = N/mm3, kN/mm3, kN/m3 1. Bulk unit weight is denoted as unit weight, ? ?? W ? Gs ? S r e ? ?? ?? w v ? 1? e ? 2) Dry unit weight (? dry) when Sr = 0, ?d ? Ws ? Gs ? ? ?? ?w ? ? v ? 1? e ? 1? w 3) Saturated unit weight (? sat) ? G ? e? ? sat ? ? s ?? w when Sr = 1, 1? e ? ? ? Gs ? 1 ? ?? w ? 1? e ? 4) Effective/bouyant/submerged unit weight (? ) ? ? ? ? sat ? ? w ? ? Prepared by:[email protected] unimas. my 12 1) Bulk Density (? bulk) total mass mass of solids ? mass of water ? total volume total volume G ? ? S r e? w Gs (1 ? w) ? s w ? ?w 1? e 1? e ?bulk ? 2) Dry Density (? dry) when Sr = 0, mass of solid Gs ? w ? d ? ? total volume 1? e These equations can be related by: 3) Saturated Density (? sat) when Sr = 1, G ? e ? sat ? s ? w 1? e 4) Submerged Density (? ‘) a. k. a as effective density of a soil & gives, ?bulk ? (1 ? w) ? dry ? ? ? ? sat ? ? w 13 Prepared by:[email protected] unimas. my

Example 1: (pg 47, Whitlow (2001)_Basic Soil Mechanics) For a soil having a void ratio of 0. 750 and percentage saturation of 85%, determine the porosity and air-voids ratio. Solution: Given: e = 0. 750, Sr = 85% = 0. 85 n = ? , Av = ? Step 1= n ? e ? 0. 750 ? 0. 429 1 ? e 1 ? 0. 750 Step 2= Av ? n(1 ? S r ) ? 0. 429(1 ? 0. 85) ? 0. 064 or percentage air voids ? 6. 4% So, the porosity, n = 0. 429 & air-voids ratio = 6. 4% Prepared by:[email protected] unimas. my 14 Example 1: (pg 48, Whitlow (2001)_Basic Soil Mechanics) An oven tin containing a sample of moist soil was weighed and had a mass of 37. 2g; the empty tin had a mass of 16. 15g. After drying, the tin and soil were weighed again and had a mass of 34. 68g. Determine the void ratio of the soil if the air-voids content is a) zero b) 5 per cent (Gs = 2. 70) Solution: Given: Soil + water + tin = 37. 82g Soil + tin = 34. 68g (after drying) Tin only = 16. 15g e = ? , if Av = 0 and 5% So, water content, w ? M w Ms = 37. 82-34. 68 = 0. 169 34. 68-16. 15 a) Av = 0, e = wGs = 0. 169 x 2. 70 = 0. 456 b) Av = 0. 05, rearranging (8), wGs ? Av (0. 169 x 2. 70) ? 0. 05 e? ? ? 0. 533 1 ? Av 0. 95 So, a) 0. 169 b) 0. 33 Prepared by:[email protected] unimas. my 15 Example 3: (pg 27, Craig (1992)_Soil Mechanics) In its natural condition a soil sample has a mass of 2290g and a volume of 1. 15E10-3 m3. After being completely dried in an oven the mass of the sample is 2035g. The value of Gs for the soil is 2. 68. Determine the bulk density, unit weight, water content, void ratio, porosity, degree of saturation and air content. Solution: Given: Msoil = 2290g, Vsoil = 1. 15E10-3 m3 Msoil = 2035g, Gs = 2. 68 ? bulk = ? , ? = ? , w = ? , e = ? , n = ? , Sr = ? , Av = ? 1) ? bulk ? 2) M 2. 90 ? ? 1990kg/m3 ? 3 V 1. 15E10 ?? Mg ? 1990 ? 9. 8 ? 19500 N /m3 ? 19. 5 kN/m3 V 3) w ? M w ? 2290 ? 2035 ? 0. 125 or 12. 5% Ms 2035 Prepared by:[email protected] unimas. my 16 Example 3: (cont’d) 4) e ? Gs (1 ? w) ?w ? 1 ? bulk 1000 ? ? ? ? 2. 68 ? 1. 125 ? ? ? 1 1990 ? ? ? 1. 52 ? 1 ? 0. 52 e 0. 52 ? ? 0. 34 or 34% 1 ? e 1. 52 5) n? 6) Sr ? wGs 0. 125 ? 2. 68 ? ? 0. 645 or 64. 5% e 0. 52 7) Av ? n(1 ? Sr ) ? 0. 34 ? 0. 355 ? 0. 121 or 12. 1% Prepared by:[email protected] unimas. my 17 Thank you let proceed with tutorials! Prepared by:[email protected] unimas. my 18