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2 ENCE 361 Soil Mechanics Lecture 4 Soil Composition

3 Soil Composition Soil is a collection of particles that do not form an totally solid substance Soil is a combination of: Weight-Volume Relationships Water or Moisture Content Unit Weight or Mass Specific Gravity Relative Density Soil material in particles Air Water The relationship between this combination defines much of what any particular soil can do to support foundations

4 Phase Diagram Assumptions and Definitions: Weight of air = 0 Dry Soil: Water weight and volume = 0, all voids filled with air Volume of voids include all non-soil volume, both air and water Equalities and Summations Vtotal=Vair+Vwater+Vsolids Wtotal=Wwater+Wsolids Mtotal=Mwater+Msolids Wx=γx*Vx or Mx=ρx*Vx

5 Saturated Soil Saturated Soil: Air volume = 0 Only water and solids appear in completely saturated soil

6 Weight and Volume Relationships

7 Weight and Volume Relationships

8 Systems of Units Used in Phase Calculations Weight and Volume Relationships In most cases, calculations in soil mechanics are done on a weight basis (SI or US units) Exceptions include wave propagation problems (earthquakes, pile dynamics, etc. U.S. Units Most common within the U.S., generally not used elsewhere Basic units of weight: 1 pound (lb.) or kip (1000 lbs.) Basic unit of length: feet Will frequently write psf and pcf instead of lb/ft2 and lb/ft3 SI Units Technically the official system everywhere else Basic units of weight: N or kn Basic unit of length: m MKS Units Commonly used in metric countries Basic unit of mass: kg Basic unit of length: m

9 Specific Gravity and Density Unit Weight of Water (γw) 62.4 pcf = kcf 9.81 kn/m3 Density of Water 1.95 slugs/ft 1000 kg/m3 1 g/cm3 = 1 Mg/m3 = 1 Metric Ton/m3 3 Typical Values for Soil Particles Quartz Sand: Silt: Clay: Chalk: Loess: Peat: Except for organic soils, range is fairly narrow

10 Example 1 Given: Total Volume = 1 cu. ft. Total Weight = 140 lb. Dry Weight = 125 lb. Find Water Content Wet Unit Weight Dry Unit Weight By Definition: Dry Unit Weight = Dry Weight = 125 lb/ft3 Wet Unit Weight = Total Weight = 140 lb/ft3 Solve for Weight of Water W T = W s + Ww 140 = 125+Ww Ww = 15 lb/ft3 Solve for Water Content w = Ww/Ws = Ww/125 = 15/125 = 0.12 = 12%

11 Example 2 Given: Convert masses to weights Total Mass = kg Wt = (18.18)(9.8) = Total Volume = m3 N = kn Dry Mass = kg Ws = (16.13)(9.8) = Specific Gravity of Solids = N = kn 2.7 Compute Weight of Water Find Wt = Ws+Ww Wet Density.178 = Ww Dry Unit Weight Ww =.02 kn Void Ratio Compute Water Content Water Content w = Ww/Ws Degree of Saturation w =.02/.158 =.127 = 12.7%

12 Example 2 Compute Wet Unit Weight Compute Volumes γwet = WT/VT = 0.178/0.009 = kn/m3 Volume of Water Vw = W w / γ w Vw =.02 kn/9.8 kn/m = m3 3 Volume of Solids Vs = Ws/ γs = ws/(gs γw) Vs = 0.158/((9.8)(2.7)) = m3 Volume of Air Va = V t V w V s Va = = m3 Compute Dry Unit Weight γdry = Ws/VT = 0.158/0.009 = kn/m3 Void Ratio e = Vv/Vs = (Vw+Va)/Vs = ( )/ =.507 Compute Degree of Saturation S = Vw/(Vw+Va) S =.00205/( ) =.677 = 67.7%

13 Example 3 Given Saturated Soil Void Ratio = 0.45 Specific Gravity of Solids = 2.65 Find Wet Unit Weight Water Content Assumptions Va = 0 Vt = 1 Vs + V w = 1 γwater = 62.4 lb/ft 3 Solve for Volumes Compute Wet Unit Weight e = Vw/Vs = 0.45 Vw = 0.31 ft3 Vs = 0.69 ft3 Weight of Solids = γwvsgs = (62.4)(0.69)(2.7) = 114 lb Weight of Water = γwvw = (62.4)(0.31) = 19.4 lb Total Weight = = lb Since volume is unity, total weight is also net unit weight = pcf Compute Water Content w = Ww/Ws = 19.4/114 = 0.17 = 17%

14 Example 4 Given Well Graded Sand Specific Gravity of Solids = 2.65 Void Ratio = 0.57 Porosity = 36.5% Saturated Soil Find and Vw =.365 ft3 Dry Unit Weight Set sample volume = 1 ft Total Volume = 1= Vw + Va + Vs Use porosity to compute volume of voids n = Vv/Vt Vv = nvt = (.365)(1) =.365 ft3 Vs = Vt Vv = =.635 ft3 Vv = Vw + Va Since soil is saturated, Va = 0 Wet and Dry Unit Weight of Soil Solution Compute volume of solids 3 Weight of Water Ws = γwgsvs = (62.4)(2.65) (.635) = 105 lb/ft3 Ww = γ wvw = (62.4)(.362) = 22.6 lb/ft3 Wet Unit Weight Wt = Ww + Wv = lb/ft3

15 Relative Density Density Both unit weight and strength of soil can vary with particle arrangement Denser soils have both higher load carrying capacity and lower settlement Definition of Relative Density emax = void ratio of the soil in its loosest condition emin = void ratio of the soil in its densest condition e = void ratio in the natural or condition of interest of the soil Convenient measure for the strength of a cohesionless soil

16 Relative Density Example Given Sand Backfill Unit Weight = 109 pcf Water Content = 8.6% Specific Gravity of Solids = 2.6 e max = (loosest state) e min = (densest state) Assume Vt = 1 ft3; thus, Wt = 109 lbs. Weight balance: 109 = W + W s w Water content w = W /W = w s Solving two previous equations: Ws = lbs; Ww = 8.6 lbs. V = W /γ = 100.4/((2.6)(62.4)) = ft3 s s s V = W / γ = 8.6/62.4 = ft3 w w w V = V V V = = a t w s ft3 e = (Va + Vw)/Vs = ( )/0.618 = Find Void Ratio Relative Density RD = ( )/( ) = = 14.4%

17 Use of Relative Density for Empirical Values

18 Phase Relationships, Relative Density and Atterberg Limits

19 Questions?