Weight-Volume Relations Soil can be considered as a 3-phased material Air, Water, Solids

Soil Structure

3-Phase Idealization

3-Phase Soil Block Weight lb g kg kN Soil Phase Volume ft 3 cc m 3 W A = 0 Air V A WTWT W W Water V W V V V T W S Solids V S

Weight Relations Water content, w w = [W W /W S ] x 100% may be > 100% for clays Total (Moist,Wet) Unit Weight ( = ( T = ( WET = W T / V T Dry Unit Weight ( d = W S / V T Table 2.2

Volumetric Relations Void Ratio, e e = V V / V S may be > 1, especially for clays Porosity, n n = [V V / V T ] x 100% 0% < n < 100% Degree of Saturation, S S = [V W / V V ] x 100% 0% < S < 100%

Inter-relationships Wet -> Dry Unit Weight ( d = ( WET / (1+w/100) W S = W T / (1+w/100) Dry Unit Saturation (Zero Air Voids) ( zav = ( W / (w/100+1/Gs)

Soil Block Analysis Use given soil data to completely fill out weight and volume slots Convert between weight and volume using specific gravity formula Known Weight: V = W / G s ( w Known Volume: W = V G s ( w ( w =1g/cc=9.81kN/m 3 =1000kg/m 3 =62.4lb/ft 3

Example Soil Block Analysis kgm3m3 0 A 4.0 W S Given: W T =4kg, V T =0.002m 3, w=20%, Gs=2.68

Example Soil Block Analysis Given: W T =4kg, V T =0.002m 3, w=20%, Gs=2.68 W S = W T / (1+w/100) W S = 4kg / (1+20/100) = kg W W = W T – W S W W = 4kg – kg = kg Check w = W W /W S x 100% w = 100% x / = 20.01%  OK

Example Soil Block kgm3m3 0 A W S

Example Soil Block Analysis V S = W S / G S ( w V S = 3.333kg / (2.68 x 1000 kg/m 3 ) = m 3 V W = W W /G S ( w V W = 0.667kg / (1 x 1000kg/m 3 ) = m 3 V A = V T – V S - V W V A = – – = m 3 V V = V A + V W V V = = m 3

Example Soil Block kgm3m3 0 A W S

Example Soil Block Analysis ( T = 4.0kg/0.002m 3 = 2000 kg/m 3 =19.62 kN/m 3 =124.8lb/ft 3 ( D = 3.333kg/0.002m 3 =1666.5kg/m 3 =16.35 kN/m 3 ( D = 19.62kN/m 3 / 1.20 =16.35 kN/m 3 e = / = n = 100x /0.002 = 38.0% S = 100x / = 88.2%

Modified Soil Block Analysis kgm3m3 0 A W 100 S Given: W T =4kg, V T =0.002m 3, w=20%, Gs=2.68

Modified Soil Block Analysis Given: W T = 4 kg, V T = m 3 W T / V T ratio must remain unchanged 4 kg / m 3 = 120 kg / X X = 0.06 m 3 = V T

Modified Soil Block Analysis V S = W S / G S ( w V S = 100kg / (2.68 x 1000 kg/m 3 ) = m 3 V W = W W /G S ( w V W = 20kg / (1 x 1000kg/m 3 ) = m 3 V A = V T – V S - V W V A = –0.0373– = m 3 V V = V A + V W V V = = m 3

Modified Soil Block kgm3m3 0 A W S

Modified Soil Block Analysis ( T = 120kg/0.06m 3 = 2000 kg/m 3 =19.62 kN/m 3 ( D = 100kg/0.06m 3 =1666.7kg/m 3 =16.35 kN/m 3 e = / = (0.613) n = 100x0.0227/0.06 = 37.8% (38.0%) S = 100x0.02/ = 88.1% (88.2%)

Saturation Assumption If a soil is partially saturated, we can get to full saturation by direct replacement of air with water. It is further assumed that there will be no increase in total volume.

3-Phase Idealization Solids Water Air

Modified Soil Block kgm3m3 2.7 AWAW W S

In Situ Comparators Relative Density, D r D r =100% x [e max – e in situ ] / [e max – e min ] O% < D r < 100% Relative Compaction, R% R% = [ ( d-in situ / ( d-max,lab ] x 100% R% may be > 100%

Consistency of Soil Atterberg Limits Liquid Limit, LL Plastic Limit, PL Shrinkage Limit, SL

Atterberg Limits

Liquid Limit

Liquid Limit Plot Shear strength of LL is approx. 2.5 kN/m 2 (0.36 psi)

Liquid Limit Europe & Asia Fall Cone Test BS1377

Plastic Limit 3mm Diameter Thread

Shrinkage Limit

Consistency of Soil Plasticity Index, PI PI = LL - PL Activity, A A = PI / % Clay Liquidity Index, LI LI = [w – PL] / [LL – PL]

Activity (Skempton, 1953) A = PI / % Clay

Clays

Liquidity Index LI = [w-PL] / [LL-PL]