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Published byAugust Lester Small Modified over 9 years ago
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Weight-Volume Relations Soil can be considered as a 3-phased material Air, Water, Solids
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Soil Structure
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3-Phase Idealization
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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
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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
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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%
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Inter-relationships Wet -> Dry Unit Weight ( d = ( WET / (1+w/100) W S = W T / (1+w/100) Dry Unit Weight @ Saturation (Zero Air Voids) ( zav = ( W / (w/100+1/Gs)
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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
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Example Soil Block Analysis kgm3m3 0 A 4.0 W 0.002 S Given: W T =4kg, V T =0.002m 3, w=20%, Gs=2.68
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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) = 3.333 kg W W = W T – W S W W = 4kg – 3.333 kg = 0.667 kg Check w = W W /W S x 100% w = 100% x 0.667 / 3.333 = 20.01% OK
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Example Soil Block kgm3m3 0 A 4.0 0.667 W 0.002 3.333 S
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Example Soil Block Analysis V S = W S / G S ( w V S = 3.333kg / (2.68 x 1000 kg/m 3 ) = 0.00124 m 3 V W = W W /G S ( w V W = 0.667kg / (1 x 1000kg/m 3 ) = 0.00067 m 3 V A = V T – V S - V W V A = 0.00200–0.00124–0.00067 = 0.00009m 3 V V = V A + V W V V = 0.00067+0.00009 = 0.00076m 3
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Example Soil Block kgm3m3 0 A 0.00009 4.0 0.667 W 0.00067 0.00076 0.002 3.333 S 0.00124
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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 = 0.00076/0.00124 = 0.613 n = 100x0.00076/0.002 = 38.0% S = 100x0.00067/0.00076 = 88.2%
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Modified Soil Block Analysis kgm3m3 0 A 120 20 W 100 S Given: W T =4kg, V T =0.002m 3, w=20%, Gs=2.68
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Modified Soil Block Analysis Given: W T = 4 kg, V T = 0.002 m 3 W T / V T ratio must remain unchanged 4 kg / 0.002 m 3 = 120 kg / X X = 0.06 m 3 = V T
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Modified Soil Block Analysis V S = W S / G S ( w V S = 100kg / (2.68 x 1000 kg/m 3 ) = 0.0373 m 3 V W = W W /G S ( w V W = 20kg / (1 x 1000kg/m 3 ) = 0.0200 m 3 V A = V T – V S - V W V A = 0.0600–0.0373–0.0200 = 0.0027m 3 V V = V A + V W V V = 0.0027+0.0200 = 0.0227m 3
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Modified Soil Block kgm3m3 0 A 0.0027 120 20 W 0.0200 0.0227 0.06 100 S 0.0373
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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 = 0.0227/0.0373 = 0.609 (0.613) n = 100x0.0227/0.06 = 37.8% (38.0%) S = 100x0.02/0.0227 = 88.1% (88.2%)
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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.
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3-Phase Idealization Solids Water Air
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Modified Soil Block kgm3m3 2.7 AWAW 0.0027 122.7 20 W 0.0200 0.0227 0.06 100 S 0.0373
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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%
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Consistency of Soil Atterberg Limits Liquid Limit, LL Plastic Limit, PL Shrinkage Limit, SL
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Atterberg Limits
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Liquid Limit
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Liquid Limit Plot Shear strength of soil @ LL is approx. 2.5 kN/m 2 (0.36 psi)
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Liquid Limit Europe & Asia Fall Cone Test BS1377
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Plastic Limit 3mm Diameter Thread
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Shrinkage Limit
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Consistency of Soil Plasticity Index, PI PI = LL - PL Activity, A A = PI / % Clay Liquidity Index, LI LI = [w – PL] / [LL – PL]
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Activity (Skempton, 1953) A = PI / % Clay
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Clays
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Liquidity Index LI = [w-PL] / [LL-PL]
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