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

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Presentation on theme: "Weight-Volume Relations Soil can be considered as a 3-phased material Air, Water, Solids."— Presentation transcript:

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

2 Soil Structure

3

4 3-Phase Idealization

5 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

6 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

7 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%

8 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)

9 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

10 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

11 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

12 Example Soil Block kgm3m3 0 A 4.0 0.667 W 0.002 3.333 S

13 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

14 Example Soil Block kgm3m3 0 A 0.00009 4.0 0.667 W 0.00067 0.00076 0.002 3.333 S 0.00124

15 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%

16 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

17 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

18 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

19 Modified Soil Block kgm3m3 0 A 0.0027 120 20 W 0.0200 0.0227 0.06 100 S 0.0373

20 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%)

21 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.

22 3-Phase Idealization Solids Water Air

23 Modified Soil Block kgm3m3 2.7 AWAW 0.0027 122.7 20 W 0.0200 0.0227 0.06 100 S 0.0373

24 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%

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

26 Atterberg Limits

27 Liquid Limit

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

29 Liquid Limit Europe & Asia Fall Cone Test BS1377

30 Plastic Limit 3mm Diameter Thread

31 Shrinkage Limit

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

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

34 Clays

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


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