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Weight-Volume Relations

Weight-Volume Relations. Soil can be considered as a 3-phased material Air, Water, Solids. Soil Structure. Soil Structure. 3-Phase Idealization. 3-Phase Soil Block. Table 2.2. Weight Relations. Water content, w w = [W W /W S ] x 100% may be > 100% for clays

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Weight-Volume Relations

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

  2. Soil Structure

  3. Soil Structure

  4. 3-Phase Idealization

  5. 3-Phase Soil Block

  6. Table 2.2 Weight Relations • Water content, w w = [WW/WS] x 100% may be > 100% for clays • Total (Moist,Wet) Unit Weight ( = (T = (WET = WT / VT • Dry Unit Weight (d = WS / VT

  7. Volumetric Relations • Void Ratio, e e = VV / VS may be > 1, especially for clays • Porosity, n n = [VV / VT] x 100% 0% < n < 100% • Degree of Saturation, S S = [VW / VV] x 100% 0% < S < 100%

  8. Inter-relationships • Wet -> Dry Unit Weight (d = (WET / (1+w/100) WS = WT / (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 / Gs(w Known Volume: W = V Gs (w (w=1g/cc=9.81kN/m3=1000kg/m3=62.4lb/ft3

  10. Example Soil Block Analysis Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68

  11. Example Soil Block Analysis Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68 WS = WT / (1+w/100) WS = 4kg / (1+20/100) = 3.333 kg WW = WT – WS WW = 4kg – 3.333 kg = 0.667 kg Check w = WW/WS x 100% w =100% x 0.667 / 3.333 = 20.01% bOK

  12. Example Soil Block

  13. Example Soil Block Analysis VS = WS / GS(w VS = 3.333kg / (2.68 x 1000 kg/m3) = 0.00124 m3 VW = WW/GS(w VW = 0.667kg / (1 x 1000kg/m3) = 0.00067 m3 VA = VT – VS - VW VA = 0.00200–0.00124–0.00067 = 0.00009m3 VV = VA + VW VV = 0.00067+0.00009 = 0.00076m3

  14. Example Soil Block

  15. Example Soil Block Analysis (T = 4.0kg/0.002m3 = 2000 kg/m3=19.62 kN/m3=124.8lb/ft3 (D = 3.333kg/0.002m3=1666.5kg/m3=16.35 kN/m3 (D = 19.62kN/m3 / 1.20 =16.35 kN/m3 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 Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68

  17. Modified Soil Block Analysis Given: WT = 4 kg, VT = 0.002 m3 WT / VT ratio must remain unchanged 4 kg / 0.002 m3 = 120 kg / X X = 0.06 m3 = VT

  18. Modified Soil Block Analysis VS = WS / GS(w VS = 100kg / (2.68 x 1000 kg/m3) = 0.0373 m3 VW = WW/GS(w VW = 20kg / (1 x 1000kg/m3) = 0.0200 m3 VA = VT – VS - VW VA = 0.0600–0.0373–0.0200 = 0.0027m3 VV = VA + VW VV = 0.0027+0.0200 = 0.0227m3

  19. Modified Soil Block

  20. Modified Soil Block Analysis (T = 120kg/0.06m3 = 2000 kg/m3=19.62 kN/m3 (D = 100kg/0.06m3=1666.7kg/m3=16.35 kN/m3 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 Air Water Solids

  23. Modified Soil Block

  24. In Situ Comparators • Relative Density, Dr Dr=100% x [emax – ein situ] / [emax – emin] O% < Dr< 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/m2 (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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