1. 1 Kh Training April 2005

2. 2 Gudiance on the selection of Kh Soil Soil to Rock Rock Soil Soil to Rock Rock

3. 3 SOIL The following slides relate Relative density and Consistency of soils to the Material Strength in Chapter 52 – Tables 52-2 and 52-3.

4. 4 Soil Massive Jointed Cohesive Cohesive Low Plastic Cohesionless Non-Plastic Cohesionless Massive Jointed Cohesive Cohesive Low Plastic Cohesionless Non-Plastic Cohesionless

5. 5 Cohesive Field Tests Lab Tests The PI of the Soils is greater than 10 Field Tests Lab Tests The PI of the Soils is greater than 10

6. 6 Lab. Data Natural moisture content % Specific Gravity Gs Liquid Limit LL Plasticity Index PI Clay Fraction (%<0.002mm) Unconfined Compression Strength (psf)

7. 7 C haracterizing Clays Plasticity Index - The numerical difference between the Liquid Limit water content and the Plastic Limit water content PI = LL - PL

8. 8 C onsistency of Clays To evaluate, clays must be saturated Saturation establishes base line for comparisons -even soft clays will become firm when dried Saturated consistency strong indicator of engineering behavior Water has zero shear strength To evaluate, clays must be saturated Saturation establishes base line for comparisons -even soft clays will become firm when dried Saturated consistency strong indicator of engineering behavior Water has zero shear strength

9. 9 S teps in Evaluating Consistency of Saturated Clay Obtain data - LL, PI, either  dry or w sat (%) Calculate PL (not usually reported) PL = LL - PI Liquid Limit Plastic Limit Establish location of wsat % on diagram v/soft soft medium stiff

10. 10 E valuating Consistency of Saturated Clay Steps to Evaluate –For saturated deposit, obtain sample and measure oven dry w(%) –For unsaturated deposit, measure dry density, assume value for G s and calculate theoretical saturated w(%)

11. 11 Example What is the saturated water content of the following clay compared to its LL and PL? Given: LL = 59, PI = 36,  dry = 1.28 g/cm3 Assume Gs = 2.70

12. 12 C onsistency and Atterberg Limits Begin Constructing a consistency diagram by locating the LL value of 59 Liquid Limit 59

13. 13 C onsistency and Atterberg Limits Next, calculate the Plastic Limit, which is equal to the LL minus the PI and plot that PL = 59 - 36 = 23 Plastic Limit 23

14. 14 C onsistency and Atterberg Limits Now, divide the range between the LL and PL into thirds (PI  3) - (36  3 = 12) Liquid Limit 59 Plastic Limit 23 v/soft 3547

15. 15 C onsistency and Atterberg Limits Now label the ranges as shown Liquid Limit 59 Plastic Limit 23 v/softsoft medium stiffv/stiff 35 47

16. 16 C onsistency and Atterberg Limits Next, Calculate the saturated water content, given that  dry = 1.28 g/cm3 and Gs = 2.70

17. 17 C onsistency and Atterberg Limits Plot the saturated water content on the diagram to identify its consistency Liquid Limit 59 Plastic Limit 23 wsat % = 41 %, medium consistency v/softsoft medium stiffv/stiff 35 47

18. 18 L iquidity Index Another method for quickly assessing the saturated consistency of a clay uses a term Liquidity Index Liquidity Index is defined from a simple equation that expresses numerically where the saturated water content is in relation to the Atterberg Limits

19. 19 L iquidity Index Equation If w sat = Liquid Limit, then LI = 1.0 If w sat = Plastic Limit, then LI = 0

20. 20 Summary of Clay Problem Soils Problem Clays are very soft clays with liquidity index  1, or Stiff to very stiff clays with liquidity index  0

21. 21 Liquidity Index and Saturated Consistency Liquidity Index < 0 0 to 0.33 0.33 to 0.67 0.67 to 1.00 > 1.00 Liquidity Index < 0 0 to 0.33 0.33 to 0.67 0.67 to 1.00 > 1.00 Saturated Consistency very stiff stiff medium soft very soft Saturated Consistency very stiff stiff medium soft very soft

22. 22 What is the liquidity index of the Example Soil? Given that wsat = 41%, LL = 59, and PI = 36, entering these terms in the LI equation:

23. 23 Conclusion - Medium Consistency

24. 24 Field Estimate of Consistency Rule of Thumb Thumb will penetrate soil more than 1-inch. Extrudes between fingers when squeezed in fist

25. 25 Field Estimate of Consistency Rule of Thumb Thumb will penetrate soil about 1-inch. Easily molded in fingers

26. 26 Field Estimate of Consistency Rule of Thumb Thumb will not penetrate soil, but will indent about 1/4 inch. Molded by finger pressure

27. 27 Field Estimate of Consistency Rule of Thumb Thumb will not indent soil, but soil can be indented with thumbnail.

28. 28 Field Estimate of Consistency Rule of Thumb Can only be marked with knife - not indented with thumbnail.

29. 29 Soft Clays - Field Estimates

30. 30 Estimating Effective  angle for clays

31. 31 SOILS WITH PI LESS THAN 10 Methods to estimate RELATIVE DENSITY for nonplastic soils

32. 32 Relative Density Studies for the Alaska pipeline established empirical estimates of relative density for various soil types, based on the measured in place density of the soils Average values of minimum and maximum index density for the project were used May be useful for preliminary estimates

33. 33

34. 34 Relative Density Because saturated water content is related to dry density, a chart can be derived relating in place saturated water content to relative density, as shown on the following slide

35. 35 Chart is for silty sands (SM) Other information on Relative Density

36. 36 Field Estimates for Relative Density Description 1/2” Reinforcing rod can be pushed easily by hand into soil

37. 37 Field Estimates for Relative Density Description Can be excavated with a spade. A wooden peg 2”x 2”can easily be drive to depth of 6 ”.

38. 38 Field Estimates for Relative Density Description Easily penetrated with a 1/2” reinforcing rod driven with a 5 pound hammer

39. 39 Field Estimates for Relative Density Description Requires a pick for excavation. Wooden peg 2”x 2”is hard to drive beyond 6 ”.

40. 40 Field Estimates for Relative Density Description Penetrated only a few centimeters with a 1/2” reinforcing rod driven with a 5 pound hammer

41. 41 Fort Worth Relative Density Study NRCS lab in Fort Worth studied 28 filter sands and used some published data Minimum and Maximum Index Densities were performed on each sample A 1 point dry Standard Proctor energy mold was also prepared for each sample. Values of 50% and 70% relative density were plotted against the 1 point Proctor value

42. 42 70 % Relative Density vs. 1 Point Proctor

43. 43 Conclusion is that 100 % of the Field 1 point Proctor dry test is about equal to 70 % relative density 70 % Relative Density Vs. 1 Point Proctor

44. 44 50 % Relative Density Vs. 1 Point Proctor

45. 45 Conclusion is that 95 % of the field 1 point Proctor dry test is about equal to 50 % relative density 50 % Relative Density Vs. 1 Point Proctor

46. 46  D 70 = 1.075 x  d 1pt -9.61, for RD 70 and  d 1pt in lb/ft3  D 50 = 1.07 x  d 1pt - 12.5, for RD 50 and  d 1pt in lb/ft 3 Relative Density Estimates from FW SML Study

47. 47 Example Relative Density Estimates –Given: 1 Point Proctor Test  d = 105.5 pcf –Estimate 70 % and 50% Relative Density from the Fort Worth equations –Given that measured  d is 98.7, evaluate state of compaction of sand. Relative Density Estimates from FW SML Study

48. 48  D 70 = 1.075 x  d 1pt -9.61, = 1.075 x 105.5 -9.61 = 103.8 pcf  D 50 = 1.07 x  d 1pt - 12.5, = 1.07 x 105.5 - 12.5 = 100.4 Solution to Example Problem

49. 49 Solution to Example Problem Conclusion is that measured dry density of 98.7 pcf is less than the estimated 50% Relative Density dry density value of 100.4 pcf

50. 50 Solution to Example Problem Measured Dry Density is 98.7 pcf and Field 1 Point Dry Density is 105.5 pcf. Measured Dry Density then is 98.7  105.5 = 93.6 percent of the 1 Point Proctor - Again, by rule of thumb < 50% RD

51. 51 Empirical Estimates for Sands and Gravels Empirical Estimates of Shear Strength Estimating effective friction angle 

52. 52

53. 53 The End