A very very practical document...

1. PE Refresher Course Geotechnical Component Class 1 Notes available at: www.ce.washington.edu/~geotech

13. Given: Sieve analysis and plasticity data for the following three soils classify the soils Example * non-plastic 77 NP* 5 PI 47 - 15 PL 124 - 20 LL 97 5 60 No. 200 99 8 78 No. 100 100 40 86 No. 40 100 90 92 No. 10 100 97 99 No. 4 Soil 3, % Finer Soil 2, % Finer Soils 1, % Finer Sieve Size

15. Soil 1 > 50% passes #200 - Fine-grained LL=20, Pl=5 - plots in CL-ML (p. 35.6) Soil 2 < 50% passes #200 - Coarse-grained > 50% passes #4 - Sand D 60 = 0.71 mm D 30 = 0.34 mm D 10 = 0.18 mm SP - SM Soil 3 > 50% passes #200 - Fine -grained LL=124 Pl=77 - Off the chart - Extrapolating gives CH Could be CH-MH

17. Common practice is to assume V s = 1, then express other volumes and weights accordingly. From definitions Gas Water Solid wG s 1 e wG s ρ w G s ρ w ρ sub = ρ sat - ρ w Buoyant unit weight ρ sat = ρ m for S=100% Saturated unit weight ρ d = W s /V t Dry unit weight (dry density) ρ m = W t /V t Moist unit weight

18. Table 35.7 - Useful for rapid calculation of phase relationships

19. Given : e = 0.62 w = 15% G s = 2.65 Calculate : a. ⍴ d b. ⍴ m c. w for S = 100% d. ⍴ sat for S = 100% Example Gas Water Solid wG s 1 e wG s ρ w= S e ρ w G s ρ w

23. Field Density Tests

24. Direct Backscattering

31. Shear Strength and Principal Stresses Ϭ 3 Ϭ 1 Ϭ Շ c Փ Failure surface is always oriented at 45 + Փ/2 angle to minor principal stress axis At failure Shear strength Shear stress failure 45+  /2

34. Example 10’ e = 0.40 w = 10% z Layer 1 Layer 2 5 ' 15' e = 0.60 S = 20% S = 100% First, calculate soil densities Then, calculate stresses

38. Example 1 Calculate vertical stress 5 ft. below and 2 ft. to the side of a surface point load of 1,000 lbs. 1000 lbs 5 ' 2 ' 5 ' P v Example 2 Calculate the vertical stress at a depth of 15 feet below the edge of a 5-foot-wide strip footing which imposes a bearing pressure of 2,000 psf on the ground surface. 15 ' P v 0.2p Chart in Appendix 40A p. A-69 (left side) 2,000 psf

39. PLOT 40.A

40. Example 3 Calculate the vertical stress at a depth of 14 feet below the center of a 4 ft. square footing that applies 10,000 psf bearing pressure to the ground surface 4 ' 14 ' P v 0.04p 10,000 psf p. 40.A Right side

41. PLOT 40.A

42. Example 3 Calculate the vertical stress at a depth of 14 feet below the center of a 4 ft. square footing that applies 10,000 psf bearing pressure to the ground surface 4 ' p. 40.A Right side Example 4 A 16 ft diameter water tank contains 20 feet of water. Calculate the vertical stress caused by the tank at a point 8 feet below the ground surface and 10 feet from the center of the tank. 8 ' P v 16 ft 10 ' 14 ' P v 0.04p 10,000 psf I = 0.2 Appendix D p. 40.B

43. PLOT 40.B

44. Determination of appropriate soil properties Compute C c or C r from e-log p curve Consolidation test C c applies to normally consolidated range C r applies to over-consolidated range Initial Conditions Final Conditions C c or C r e e 1 e 2 p 1 p 2 Log p

46. Pre-consolidation Pressure, P p Disturbance Effects

47. Pre-consolidation Pressure, P p Casagrande Method

49. First, calculate initial effective stress at center of soft clay layer  before new fill placed Next, calculate final stress after placement of new fill Then, calculate ultimate settlement as Example 5 Calculate the ultimate settlement of the soft clay layer due to placement of the new fill 4 ' 3 ' 2 ' 5 ' New Fill  = 125 pcf; w= 10% Old Fill Same properties as new fill Soft Clay C c =1.06 e o =2.53   sub = 30 pcf Dense Sand

50. Now, what would happen if half of the new fill was removed ? Since effective stress is decreasing, use C r Assuming C r = 0.10 rebound

51. Let’s now assume that 4 more feet of new fill is placed, bringing the total thickness Of new fill to 6 ft. Then

53. Degree of Consolidation curves

54. Example 6 If C v for the soft clay of Example 5 was 10ft 2 /yr, how long would it take for 2 in of settlement to occur? What if the soft clay was underlain by impermeable bedrock? Then z= 5 ft Double drainage Single drainage