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The design of earth-retaining structures - Lecture 2
- 1. The Design of Earth-retaining Structures ENB485 Chris Bridges May 2016 Lecture 2 The Design of Earth-retaining Structures
- 2. Outline of Presentation Lateral Earth Pressures Wall Geometry / Ground Model – Surface slope – Surcharge – Groundwater conditions – Geological Profile Soil Properties – Unit Weight – Strength parameters – Earth pressure coefficients – Deformation parameters – Wall friction/adhesion Analysis – Gravity Walls – Bearing – Sliding – Overturning 5 May 2016 ENB485 2
- 4. Earth Pressures • At-rest – no movement of support • Active pressure – support moves AWAY from soil • Passive pressure – support moves INTO soil 5 May 2016 ENB485 4
- 5. Lateral earth pressures • The way that a wall is installed affects the way pressures are applied to it • Unyielding wall – the horizontal pressures in the soil before excavation are retained • Yielding wall – horizontal pressures in supported soil are reduced • Soil placed after wall is built – method of placing soil will control horizontal pressures 5 May 2016 ENB485 5
- 9. Effective Stress sv’ = sv - u Effective Stress Total Stress (gh) Porewater Pressure (gwhw) 5 May 2016 ENB485 9
- 10. What are the total and effective stresses at the top of the clay layer? 5 May 2016 ENB485 10 sv = gh = u = gwhw = sv’ = sv - u = sv = u = sv’ = sv = u = sv’ = sv = u = sv’ = sv = u = sv’ = 3 x 19 = 57 kPa 10 x 1 = 10 kPa 57 – 10 = 47 kPa Silty sand Silty sand Silty sand Silty sand Silty sand g=19kN/m3 1 2 3 4 5
- 11. Answers 5 May 2016 ENB485 11 sv = u = sv’ = sv = u = sv’ = sv = u = sv’ = sv = u = sv’ = sv = u = sv’ = 3 x 19 = 57 kPa 10 x 1 = 10 kPa 57 – 10 = 47 kPa 3 x 19 = 57 kPa 10 x 2 = 20 kPa 57 – 20 = 37 kPa 3 x 19 = 57 kPa 10 x 3 = 30 kPa 57 – 30 = 27 kPa 3 x 19 + 2 x 10 = 77 kPa 10 x 5 = 50 kPa 77 – 50 = 27 kPa 3 x 19 + 3 x 10 = 87 kPa 10 x 6 = 60 kPa 87 – 60 = 27 kPa 1 2 3 4 5
- 12. Q.1 5 May 2016 ENB485 12 g=19kN/m3 Silty sand Depth, h (m) Vertical effective stress, sv’ (kPa) h1 g = 2 x 19 = 38 kPa h1 g + hw (g - gw ) = 2 x 19 + 1 x (19 - 10) = 47 kPa h1 hw Depth, h (m)Depth, h (m) Porewater pressure, u (kPa)Total vertical stress, sv (kPa) h2 h2 g = 3 x 19 = 57 kPa hw gw = 1 x 10 = 10 kPa 0 1 2 3 0 1 2 3 0 1 2 3
- 13. Earth Pressure Coefficients K = s’h / s’v s’h Horizontal effective stress s’v Vertical effective stress K can be either: Ka Coeff. of Active Earth Pressure Kp Coeff. of Passive Earth Pressure Ko Coeff. of Earth Pressure at Rest 5 May 2016 ENB485 13
- 14. Coefficient of Earth Pressure at Rest (K0) Empirical formulae Jaky’s formula (simplified) K0 = 1- sin f For sloping ground K0 = {1- sin f}/ {1- sin b} (where b is the slope angle) The above equations are generally valid for normally consolidated soils and lightly over-consolidated soils. Higher values can be expected for heavily over-consolidated soils. K0(OC) = (1- sin f) OCRsinf (OCR = over-consolidation ratio) 5 May 2016 ENB485 14
- 15. Typical K0 values 5 May 2016 ENB485 15 Soil Type Typical Ko Dense sand (not highly preconsolidated) 0.35-0.40 Loose sand 0.55-0.60 Soft clay 0.50-0.60 Overconsolidated London Clay 1.0-2.8 Compacted clay 1.0-2.0 Compacted sand 1.0-1.5
- 16. Earth pressure theories Rankine theory (1862) – Assumes failure in entire soil mass. Very simplified. Coulomb wedge (1776) – Assumes wedge failure for both active & passive cases. Very simplified. Numerical Analyses – More versatile – can incorporate many factors – Generally part of a more complete system analysis 5 May 2016 ENB485 16
- 17. Rankine Coefficient of Active/Passive Earth Pressure Ka = {1- sin (f)}/ {1+ sin (f)} = tan2(45 - f/2) for horizontal ground surface and smooth wall and Kp = 1/ Ka = tan2(45 + f/2) 5 May 2016 ENB485 17 In total stress, undrained short-term conditions use undrained cohesion, Cu and fu = 0, therefore, Ka = Kp = 1
- 18. Earth Pressures – non-cohesive soil 5 May 2016 ENB485 18 g=19kN/m3 Silty sand Depth, h (m) Vertical effective stress, sv’ (kPa) h1 g = 2 x 19 = 38 kPa h1 g + hw (g - gw ) = 2 x 19 + 1 x (19 - 10) = 47 kPa h1 hw Depth, h (m) Porewater pressure, u (kPa) h2 pw = hw gw = 1 x 10 = 10 kPa Depth, h (m) Horizontal effective stress, sh’ (kPa) pa1’ = Kasv’ = Ka h1 g = 0.33 x 38 = 12.5 kPa pa2’ = Ka (h1 g + hw (g - gw )) = 0.33 x 47 = 15.5 kPa Assume f’=30 then Ka = 0.33 pa’ = effective active pressure = Ka sv’ If we have a retaining wall supporting the top 3m of soil what are the active earth pressures acting on the wall? 0 1 2 3 0 1 2 3 0 1 2 3
- 19. Forces 5 May 2016 ENB485 19 Depth, h (m) Horizontal effective stress, sh’ (kPa) P1 = ½ x pa1’ x h1 = 0.5 x 12.5 x 2 = 12.5 kN P3 = ½ x (pa2’-pa1’) x hw = 0.5 x (15.5-12.5) x 1 = 3 kN Depth, h (m) Porewater pressure, u (kPa) pa1’ = 12.5 kPa pa2’ = 15.5 kPa Depth, h (m) Porewater pressure, u (kPa) pw = 10 kPa Depth, h (m) Horizontal effective stress, sh’ (kPa) P2 = pa1’ x hw = 12.5 x 1 = 12.5 kN The forced are calculated from the areas under the graphs Pw = ½ x hw x pw = 0.5 x 1 x 10 = 5 kN 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 h1 hw
- 20. Effective pressures • Non-cohesive soils • pa’ = Ka (gh – gwhw) = Ka sv’ • pp’ = Kp (gh – gwhw) = Kp sv’ • Cohesive soils • pa’ = Ka (gh – gwhw) – 2c’ √Ka • pp’ = Kp (gh – gwhw) + 2c’ √Kp 5 May 2016 ENB485 20 Depth, h (m) Horizontal effective stress, sh’ (kPa) 0 – 2c’ √Ka Ka (gh – gwhw) – 2c’ √Ka h0 c’ means reduced active pressure and increased passive earth pressure as well as effectively zero active earth pressure over height h0
- 21. Ref.: Geoguide 1 Example 5 Clay soils only, depth of tension crack (long term) – zo or ho = 2c’ / g √Ka depth of tension crack (short term) – zo or ho = 2Cu / g as Ka=1 5 May 2016 ENB485 21 Soil1Soil2 We assume a water pressure over the depth of the tension crack
- 22. Ref.: Geoguide 1 5 May 2016 ENB485 22
- 25. Ref.: Geoguide 1 5 May 2016 ENB485 25
- 26. Ref.: Geoguide 1 5 May 2016 ENB485 26
- 27. Ref.: Geoguide 1 5 May 2016 ENB485 27
- 28. Ref.: AS4678:2002 5 May 2016 ENB485 28
- 29. Displacements Required to Develop Active & Passive Pressures – AS4678-2002 Displacement (translation or rotation) required for active conditions Sand 0.001H Clay 0.004H Displacement required for passive conditions Sand – translation 0.05H Sand – Rotation about base >0.1H Greater movement is required to activate passive resistance than active. 5 May 2016 ENB485 29
- 30. Passive Pressures So, although the wall may move enough to generate active earth pressure conditions, the movement may not be enough to fully develop passive resistance. Therefore, a reduction to Kp may be applied (i.e. use Kp/2 or Kp/3 in analysis). Should consider the effects of unplanned excavations in front of wall and reduction to Kp (typically H/10 or 0.5m) 5 May 2016 ENB485 30
- 31. 07 Wall Geometry & Ground Model
- 32. Wall Geometry & Ground Model 5 May 2016 ENB485 32
- 33. Wall Definitions 5 May 2016 ENB485 33
- 34. Gravity Walls 5 May 2016 ENB485 34 Cost effective height 1m to 3m Cost effective height 2m to 6m
- 36. Ref: Clayton, Milititsky & Woods (1993). Earth Pressure and Earth-retaining Structures 5 May 2016 ENB485 36
- 37. Embedded Cantilever Walls 5 May 2016 ENB485 37
- 38. Groundwater Existing groundwater monitoring data Existing water bearing services. Leaking? 1/3 retained wall height Leaking services 5 May 2016 ENB485 38
- 39. Surcharges Ref.: Geoguide 1, 1993 5 May 2016 ENB485 39
- 40. a. Bearing capacity b. Sliding & overturning 5 May 2016 ENB485 40
- 41. Use elastic theory for lateral pressure due to surface loadings 5 May 2016 ENB485 41 QL = line load QP = point load
- 43. Soil Properties - General Ranges of soil properties are given in AS4678:2002 in the Appendices. 5 May 2016 ENB485 43
- 44. Determination of Shear Strength Parameters There are a number of different shear strengths used in retaining wall design: Peak Values: c’ & f’ – should be used normally Residual Values: fr’ – should be used when previous shear surface are present and may be reactivated Critical State: f’crit - sliding, wall friction 5 May 2016 ENB485 44
- 45. Determining Shear Strength Parameters Plasticity Index (%) f’crit (degrees) 15 30 50 80 30 25 20 15 Clay Soils: (c’=0 kPa) Ref: BS8002:1994 5 May 2016 ENB485 45
- 46. Determination of Shear Strength Parameters Granular Soils: f’max = 30 + A + B + C f’crit = 30 + A + B Ref: BS8002:1994 & AS4678:2002 5 May 2016 ENB485 46
- 47. Determination of Shear Strength Parameters Granular Soils: Ref: BS8002:1994 (Table 3) & AS4678:2002 (Table D1) 5 May 2016 ENB485 47
- 48. Wall Friction & Adhesion BS8002 EC7 (Draft) CP2 Geoguide 1, 1st Ed Geoguide 1, 2nd Ed Wall friction, dw 2/3f’ 75% of design shear strength of soil (or representative value determined by test) Clause 3.2.6 d = fcrit’ for rough surfaces d = 20 for smooth surfaces Clause 2.2.8 2/3 fcrit’ Clause 8.5.1(4) dsoil/concrete = 20 Clause 1.4321 Design of base and stem d = 0 Sliding or tilting, d = f Clause 1.435 dpassive = ½ dsoil/concrete Clause 1.435 dmax = f’/2 Precast concrete dmax = 2/3 f’ Cast insitu concrete Section 2.7 f’/2 or backfill slope angle, whichever is smaller for virtual back of wall Section 5.11.2 Wall adhesion, cw 0 Clause 2.2.8 0 Clause 8.5.1(4) 0 (cohesionless soil) 0 0 Section 5.11.2 Base friction, db 2/3 f’ 75% of design shear strength of soil (or representative value determined by test) Clause 3.2.6 d = fcrit’ for rough surfaces d = 20 for smooth surfaces Clause 2.2.8 f’design for cast insitu concrete 2/3 f’design for smooth precast foundations Clause 6.5.3(8) f for cast insitu concrete when precast db = dw = 20 Clause 1.4922 d = 2/3 f’ without shear key d = f’ with shear key Section 2.6 d = 0.9 – 1 fcrit’ Rough surfaces d = 0.8 – 0.9 fcrit’ Smooth surfaces Section 5.12 Base adhesion, cb 0 Clause 2.2.8 0 Clause 6.5.3(8) 0 (cohesionless) 0 0 (cohesionless soil) 5 May 2016 ENB485 48
- 49. Values of d from CP2 Material/Condition Values of d Concrete/brick Steel sheet piling coated with tar/bitumen Uncoated steel sheet piling Structure subject to vibration or downward movement 20 30 15 0 Notes: 1. In the case of reinforced concrete cantilever walls d can be considered as being equal to f (Cl. 1.435). 2. Bitumen coating is usually applied to reduce skin friction and other references state d = 0. Note: in the tutorials and exam a “smooth wall” means d = 0. 5 May 2016 ENB485 49
- 51. Wall design process Step 1: establish ground model 5 May 2016 ENB485 51
- 52. Wall design process Step 2: determine earth and water pressures 5 May 2016 ENB485 52
- 53. Wall design process Step 3: determine forces on the back of wall 5 May 2016 ENB485 53
- 54. Wall design process Step 4a: determine factors of safety on sliding & overturning Typical required FOS: Sliding – 1.5 Overturning – 2.0 Bearing – 3.0 5 May 2016 ENB485 54
- 55. Ref.: Geoguide 1 Wall design process 5 May 2016 ENB485 55
- 56. Ref.: Geoguide 1 Wall design process 5 May 2016 ENB485 56
- 57. Sliding Failure 5 May 2016 ENB485 57
- 59. Ref.: Geoguide 1 Step 4b: determine factor of safety - bearing capacity 5 May 2016 ENB485 59
- 60. Ref.: Geoguide 1 5 May 2016 ENB485 60
- 61. Ref.: Geoguide 1 5 May 2016 ENB485 61
- 62. Bearing Capacity Failure 5 May 2016 ENB485 62
- 63. Summary Process 5 May 2016 ENB485 63
- 64. Fill walls 5 May 2016 ENB485 64
- 65. Cut walls 5 May 2016 ENB485 65