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Pe Test Geotechnical Rerview

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A very very practical document...

A very very practical document...

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  • 1. PE Refresher Course Geotechnical Component Class 1 Notes available at: www.ce.washington.edu/~geotech
  • 2.
    • Organization
    • Lecture No. 1
      • Basics ( Chapter 35 )
        • Soil classification
        • Phase diagrams
        • Soil properties
          • Compaction
          • Permeability
          • Consolidation
          • Shear strength
      • Applications ( Chapter 35, 40 )
        • Settlement problems
        • Magnitude of settlement
        • Rate of settlement
  • 3.
    • Organization
    • Lecture No. 2
    • Applications ( Chapters 36, 37, 38, 39, 40 )
      • Seepage problems
      • Slope stability problems
      • Foundations
        • Shallow Foundation
        • Deep foundations
      • Retaining structures
        • Retaining walls
        • Braced excavations
  • 4.
    • Grain Size and Plasticity Characteristics
    • Grain Size Characteristics
      • Sieve Analysis
      • Coefficient of Uniformity
        • C u = D 60 /D 10
      • Coefficient of Curvature
        • C z = (D 30 ) 2 / (D 60 x D 10 )
    1-3 1-3 >4 5-10 4-6 15-300 25-1000 Gravel Fine sand Coarse sand Mixture of silty sand and gravel Mixture of clay, sand, silt and gravel Cz Cu Soil
  • 5.
    • Grain Size and Plasticity Characteristics
    • Hydrometer Analysis
      • Relates particle size to settling velocity
      • Used to determine size of -#200 fraction
    • Plasticity Characteristics
      • Plastic Limit - lowest water content at which soil exhibits plastic behavior
      • Liquid limit - highest water content at which soil exhibits plastic behavior
      • Plasticity Index
        • Pl = LL - PL
      • Classification of fine-grained soils often based on plasticity characteristics as described by liquid limit and plasticity index
  • 6.
    • Initial classification generally based on grain size
      • Gravel Large grain size ( 4.75mm – 75mm)
      • Sand
      • Silt
      • Clay
      • Organics small grain size (.075mm – 4.75mm)
  • 7.
    • USDA (US Department of Agriculture)
      • Triangle identification chart - easy to use
      • Good for gardening (plant in loam)
    • AASHTO (Am Assoc of State Highway Trans Officials)
    • Based on suitability of soil for use as pavement base
      • Divides soil types into 8 groups, A-1 through A-8
      • Granular soils (gravels and sands) fall into A-1 through A-3
      • Differentiated primarily on basis of grain size distribution
      • Fine-grained soils (silts and clays) fall into A-4 through A-7.
      • Differentiated primarily on basis of plasticity characteristics
      • Highly organic soils fall into A-8
    • Subgroups depend on grain size and plasticity characteristics - See Table 9.2
    • Group index added in parentheses after group and subgroup classification.
    • Group index calculated by Eq. 35.3 ( sub-grade suitability decreases with
    • increasing group index).
    Soil Classification (Section 9.3)
  • 8.  
  • 9.
    • USCS (Unified Soil Classification System)
    • soils are classified on basis of parameters which influence their engineering properties .
    • Coarse – grained soils (gravels and sands) classified on basis of grain size characteristics
    • Fine-grained soils (silts and clays) classified on basis of plasticity characteristics .
    • Symbols:
      • G Gravel
      • S Sand
      • M silt
      • C Clay
      • O Organic
    • Modifiers:
      • W Well Graded
      • P Poorly Graded
      • H High Plasticity
      • L Low Plasticity
    • Examples:
      • GW Well-graded gravel
      • SP Poorly-graded (uniform) sand
      • MH Highly plastic silt
      • CL Low plasticity clay
      • GM Silty gravel
  • 10.  
  • 11.  
  • 12.  
  • 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
  • 14.  
  • 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
  • 16. Aggregate Soil Properties (Phase Diagrams)
    • Phase Diagrams
      • Solid, Water, and Gas phases shown separately
      • Volumes indicated on left side of phase diagram
      • Weights indicated on right side of phase diagram
    • Definitions
      • Void Ratio e = V v /V s
      • Porosity n = V v /V t
      • Water Content w = W w /W s
      • Degree of Saturation S = V w /W v
      • Density* ρ= Mass/Volume
      • Unit Weight* γ= weight/Volume
      • Specific Gravity G = ρ s /ρ w
      • *Review text lumps density and unit weight together and uses symbol ρ
    Gas Water Solid V g V w V s V v V t O W w W s W t
  • 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
  • 20.
    • Standard Penetration Test
      • 140lb hammer dropped 30&quot; to drive standard sampler. Number of blows
      • required for 12&quot; penetration measured as standard penetration resistance, N.
      • Crude test but useful index of soil characteristics.
      • More useful in sands than in fine-grained soils.
    • Moisture-Density Tests and Relationships
      • Compaction Tests
        • Proctor Test
        • Modified Proctor Test
      • Density of soil for given compactive effort Influenced by water content
      • Density of soil for given water content influenced by level of compactive effort
    Soil Testing and Mechanical Properties ⍴ d w Increasing E ⍴ d w opt w ( ⍴ d ) max
  • 21.  
  • 22.  
  • 23. Field Density Tests
  • 24. Direct Backscattering
  • 25.  
  • 26.  
  • 27. Consolidation Test
    • Procedure:
    • Apply vertical load in increments.
    • During each increment, measure change in
    • height of specimen as function of time .
    • At end of each increment when settlement stops,
    • measure change in height of specimen as function of vertical stress.
  • 28.
    • Measure deformation of sample with time
    • Plot:
    • Change in equilibrium void ratio w/ stress
      •  settlement magnitude information
    e Change in void ratio w/time for stress Increment  settlement rate information
    • Apply increment of stress
    e 0 e f P 0 P f Log p Initial equilibrium Final equilibrium e 0 e f time Initial equilibrium Final equilibrium Fast rate Slow rate
  • 29. Consolidation Parameters
    • Compression Index, C c
      • Given by slope of e-log p curve (NC portion)
    • Recompression Index, C r
      • Given by slope of rebound portion of curve (OC portion)
    Coefficient of Consolidation, C v e i e f High C v (fast settlement) Low C v (slow settlement) time e C c C r Log p e Normally consolidated Over-consolidated
  • 30.
      • Shear strength influenced by pore fluid drainage
        • Free drainage during loading  drained
        • No drainage during loading  undrained
      • Mohr – Coulomb Failure Criterion
    Shear Strength of soils friction cohesion c s For drained loading, c = 0 S Typical for sands For un-drained loading , S Typical for clays c S nc
  • 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
  • 32.
      • Generally fall into one (or both) of two categories:
      • Magnitude of settlement
      • Rate of settlement
    • Must be able to :
    • Evaluate initial effective stress conditions
    • Evaluate change in effective stress due to imposed loading
    • Determine appropriate soil properties
    • Perform calculations
    APPLICATIONS Settlement Problems
  • 33. Evaluation of Initial Effective Stresses
    • For effective stresses, use ρ m above water table
    • ρ sub below water table
    • or calculate total stress and subtract water pressure
    • For total stresses, use ρ m above water table
    • ρ sat below water table
    • For water pressure, take product of ρ w and depth below water table
    Groundwater level Density of soil layers Thickness of soil layers Need to know
  • 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
  • 35.
    • Change in effective stresses can be caused by:
    • External loading
        • Placement of fill ( Ϭ ‘ v up )  settlement
        • Construction of structure ( Ϭ ‘ v up )  settlement
        • Excavation ( Ϭ ‘ v down )  rebound
    • Change in groundwater conditions
        • Drawdown of water level – ( Ϭ ‘ v up )  settlement
        • Rising water level ( Ϭ ‘ v down )  rebound
    • Calculation of final effective stresses  after u excess dissipates
      • Based on assumption of hydrostatic water pressures, u = ρ w (z-z w )
      • Proceed in same way as for initial effective stresses
    • Two important cases:
    • 1. Areal loads – vertical stress = f (z) only
    • (large areal extent w /r /t thickness of soil layer)
    • 2. Local loads – vertical stress = f (x, y, z)
      • Must compute stress distribution
    Evaluation of Change in Effective Stresses
  • 36.
      • Areal Load
        • Assume 5-ft-thick fill placed on top of
        • previous two-layered soil deposit.
        • Tests show ρ m =120 pcf.
    The increase in stress produced by an areal load is constant with depth Local Load Spread footing imposes uniform load of 1,000 psf over 10 ft x 10 ft area What is σ v '  different below edge of footing than below center. Different at depth than shallow Examples Z=20’ 5 ‘ ρ m = 120 10 ‘ z ρ m = 132 5 ‘ ρ m = 110 15 ‘ ρ sub = 66
  • 37. Stress Distribution
    • Important to be able to calculate subsurface stresses caused by loads or loaded areas on the ground surface.
    • Usually interested for settlement calculation problems.
    • Generally accomplished by stress distribution methods based on theory of elasticity.
    • Can use principle of superposition very useful.
    • Boussinesq – stresses caused by point load on surface.
    • Boussinesq solution widely used
    • For point load, use Eq. 40.1
      • Example 1
    • For strip footing loads, use Appendix 40a (left)
      • Example 2
    • For square footing loads, use Appendix 40a (right)
      • Example 3
    • For circular loaded areas, use Appendix 40b
      • Example 4
    • For loaded areas of arbitrary shape, use (Newmark)
      • Influence chart method – see Fig 40 3
    • Influence Chart
      • Represents entire ground surface
      • Divided into number of “squares” – see Fig 40.3
      • Squares set up so that uniform load on each
      • would cause same stress on subsurface
      • point below center of chart
  • 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
  • 45. Pre-consolidation Pressure, P p
    • Maximum effective stress under which soil has ever been in equilibrium
    • Soil is normally consolidated when current effective stress is equal to current value of pre-consolidation pressure.
    • Settlement behavior controlled by C c
    • Soil is over-consolidated when current effective stress is less than current value of pre-consolidation pressure.
    • Settlement behavior controlled by C r .
    Pre-consolidation Pressure, P ' p P ' 1 P ' 2 Log p e e 1 e 2 C r C c
  • 46. Pre-consolidation Pressure, P p Disturbance Effects
  • 47. Pre-consolidation Pressure, P p Casagrande Method
  • 48. Calculation of Settlement Magnitude
    • Need:
    • Initial and final effective stresses
    • Definition of C c and C r
    3. Definition of vertical strain P ‘ 1 P ‘ 2 Log p e e 1 e 2 P ‘ p e p OC NC initial final
  • 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
  • 52. Time Rate of Primary Consolidation
    • Rate controlled by coefficient of consolidation, C v
      • High C v  rapid consolidation
      • Low C v  slow consolidation
    • Degree of consolidation
      • Fraction of ultimate settlement which has occurred by time t
    Fraction of ultimate settlement which has occurred by time t Time required to reach given degree of consolidation Dimensionless time factor Settlement at given time t Where T v (t) and U(t) are related by Eq 40.23 and Table 40.1 50 90 100 .2 .85 T v 0% U Length of longest drainage path
  • 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

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