Pe Test Geotechnical Rerview

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

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

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

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

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

Initial classification generally based on grain size Gravel Large grain size ( 4.75mm – 75mm) Sand Silt Clay Organics small grain size (.075mm – 4.75mm)

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)

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

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

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

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

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

Table 35.7 - Useful for rapid calculation of phase relationships

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

Standard Penetration Test 140lb hammer dropped 30" to drive standard sampler. Number of blows required for 12" 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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Pre-consolidation Pressure, P p Disturbance Effects

Pre-consolidation Pressure, P p Casagrande Method

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

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

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

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

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

Degree of Consolidation curves

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

Pe Test Geotechnical Rerview

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