PE Refresher CoursePE Refresher CourseGeotechnical ComponentGeotechnical Component

Class 1Class 1

Notes available at:www.ce.washington.edu/~geotech

OrganizationLecture No. 1

Basics (Chapter 35)Soil classificationPhase diagramsSoil properties

CompactionPermeabilityConsolidationShear strength

Applications (Chapter 35, 40)Settlement problemsMagnitude of settlementRate of settlement

OrganizationLecture No. 2 Applications ( Chapters 36, 37, 38, 39, 40)

Seepage problemsSlope stability problemsFoundations

Shallow FoundationDeep foundations

Retaining structuresRetaining walls Braced excavations

Grain Size and Plasticity Characteristics

Grain Size CharacteristicsSieve Analysis

Coefficient of UniformityCu = D60/D10

Coefficient of CurvatureC z = (D30)2 / (D 60 x D10)Soil Cu Cz

GravelFine sandCoarse sandMixture of silty sand and gravelMixture of clay, sand, silt and gravel

>45-104-615-30025-1000

1-31-3

Grain Size and Plasticity Characteristics

Hydrometer AnalysisRelates particle size to settling velocity Used to determine size of -#200 fraction

Plasticity CharacteristicsPlastic Limit - lowest water content at which soil exhibits plastic behaviorLiquid limit - highest water content at which soil exhibits plastic behaviorPlasticity Index

Pl = LL - PLClassification 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)SandSiltClayOrganics 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-8Granular soils (gravels and sands) fall into A-1 through A-3

Differentiated primarily on basis of grain size distributionFine-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 GravelS SandM siltC ClayO Organic

Modifiers: W Well GradedP Poorly GradedH High PlasticityL Low Plasticity

Examples: GW Well-graded gravelSP Poorly-graded (uniform) sandMH Highly plastic siltCL Low plasticity clayGM Silty gravel

Sieve Size Soils 1, % Finer

Soil 2, % Finer

Soil 3, % Finer

No. 4 99 97 100No. 10 92 90 100No. 40 86 40 100No. 100 78 8 99No. 200 60 5 97LL 20 - 124PL 15 - 47PI 5 NP* 77* non-plastic

Given:Sieve analysis and plasticity data for the following three soils classify the soils

Example

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 - SandD60 = 0.71 mmD30 = 0.34 mmD10 = 0.18 mm

SP - SMSoil 3> 50% passes #200 - Fine -grainedLL=124 Pl=77 - Off the chart - Extrapolating gives CHCould be CH-MH

0 71 3 90 18

u.C ..

0 342 10 18 0 71

c.C

( . )( . )

Aggregate Soil Properties (Phase Diagrams)

Phase DiagramsSolid, Water, and Gas phases shown separatelyVolumes indicated on left side of phase diagram Weights indicated on right side of phase diagram

GasWaterSolid

Vg

Vw

Vs

VvVt

O

Ww

Ws

Wt

DefinitionsVoid Ratio e = Vv/Vs

Porosity n = Vv/Vt

Water Content w = Ww/Ws

Degree of Saturation S = Vw/Wv

Density* ρ= Mass/VolumeUnit Weight* γ= weight/VolumeSpecific Gravity G = ρs/ρw

*Review text lumps density and unit weight together and uses symbol ρ

Common practice is to assume Vs = 1,

then express other volumes and weights accordingly.

Gas

Water

Solid

wGs

1

ewGsρw

Gsρw

Moist unit weight

ρm = Wt/Vt

Dry unit weight (dry density)

ρd = Ws/Vt

Saturated unit weight

ρsat = ρm for S=100%

Buoyant unit weight

ρsub = ρsat - ρw

From definitions

1en

es w

mG ρ( 1+w)ρ = 1+e

md

ρρ =1+w

swGS

es w

dG ρρ = 1+e

Table 35.7 - Useful for rapid calculation of phase relationships

Given: e = 0.62 w = 15% Gs = 2.65Calculate: a. ⍴d

b. ⍴m

c. w for S = 100%d. ⍴sat for S = 100%

s wmm

m

G ρ( 1+w )W ( 2.65 ) ( 62.4 pcf ) ( 1.15 )b.) ρ = = = =117pcfV 1+e 1.62

s s wd

m

W G ρa.) ρ = =V 1+e( 2.65 ) ( 62.4 pcf ) = 1.62

=102 pcf

Example

Gas

Water

Solid

wGs

1

ewGsρw= Seρw

Gsρw

ss

e 0.62c.) For S =100%,e =w G w= = =23.3%G 2.65

s w w s wsat

G ρ +S e ρ ( G +S e ) ρ ( 2.65+( 1.00 ) ( 0.62 ) )( 62.4 pcf )d.) ρ = = =1+e 1+e 1.62 =126 pcf

Standard Penetration Test140lb hammer dropped 30" to drive standard sampler. Number of blowsrequired 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 RelationshipsCompaction 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

woptw

(⍴d)max

Field Density Tests

Direct

Backscattering

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 informatione

Change in void ratio w/time for stressIncrement settlement rate information

Apply increment of stress

e0

ef

P0 Pf Log p

Initial equilibrium

Final equilibrium

e0

ef

time

Initial equilibrium

Final equilibrium

Fast rate

Slow rate

Consolidation Parameters

Compression Index, CcGiven by slope of e-log p curve (NC portion)

Recompression Index, CrGiven by slope of rebound portion of curve (OC portion)

Coefficient of Consolidation, Cv

ei

ef

High Cv (fast settlement)Low Cv (slow settlement)

time

e

Cc

Cr

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

friction

cohesion

c

s

For drained loading, c = 0

S

Typical for sands

S =c +σtan

Shear Strength of soils

For un-drained loading,

0

S

Typical for clayscSnc

Shear Strength and Principal Stresses

Ϭ3 Ϭ1 Ϭ

Շ

c

Փ

Failure surface is always oriented at 45 + Փ/2 angleto minor principal stress axis

21 3σ =σ tan( 45+ /2) + 2c tan( 45+ /2)

23 1σ =σ tan( 45+ /2) + 2c tan( 45+ /2)

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 :

1. Evaluate initial effective stress conditions

2. Evaluate change in effective stress due to imposed loading

3. Determine appropriate soil properties

4. Perform calculations

APPLICATIONS

Settlement Problems

Evaluation of Initial Effective Stresses

Need to know Thickness of soil layersDensity of soil layersGroundwater level

For effective stresses, use ρm above water table ρsub below water table

or calculate total stress and subtract water pressure

For total stresses, use ρm above water table ρsat below water table

For water pressure, take product of ρw and depth below water table

''

uu

Example

10’e = 0.40w = 10%zLayer 1

Layer 25'

15'

e = 0.60S = 20%S = 100%

First, calculate soil densities

s wm1

s

s wm2

s wsub

G ρ ( 1+w ) ( 2.7 ) ( 62.4 ) ( 1.1 )Layer1:ρ = = =132pcf1+e 1.4S e ( 0.2 ) ( 0.6 )Layer 2( abovewt): w= = =0.044=4.4%G 2.7

G ρ ( 1+w )ρ = =110pcf1+e( G -1 ) ρ ( 1.7 ) ( 62.4 )Layer2( belowwt ):ρ = = =66pcf1+e 1.6

Then, calculate stresses

At

v m1

v m1

v v m2

At z =5ft. σ =ρ z =( 132pcf )( 5ft )=660psfAt z =10ft. σ =ρ z =( 132pcf )( 10ft )=1,320psfAt z =15ft. σ =σ( z =10 ft )+ρ ( z -10 ft )

=1,320psf +( 110pcf )( 15 -10 )=1,870psfz =

v v sub20ft σ =σ ( z =15 ft )+ ρ ( z -15 ft )=1,870psf +( 66pcf )( 20 -15 )=2,200psf

Change in effective stresses can be caused by:

1. External loading Placement of fill (Ϭ‘v up ) settlementConstruction of structure (Ϭ‘v up ) settlementExcavation (Ϭ‘v down ) rebound

2. Change in groundwater conditions Drawdown of water level – (Ϭ‘v up ) settlementRising water level (Ϭ‘v down ) rebound

1. Calculation of final effective stresses after u excess dissipates

Based on assumption of hydrostatic water pressures, u = ρw (z-zw)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 ρm=120 pcf.

5‘ ρm= 120

10‘ z ρ m= 132

5‘ ρ m= 110

15‘ ρ sub= 66

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 σv' different below edge of footing than below center. Different at depth than shallow

v

At z =20σ =( 5 )( 120 pcf ) +( 10 )( 132 pcf ) +( 5 )( 110 pcf ) +( 5 )( 66 pcf ) =2,800 psf

Examples

Z=20’

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 1For strip footing loads, use Appendix 40a (left)

Example 2For 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 1Calculate vertical stress 5 ft. below and 2 ft. to the side of a surface point load of 1,000 lbs.

1000 lbs

2'

5'P v

2

2

11

11 0 4

( )

( . )

r z

5/2

v 2

5/2

2

3PP =2π z3( 1000)=2π ( 5 )

=13.2 psf

Example 2Calculate 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'Pv 0.2p

Chart in Appendix 40A p. A-69 (left side)

2,000 psf

vP =0.2( 2000psf )=400 psf

15 =3B5

5'

PLOT 40.A

14'Pv

Example 3Calculate 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 '

0.04p

10,000 psf

B.534

14=

vP =( 0.04 ) ( 10,000 psf)=400 psf

p. 40.ARight side

PLOT 40.A

14'Pv

Example 3Calculate 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 '

0.04p

10,000 psf

B.534

14=

vP =( 0.04 ) ( 10,000 psf)=400 psf

p. 40.ARight side

Example 4A 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' Pv

16 ft

10'

p=( 20 ) ( 62.4 pcf )=1248 psf

x =1.25rz =1.0r

20= .I

I = 0.2

01= .rz

251= .rx

vP =Ip= 0.2 1248 psf =250 psf

Appendix Dp. 40.B

PLOT 40.B

Determination of appropriate soil properties

Initial Conditions

Final Conditions

Cc or Cr

e

e1

e2

p1 p2 Log p

Compute Cc or Cr from e-log p curveConsolidation test

Cc applies to normally consolidated range

Cr applies to over-consolidated range

Pre-consolidation Pressure, Pp

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 Cc

Soil is over-consolidated when current effective stress is less than current value of pre-consolidation pressure.

Settlement behavior controlled by Cr.

Pre-consolidation Pressure, P'p

P'1 P'2 Log p

ee1

e2

Cr

Cc

Pre-consolidation Pressure, PpDisturbance Effects

Pre-consolidation Pressure, PpCasagrande Method

Calculation of Settlement MagnitudeNeed: 1. Initial and final effective stresses

2. Definition of Cc and Cr

p 1r

p1

e -eC = Plog( )P

ultv

1

ΔeΔH∈= =H 1+e

3. Definition of vertical strain

cr p 2ult

1 p1 1 1

CC PΔe PΔH = H= Hlog( )+ Hlog( )P P1+e 1+e 1+e

2 pc

2p

e -eC = Plog( )P

poc p 1 r

1

PΔe =e -e =C log( )P

nce

2NC 2 p c

p

oc

PΔe =e -e =C log( )Pe = e

P‘1P‘2

Log p

ee1

e2

P‘p

ep

OC NCinitial

final

d

s w

d

ssub w

125 pcf= =113.6pcf1+0.10Ge = -1=0.48

G -1= =71.7pcf1+e

oP =( 3 ft )( 125pcf ) +( 2 ft )( 71.7pcf ) +( 2.5 ft )( 30pcf )=593psf

First, calculate initial effective stress at center of soft clay layer before new fill placed

f oP =P +( 4 ft )( 125pcf ) = 593psf +500psf =1093psfNext, calculate final stress after placement of new fill

Then, calculate ultimate settlement as

c fult

oo

C PΔH = Hlog P1+e1.06 1093= 5ft log3.53 593

=0.40ft=4.8in

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 FillSame properties as new fill

Soft ClayCc=1.06 eo=2.53 sub = 30 pcf

Dense Sand

Now, what would happen if half of the new fill was removed?

'i

o

'f

Initial effective stress , psf PFinal effective stress P p

, pcf ft

psf P

1093

1093 125 2843

Since effective stress is decreasing, use CrAssuming Cr = 0.10

10 10 84353 53 10930 016 0 2

r fult

oo

C PH H log( )Pe. ( ft) log( ).. ft . in rebound

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

843

1 093

843 4 1251343

'i

'o f

Initial effective stress psf P

Precondolidationpressure , psf

Final effective stress P p Ppsff t pcfpsf

Then

1 1

0 10 1093 1 06 13435 53 53 843 3 53 10930 016 0 134

0 1501 8

cr p fult

o po o

oc nc

CC P PH H log H logP Pe e

. .log log

. .. ft . ftH H. ft. in

Time Rate of Primary Consolidation

Rate controlled by coefficient of consolidation, CvHigh Cv rapid consolidationLow Cv slow consolidation

Degree of consolidation Fraction of ultimate settlement which has occurred by time t

t ultU( t) H / H

2 v Vt T ( t )z /C

Fraction of ultimate settlement which has occurred by time t

Time required to reach given degree of consolidationDimensionless time factor

Settlement at given time t

2

t ult

v

v v

U(t) H / Hwhere U(t) f(T (t))

T (t) c t / z

Where Tv(t) and U(t) are related by Eq 40.23 and Table 40.1

5090100 .2 .85 Tv

0%

U

Length of longest drainage path

Degree of Consolidationcurves

Example 6

If Cv for the soft clay of Example 5 was 10ft2/yr, how long would it take for 2 in of settlement to occur?

2 2

2 2

2

2 100 424 8

40 23

0 42 0 144 4

40 20

0 14 2 5 0 088 110

v

v

v

inU x % %. in

From Eq. . (or from a U vs. T chart)

T U ( . ) .

From Eq. .

T z ( . )( . ft)t . yr mo.C ft / yr

What if the soft clay was underlain by impermeable bedrock? Then z= 5 ft

2

2

0 14 5 0 35 4 210

( . )( ft)t . yr . months

ft / yr

Double drainage

Single drainage