Shear Strength of Soil

Hsin-yu ShanDept. of Civil Engineering

National Chiao Tung University

Principal Stressσ1 > σ2 > σ3

Major, Intermediate, Minor principal stress

ασσσσασασσ 2cos22

sincos 313123

21

−+

+=+=

ασσαασστ 2sin2

)(cossin)( 3131

−=−=

σy

100 kN/m2

50 kN/m2

60°

B

A

150 kN/m2

50 kN/m2

σx

τ

Pole Point

The point on the Mohr’s circle from which any extended straight line will intersect the circle at a point corresponding to stress on a plane parallel to the extended line

σxσy

σ1σ3

pole

σ3

α

σ1

σ

τ

α

pole

σ1σ3

Stress and Strainσ1ε1

σ2

σ3ε2

ε3

3 Dimension

0321 ≠≠≠ σσσσ1ε1

σ2

σ3 ε2ε3

ε ε ε1 2 3 0≠ ≠ ≠

Plane Strain

σ σ σ σ1 3 20≠ ≠ =; x

ε ε ε1 3 20 0≠ ≠ =; σ1ε1

σ2

σ3 ε2ε3

If the material is linearly isotropic:

σ ν σ σ2 1 3= +( )

Axisymmetric Conditions

AxisymmetricTriaxialUniaxial

σ1ε1

σ2

σ3 ε2ε3

Axisymmetric

σ σ σ σθ1 3 20≠ ≠ =;

ε ε ε1 3 20≠ ≠ = − ; srr

sr = redial displacementr = radial coordinate of the point

σ ν σ σ ε2 1 3 2= + + ( ) E

σ1ε1

σ2

σ3 ε2ε3

For linearly isotropic materials:

TriaxialCompression

σ σz r> σ σ σ σ σ1 2 3= = =z r ;ε ε ε ε ε1 2 3= = =z r ;

Extension

σ σz r<σ σ σ σ σ1 2 3= = =r z ;ε ε ε ε ε1 2 3= = =r z;

σ1ε1

σ2

σ3 ε2ε3

Uniaxial

Same as triaxial condition except σr = 0

Compression

σ σz r> σ σ σ σ σ1 2 3= = =z r ;ε ε ε ε ε1 2 3= = =z r ;

σ1ε1

σ2

σ3 ε2ε3

Ko Stress State σ σz r> σ σ σ σ σ1 2 3= = =z r ;

0 ; 321 === εεεε z

σ σz r< σ σ σ σ σ1 2 3= = =r z ;

σ1ε1

σ2

σ3 ε2ε3ε ε ε ε1 2 30= = =; z

σ ννσr z =

−1If the material is linearly isotropic

ε2 0=

ε2 0= − ≠ur

σ σ1 = z

σ σ σ2 3= = r ε ε ε2 3= = r

σ σ2 3 0= =σ σ1 = z ε ε ε2 3= = r

σ σ1 = z ε ε2 3 0= =

σ σ σ2 3= = r

No horizontal strain, ratio of horizontal stress to vertical stress is K0

Areal fill, large tankKo state

Symmetric to axisUC testUniaxial(Unconfined)

Symmetric to axisBelow the center of circular tank

Triaxial (Comp.)

Symmetric to axisCircular foundation, pileσ2 horiz.Axisymmetric

Same stress condition along long axis, no lateral strain

Strip foundation, retaining wall, slope

σ2 horiz.Plane strain

No symmetryNoneNoneTriaxial

NoteApplicationStrain condition

Stress condition

State of stress

Rendulic Plotaσ ′

rσ ′

2 rσ ′ rσ ′

Rendulic PlotAxial Stress

'aσRadial Stress 2 ′σr

'aσ

2 ′σr

q-p Plot Roscoe, Schofield, and Wroth (1958)

q = − = −σ σ σ σ1 3 1 3

p' ( )= + +1 σ σ σ q3 1 2 3

Lambe and Whiman (1969)

q = − = −12

121 3 1 3( ) ( )σ σ σ σ

p or p’

p' ( )= +12 1 3σ σ

τ - σ Plot

τ

σ or σ’

Normal Stress and Shear Stress

N

Area=A T

T

RNormal stress, σ = N/AShear stress, τ = T/AStress resultant, p = R/AObliquity = τ/σAngle of obliquity, i = tan-1(τ/σ)

N

τfτf

Coefficient of friction, µ = Tf /N = τf /σFriction angle, φ = tan-1(τf /σ)If there is cohesion, when σ = 0, c = τf

µ = ∆Tf /∆N = ∆τf/∆σφ = tan-1(∆τf /∆s)

φ φ

c

σσ

Enlarged contact surfaces

Acontact << Aσcontact >> σ

Yield at contact area?Crushing?Jumping?Moving over?

Ideal Shear Strength Test

Independent control of stress and strain in each directionDirection of principal stressRate of stress or strain applicationUniform distribution of applied stressControl over pore (air or water) pressure

Parameters Obtained

Shear strength parametersPore pressure parametersDeformation parameters

Laboratory Shear Strength Tests

Direct shear testRing shearSimple shear, direct simple shear

Unconfined compression testTriaxial test

CompressionExtension

Plane strain test seldomDynamic strength tests

Direct Shear ASTM D3080

Normal Force, N

Shear Force, T

Soil SpecimenArea=A

Normal stress, σ = N/AShear stress, τ = T/A

Stress Path

Failure envelopeτ

Shear

Failure, continuous deformation without increasing shear stress

σ or σ’Consolidation

Problems with Direct Shear Test

Nonuniform stressNormal stressShear stress

Changing area of shear planeContact between soil and shear box

Squeezing of soil

Contact between soil and shear box

Changing area of shear plane

Direct Shear

AdvantagesEasy to performLow costA lot of experience

DrawbackSmall-size specimensLimited shear displacementForced failure plane

σ σ

poleτ

σ3 1

Ring Shear

AdvantagesCost is not very highUnlimited shear displacement

Drawbacks•Anisotropy•Small specimen size•Forced failure plane•Limited experience

Unconfined Compression Test

Soil Specimen

σ3 σ1

Triaxial Test

Soil Specimen

No drainage

∆u

∆σ1

∆σ3

∆u

∆σ1∆σ1- ∆σ3∆σ3

∆ud∆ua

∆σ3 ∆σ3

+ =

Triaxial Compression

σ3σ1

Triaxial Extention

σ3 σ1

Determination of Shear Strength from Test Results

Define failure criteriaPeak stress differenceMaximum stress ratioMaximum shear stress τmax

Specific strain levelDetermine principal stresses and shear stressDrained tests: 1. and 2. give same c, φUndrained tests: 1. and 2. give different sets of c, φ

max31 )( σσ −

max3

31

−σσσ

Selection of failure criteria depends on:Testing conditionField condition (Drained vs. Undrained)Method of analysis (Total stress vs. Effective stress)

Determining c and φMohr-Coulomb diagram, τ and σ

Tangent line of Mohr circles

τ

231 σσ −

φ

c

σ or σ’

q-p’ plotLinear regression of (q’, p) sets

q = − = −12

121 3 1 3( ) ( )σ σ σ σ

p' ( )= +12 1 3σ σ

q

)(21

31 σσ −

φφ sin'cos pcq +=ψ

φφψ

cossintan

cd ==

d

)(21

31 σσ +p’

Unconsolidated UndrainedTest

UU test or Q (quick) testσ1 = 0

+ +∆u1u0

σ3 = 0 ∆σ3

∆σ3

030133031 )( uu −==∆−∆+= σσσσ

σ1- σ3

∆u2

Reason for the Confining Stress

More similar to actual field condition where all around pressure existsIncrease the degree of saturationCan make the fissures (formed during the extrusion of tube samples) in the specimens close up

σ1σ1ff3σσ3

02133 uuuf −∆−∆−=σσ

cc u−= 33 σσ

)( 31 σσ −

d

3σ

Consolidated Drained Test

CD test or S (slow) testσ1- σ3σ3 = 0

+ ∆u=0ucu0

σ3 = 0 σ3c

σ3c

uc = back pressure

cc u−= 33 σσ

u∆−= 33 σσ

)( 31 σσ −

d

)( 33 σσ

CD, S Tests Stress Path

)( 31 σσ −

Effective stress failure envelope

failure

σ33σ

uc

σ3 or σ 3

σ1

σ3

σ1f

cf u−= 33 σσ

f3σ

uc = backpressure

Consolidated Undrained Test

Without pore pressure measurement CU test or R (rapid) testWith pore pressure measurement or testIsotropic consolidation CIUAnisotropic consolidation CAU

CU R

∆u3

σ1- σ3

∆u2

∆σ3+

∆σ3σ3 = 0

uc

σ3c

u0 +σ3cσ3 = 0

uc = back pressure

Pore Pressure CoefficientsIn an undrained test, the change of pore pressure can be related to the changes of stresses as:

∆ ∆ ∆ ∆u B A= + −[ ( )]σ σ σ3 1 3

A and B are the pore pressure coefficients

3σ∆=∆ Bua

Change of total volume:

33 )( σσ ∆⋅⋅−=∆−∆⋅⋅−=∆ VcuVcV cacc

Change of volume of voids:

avv unVcV ∆⋅⋅−=∆

3σε

∆∆

= vcc

uc v

v ∆∆

=ε

cc = compressibility of the soil skeletoncv = compressibility of the fluid in voids

vc VV ∆=∆

c

v

ccnB⋅

+=

1

1

For saturated soils, if cc >> cv, B 1(soil skeleton is much more compressible than the fluid)

For very stiff clays whose cc is also very small, Bmay not be close to 1 even if the soil is almost saturated

→

⋅ 0c

v

ccn

A and B can only be determined from test resultsWhen (σ1-σ3) remains 0, B can be determined from the relationship between the increasing σ3 and u

For saturated specimens, B is close to 1Get A by from the changes of u and (σ1-σ3)A is not a constant, it changes through out the test

BuA d 1

31

⋅∆−∆

∆=

σσ

Elastic Material

[ ])(13211 σσυσε ∆+∆−∆=∆

E

[ ])(13122 σσυσε ∆+∆−∆=∆

E

[ ])(12133 σσυσε ∆+∆−∆=∆

E

If saturated no volume change:

0321 =∆+∆+∆=∆ εεεε v

Rewrite as effective stresses:

[ ] 0)(21321321 =∆+∆+∆−∆+∆+∆ σσσυσσσ

E

[ ] 0))(21(1321 =∆+∆+∆− σσσυ

E

ν ≠ 0.5

32 σσ ∆=∆

02 31 =∆+∆ σσ

u∆−∆=∆ 11 σσ

u∆−∆=∆ 33 σσ

032 31 =∆−∆+∆ uσσ

u∆=∆−∆+∆ 3)(3 313 σσσ

)(31

313 σσσ ∆−∆+∆=∆u

31

=A

Use B to check the saturation of soil specimens

Incremental increase of ∆σ3 check the increase of ∆u compute BIf B is close to 1, the specimen is near saturation

Although B is an indication of the degree of saturation, there is no way of knowing the degree of saturation from values of B

∆ ∆ ∆ ∆u B A= + −[ ( )]σ σ σ3 1 3

)]( 313 σσσ ∆−∆⋅+∆=∆ BABu

BAA ⋅=

31 σσ ∆−∆∆

=uAFor B = 1

During shearing ∆σ3 = 0

1σ∆∆

=uA

A depends on:the state of the specimen (soil type),the stress history (overconsolidation),the stress state (anisotropy),the stress level,loading conditions (loading or unloading), and other conditions

A can vary from negative values to greater than 1A at failure is defined as Af fA

Back PressureIncrease the degree of saturationIncrease pore water pressure, u, in the specimen while maintaining the effective confining (consolidation) stressFor example:

σ3 100 kPa 150 kPa 200 kPa 250 kPau 50 kPa 100 kPa 150 kPa 200 kPa

50 kPa 50 kPa 50 kPa 50 kPaCheck B at every stage

3σ

3σ

How Does Backpressure Works?

Elevated pore water pressure can squeeze on the trapped air bubbles and make them decrease in sizeElevated pore water pressure would make the trapped air more likely to dissolve in water

Some would apply a small hydraulic gradient to help flush out the air during the saturation process

σ1σ3f3σ σ1f

02133 uuuf −∆−∆−=σσ

Eff. stress path for undrainedtest, u decreases

Effective stress path for undrained testu increases

Effective stress failure envelope

)( 31 σσ −Total stress failure envelope

σ3 or 3σ

Increased u

)( 31 σσ −

Effective stress path for undrained testu increases

Effective stress failure envelope

Total stress failure envelope

σ3 or 3σ

Decreased u

Eff. stress path for undrainedtest, u decreases

Effective stress failure envelope

)( 31 σσ −

Total stress failure envelope

σ3 or 3σ

RCU ,CU, R Tests

Effective stress path for undrained testu increases

)( 31 σσ −

max31 )( σσ −max

3

1 )(σσ

Low sensitivity

High sensitivity

σ3 or 3σ

UU, Q Tests (CU, R Tests)

)( 31 σσ −

Q envelopemax31 )( σσ −

R envelope

Failure

σ3 σ3

Energy Corrections

Direct shear testsTriaxial tests

Direct Shear Testsσ

vdAhdAhdA

∆⋅⋅+∆⋅⋅=∆⋅⋅=

σττ

*

∆h ∆v > 0 for expansion

τ

∆v

Total work done by applied shear stress

Energy used to expand (+) or contract (-) against normal stress σ

Shear stress used to shear the soil to failure, τ*:

hdvd

∆∆

−= σττ*

Triaxial Tests

Work done by axial stress:

Work done to expand the specimen laterally against the confining stress:

dW=work required to shear the specimen(subject to σ3dε3)

3333 εσεσ dd ⋅−⋅−

11 εσ d⋅

333311 εσεσεσ dddWd ⋅−⋅−=⋅ Dilate if dε3>0

)()( 1333131 εεεσεσσ ddddWd ++−=−

321strain Volumetric εεεε dddd v ++==

Bishop assumed:

131 *)( εσσ ddW ⋅−=

vddd εσεσσεσσ 3131131 *)()( −−=−

133131 )()*(

εεσσσσσ

dd v+−=− < 0 expand

> 0 contract

Say c = 0

331

31

2)()(sinσσσ

σσφ+−

−=

31

31sinσσσσφ

+−

=

331

31

2*)(*)(*sinσσσ

σσφ+−

−=

Measured φ

Particle orientation effect

Dilatancy effect

φuφcv: constant volume

φ

34%46% Initial porosity, ni DenseLoose

φu = mineral-to-mineral friction angle

1

)1(*sin

3

1

3

1

+

−−=

σσ

εε

σσ

φ a

v

ddBishop

)1(

)1(*sin

3

1

3

1

a

v

a

v

dddd

εε

σσ

εε

σσ

φ−+

−−=

< 0 expand> 0 contract

Rowe

If soil tends to dilate, φ*Rowe < φ*

Bishop(Rowe considers particle rearrangement effect)

Effect of Energy Corrections

S testUncorrected φ = 42°Bishop’s correction φ = 38° more appropriateRowe’s correction φ = 24°

Effect of Confining Pressure

ffc )( 03

1σ

σφ ==

Pourse and Bell (1971) – sand, Dr=94%

3σ φ0.2 psi 52°10 psi 28.5°

Lee and Seed (1967) – Sacramento River sand

3σ φ40 ksc 42°140 ksc 33°

Dr =100%Some particle breakage

3σ φLow stress level 34°High stress level 32°

Dr =38%

Before shear 5% passed through #200 sieveAfter shearing at 140 ksc, 50% passed through #200 sieve

τ

DenseLoose

σ≈ 40 ksc

Measured φ

Particle orientation effect

Dilatancy effect

φuφcv: constant volume

φ

3σ

Some Other “Corrections” or “Considerations”

Measured stressMembrane correctionEffect of friction between soil specimen and filter paper/porous disk/cap and pedestal

B valueFlexibility of drainage tubingUse metal tubing

Stress-Strain Properties

At low to medium confining pressure

Dense sand

3

1

σσ

Loose sand

εa~20%~4%expansion

εv εa

contraction

At high confining stress

3

1

σσ

Dense sand

Loose sand

εa~20%~4%expansion

εv εa

contraction

At larger strains dense sand and loose sand tend to have the same void ratio They tend to have the same strength

Critical Void Ratio

A. Casagrande

Loose sandContraction

Dense sandDilate

Critical void ratio line

Where shearing starts

Where shearing starts

Drained shear

Void ratio,e

Confining stress

Undrained shear

Loose sand

Where shearing starts

Where shearing starts

Change of pore water pressure

Higer shear strength

Void ratio,e

Critical void ratio lineDense sand

Confining stress

Dense Sand

εa

3

1

σσ

)( 31 σσ −0 test, =∆VR

~4% ~20%increase

u εa

decrease

Loose Sand0 test, =∆VR )( 31 σσ −

3

1

σσ

εa~4% ~20%increase

u εa

decrease

For normally consolidated clay, when (σ1-σ3) reaches maximum, its shear strength has not been fully mobilized yet the shear strength is still increasingFor overconsolidated clay, although the shear strength has been fully mobilized, (σ1-σ3)keeps increasing due to the increase of effective confining stress the difference is not as significant as for N.C. clay

)( 31 σσ −

max3

1 )(σσ

Stress path tangency

Dense sand, OC clay

Loose sand, NC clay

3σ

)( 31 σσ −

Effive and total stress path of drained testTotal stress path for undrained test

Eff. stress path for undrained testu decreases

Effective stress path for undrained testu increases

Stress path tangency

σ3 or σ’3

)( 31 σσ −

max3

31

−σσσ

Normally Consolidated and Lightly Overconsolidated ClayLoose Sand

max31 )( σσ −

εa

max31 )( σσ − Overconsolidated ClayDense Sand

)( 31 σσ −

max3

31

−σσσ

εa

Steady-State Strength

The steady-state deformation for any mass of particles is that the state in which the mass is continuously deforming at constant volume, constant normal effective stress, constant shear stress, and constant velocity

Grain Size

Diameter of specimens should be at least 6 x (largest particle size)Can use specimens with the same shape of particle distribution curve for testingAlso they have to have the same relative densityφ increases as average effective stress increases and as max. particle size decreases

100

Per

cent

pas

sing

0Particle size

Effect of State of Stress

φ

~4° Plane strain

Triaxial

LooseDense Initial porosity, ni

Direct Shear Test

Direct shear is a “plane strain” test higher friction angleThe stress on the failure plane is not uniform, not all parts reached peak stress at “failure”

εf, tx is about 3 x εf, ps

φf, TC is about the same as φf, TE

)( 31 σσ −

εa

Triaxial

Plane strain

~1.5% ~4%

Dense sand)( 31 σσ −

Plane strain

Triaxial

Loose sand

εa