Shear Strength of Soil - Geotecnica e · PDF fileShear failure of soils Soils generally fail...

Shear Strength of SoilsShear Strength of Soils

CEL 610 – Foundation Engineering

Strength of different materials

Steel Concrete Soil

Tensilestrength

Compressivestrength

Shearstrengthstrength strength strength

Presence of pore waterComplexbehaviorbehavior

Shear failure of soilsSoils generally fail in shear

EmbankmentEmbankment

Strip footing

Failure surface

Mobilized shear resistance

At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.


Retaining wall


Retaining wall

Mobilized shear resistance

Failure surface

resistance

At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.

Shear failure mechanismShear failure mechanism

failure surface

The soil grains slideover each other alongthe failure surface.

No crushing ofNo crushing ofindividual grains.

Shear failure mechanismShear failure mechanism

At failure, shear stress along the failure surface ()reaches the shear strength (f).g ( f)

Mohr-Coulomb Failure Criterion( )(in terms of total stresses)

tan cf

Cohesion Friction angle

cf

is the maximum shear stress the soil can take without

f is the maximum shear stress the soil can take without failure, under normal stress of .

Mohr-Coulomb Failure Criterion( )(in terms of effective stresses)

'tan'' cf'

’u '

u = pore waterEffective cohesion Effective

f i ti l

u pore water pressure

’

c’friction anglef

’

is the maximum shear stress the soil can take without

’’

f is the maximum shear stress the soil can take without failure, under normal effective stress of ’.

Mohr-Coulomb Failure CriterionShear strength consists of two components: cohesive and frictional.p

'tan'' c tan ff c f

’f tan ’ frictional ’

c’ c’

f tan component

’f '

c and are measures of shear strengthc and are measures of shear strength.

Higher the values higher the shear strengthHigher the values, higher the shear strength.

Mohr Circle of stress’1

’3’3

’

Soil element

’1

Resolving forces in and directions

2'3

'1 Sin

Resolving forces in and directions,

22

2

22

'3

'1

'3

'1' C

Sin

2'3

'1

2'3

'1'2

22

2

223131 Cos 22


’

’1’1

’’

Soil element

’3’3

Soil elementSoil element

’3’3 ’3’3

Soil element

’1


’1’1

22

2'3

'1

2'3

'1'2

22

2

'3

'1

22

’

2

'3

'1 '

3'1


’

’1’1

’’

Soil element

’3’3


’3’3 ’3’3

Soil element

’1


’1’1

22

’, 2'

3'1

2'3

'1'2

22

2

'3

'1

22

’

2

'3

'1 '

3'1

PD = Pole w.r.t. plane

Mohr Circles & Failure Envelopep

Failure surface

'tan'' cf f

X XYY

Soil elements at different locations

’

X ~ failure

Y ~ stable

X ~ failure

Mohr Circles & Failure EnvelopepThe soil element does not fail if the Mohr circle is containedthe Mohr circle is contained within the envelope

GL

GL

Y

c

cc

Initially, Mohr circle is a pointc+

Mohr Circles & Failure EnvelopepAs loading progresses, Mohr circle becomes larger…

GLGL

Y

c

c

c

.. and finally failure occurs when Mohr circle touches the envelopeenvelope

Orientation of Failure Plane’1’1 Failure envelope

’, f’3’3

’

’3’3

’

’1

’1

–

’'3

'1 '

3'1

’

2

PD = Pole w.r.t. plane

Therefore,45 + ’/2

– ’ = 45 + /2

Mohr circles in terms of total & effective stresses

v’ uv

= Xh’

X u+Xh +

effective stressestotal stresses

v’h’ vh or ’vhu

vh

Failure envelopes in terms of total & effectivestressesstresses

v’ uv

= Xh’

X u+Xh +

If X is on failure

Failure envelope interms of total stresses

’

Failure envelope in termsof effective stresses

effective stressestotal stresses

v’h’ vh or ’cc’vh

uvh

Mohr Coulomb failure criterion with Mohr circle of stressof stress

’v = ’1 Failure envelope in termsof effective stresses

X’h = ’3

effective stresses

’ c’’ ’

X is on failure ’1’3 ’ c

c’ Cot’ ’’c Cot

''''

Therefore,

2'

2''

'3

'1

'3

'1 SinCotc

Mohr Coulomb failure criterion with Mohr circle of stressof stress

''''3

'1

'3

'1 SinCotc

22

SinCotc

''2''''' CS ''2'3131 CoscSin

''2'1'1 '' CoscSinSin 211 31 CoscSinSin

''2'1'' CoscSin

'1

2'131

Sin

cSin

'45'2'452'' TancTan

2

4522

4531 TancTan

Determination of shear strength parameters ofsoils (c or c’ ’soils (c, or c

Laboratory tests on Field testsLaboratory tests onspecimens taken fromrepresentative undisturbed

Field tests

samples

M t l b t t t 1 V h t tMost common laboratory teststo determine the shear strengthparameters are,

1. Vane shear test2. Torvane3. Pocket penetrometer

1.Direct shear test2.Triaxial shear test

4. Fall cone5. Pressuremeter6. Static cone penetrometer

Other laboratory tests include,Direct simple shear test, torsionali h t t l t i t i i l

p7. Standard penetration test

ring shear test, plane strain triaxialtest, laboratory vane shear test,laboratory fall cone test

Laboratory tests

Field conditions

A representative

z vc

A representative soil sample z

vc +

hchc hchc

hc hc

+ vc

Aft d d i

vc +

Before construction After and during construction

Laboratory testsvc +

Simulating field conditions in the laboratory

hchc

vc0 vc +

hchc00vc

vc0

Step 1Representativesoil sample St 2

vc

Set the specimen inthe apparatus and

l th i iti l

soil sampletaken from thesite

Step 2

Apply theapply the initialstress condition

corresponding fieldstress conditions

Direct shear testS h ti di f th di t h tSchematic diagram of the direct shear apparatus

Direct shear testDi t h t t i t it bl f lid t d d i d t t

Preparation of a sand specimen

Direct shear test is most suitable for consolidated drained testsspecially on granular soils (e.g.: sand) or stiff clays

Preparation of a sand specimen

PorousPorous plates

Components of the shear box Preparation of a sand specimen

Direct shear testPreparation of a sand specimen Pressure plate

L li h fLeveling the top surface of specimen

Specimen preparation completed

Direct shear testSt l b llTest procedurePressure plate

Steel ballP

Porous plates

S

Proving ring tto measure shear force

Step 1: Apply a vertical load to the specimen and wait for consolidation

Direct shear testSt l b llPTest procedurePressure plate

Steel ball

Porous plates

S

Proving ring tto measure shear force

Step 1: Apply a vertical load to the specimen and wait for consolidation

Step 2: Lower box is subjected to a horizontal displacement at a constant rate

Direct shear test

Shear box

Dial gauge tomeasure verticaldisplacementdisplacement

Proving ringto measure

Loading frame to Dial gauge to

shear force

gapply vertical load

g gmeasure horizontaldisplacement

Direct shear testfAnalysis of test results

sampletheofsectioncrossofArea(P) force Normal stress Normal

sampletheofsection crossofArea

(S)flidihd l diShsample theofsection cross of Area

(S) surfaceslidingat thedevelopedresistanceShear stressShear

Note: Cross-sectional area of the sample changes with the horizontal displacementd sp ace e t

Direct shear tests on sandsStress-strain relationship

ess,

Dense sand/ OC l

p

ear s

tre OC clay

fLoose sand/

She

NC clayf

Shear displacement

ion

Dense sand/OC Clay hei

ght

mpl

e

Expa

nsi

Loose sand/NC Clayhang

e in

the

sam

ssio

n Shear displacementy

Ch

of

Com

pres

Direct shear tests on sandsHow to determine strength parameters c and How to determine strength parameters c and

s,

Normal stress = 1ear s

tres

s

Normal stress = 2

Normal stress = 3

f1

Normal stress 1

She f2

f3

Shear displacement

re,

fs

at fa

ilu

Mohr – Coulomb failure envelope

ar s

tres

s

Shea

Normal stress,

Direct shear tests on sandsSome important facts on strength parameters c and of sandSome important facts on strength parameters c and of sand

Direct shear tests areSand is cohesionlesshence c = 0

Direct shear tests aredrained and pore waterpressures aredissipated hence u = 0dissipated, hence u = 0

Therefore,

’ = and c’ = c = 0

Direct shear tests on clays

In case of clay, horizontal displacement should be applied at a veryslow rate to allow dissipation of pore water pressure (therefore, onetest would take several days to finish)

Failure envelopes for clay from drained direct shear tests

test would take several days to finish)

e, f Overconsolidated clay (c’ ≠ 0)

at fa

ilure

Normally consolidated clay (c’ = 0)

y ( )

rstr

ess

a

’

Shea

r

Normal force,

Interface tests on direct shear apparatusI f d ti d i bl d t i i ll bl itIn many foundation design problems and retaining wall problems, itis required to determine the angle of internal friction between soiland the structural material (concrete, steel or wood)

PP

Soil SSoil S

Foundation materialFoundation material

tan' af c Where,

dh iaf ca = adhesion,

= angle of internal friction

Triaxial Shear TestPiston (to apply deviatoric stress)

Failure planeO-ring

Soil sample

impervious membrane

SoilSoil sample at failure Porous

stonePerspex

Soil sample

cellWater

pedestal

Cell pressureBack pressure Pore pressure or

volume change

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Sampling tubes

Sample extruder


Edges of the sample are carefully trimmed

Setting up the sample in the triaxial celly


Sample is coveredwith a rubber Cell is completelywith a rubbermembrane and sealed

p yfilled with water


Proving ring toProving ring tomeasure thedeviator load

Dial gauge tomeasure verticalmeasure verticaldisplacement

Types of Triaxial Tests deviatoric stress ( = q)cStep 1( q)

Step 2

ccc c

c+ q

Under all-around cell pressure c

c

Shearing (loading)

c q

Is the drainage valve open? Is the drainage valve open?

yes no

Consolidated U lid t d

yes no

D i d U d i dConsolidatedsample

Unconsolidatedsample

Drained loading

Undrainedloading

Types of Triaxial Tests

Under all aro nd cell press re

Step 1

Sh i (l di )

Step 2

Under all-around cell pressure c Shearing (loading)

Is the drainage valve open?

yes no

Is the drainage valve open?

yes no

Consolidatedsample

Unconsolidatedsample

yes no

Drained l di

Undrainedloadingsample sample loading loading

CD test

CU test

UU test

CU test

Consolidated- drained test (CD Test)Total, = Neutral u Effective ’+

Step 1: At the end of consolidationVC

Total, = Neutral, u Effective, +

’VC = VCVC

hC 0

VC VC

’hC =Drainage 0

Step 2: During axial stress increase

hC

’

Drainage

VC +

’V = VC + =’1

’ ’hC 0 ’h = hC = ’3Drainage

Step 3: At failureStep 3: At failureVC + f ’Vf = VC + f = ’1f

hC 0 ’hf = hC = ’3fDrainage

Consolidated- drained test (CD Test)

1 = VC + 1 VC +

3 = hC

Deviator stress (q or d) = 1 – 3

Volume change of sample during consolidationConsolidated- drained test (CD Test)

g p g

of th

e

xpan

sion

chan

ge o Ex

Time

Volu

me

csa

mpl

e

ssio

n

V s

Com

pres

Stress-strain relationship during shearingConsolidated- drained test (CD Test)

s, d Dense sand

or OC clay

p g g

r str

ess or OC clay

d)fLoose sand

Dev

iato or NC Clayd)f

D

Axial strain

ion

Dense sand or OC clayha

nge

mpl

e

Expa

nsi

y

L dolum

e ch

the

sam

ssio

n Axial strain

Loose sand or NC clay

Vo of

Com

pres

CD tests How to determine strength parameters c and

d d)fc = + ( )re

ss,

d

C fi i t

d)fc

Confining stress = 3c

1 = 3 + (d)fvi

ator

str

)Confining stress = 3ad)fb

Confining stress = 3b

3

Dev

Axial strain

d)fa

ess,

Mohr – Coulomb failure envelope

ear s

tre failure envelope

She

or ’ 3c 1c3a 1a

(d)fa

3b 1b

(d)fb

CD testsSt th t d bt i d f CD t tStrength parameters c and obtained from CD tests

Since u = 0 in CD tests = ’

Therefore, c = c’ and = ’ tests, =

cd and d are used to denote them

CD tests Failure envelopes

For sand and NC Clay, cd = 0s, d

M h C l b

stre

ss Mohr – Coulomb failure envelope

Shea

r S

or ’3a 1a

( )(d)fa

Therefore, one CD test would be sufficient to determine dTherefore, one CD test would be sufficient to determine d of sand or NC clay

CD tests Failure envelopes

For OC Clay, cd ≠ 0

OC NC

OC NC

c or ’3 1

(d)f

cc(d)f

Some practical applications of CD analysis forclaysclays1. Embankment constructed very slowly, in layers over a soft clay

deposit

Soft clay

= in situ drained shear strength

Some practical applications of CD analysis forclaysclays2. Earth dam with steady state seepage

Core

= drained shear strength of clay corestrength of clay core

Some practical applications of CD analysis forclaysclays3. Excavation or natural slope in clay

= In situ drained shear strength

Note: CD test simulates the long term condition in the field.Thus, cd and d should be used to evaluate the longterm behavior of soils

Consolidated- Undrained test (CU Test)Total, = Neutral u Effective ’+

Step 1: At the end of consolidationVC

Total, = Neutral, u Effective, +

’VC = VCVC

hC 0

VC VC

’hC =Drainage 0

Step 2: During axial stress increase

hC

Drainage

’ = + ± u = ’VC +

No

V = VC + ± u = 1

’ ± ’hC ±u

Step 3: At failure

drainage ’h = hC ± u = ’3

Step 3: At failureVC + f

N

’Vf = VC + f ± uf = ’1f

hCNo drainage ±uf ’hf = hC ± uf = ’3f

Volume change of sample during consolidationConsolidated- Undrained test (CU Test)

g p g

of th

e

xpan

sion

chan

ge o Ex

Time

Volu

me

csa

mpl

e

ssio

n

V s

Com

pres

Stress-strain relationship during shearingConsolidated- Undrained test (CU Test)

s, d Dense sand

or OC clay

p g g

r str

ess or OC clay

d)fLoose sand

Dev

iato or NC Clayd)f

D

Axial strain

+

Loose sand /NC Clay

+

Dense sand

u

-

Axial strain

Dense sand or OC clay

CU tests How to determine strength parameters c and

d d)fb 1 = 3 + (d)fre

ss,

d

d)fb

Confining stress = 3b

1 3 + (d)f

Confining stress =

viat

or s

tr

3

)

Confining stress = 3a

Dev

Axial strain

Total stresses at failured)fa

stre

ss, cuMohr – Coulomb

failure envelope interms of total stresses

hear

sS

or ’

ccu3b 1b3a 1a

(d)fa

CU tests How to determine strength parameters c and ’1 = 3 + (d)f - uf

Mohr – Coulomb failureenvelope in terms of

’ = 3 - ufuf

ss,

Mohr – Coulombf il l i

peffective stresses

’Effective stresses at failure

ar s

tres cu

failure envelope interms of total stresses

Shea

or ’

ccu ’3b ’1b’ ’

C’ ufa

ufb

(d)fa

3b 1b3a 1a(d)fa

’3a ’1a

CU testsSt th t d bt i d f CD t tStrength parameters c and obtained from CD tests

Shear strengthShear strengthparameters in terms

Shear strengthparameters in termsof effective stressesp

of total stresses areccu and cu

are c’ and ’

c’ = cd and ’ = dd d

CU tests Failure envelopes

F d d NC Cl d ’ 0For sand and NC Clay, ccu and c’ = 0

Mohr – Coulomb failurel i t f

Mohr – Coulomb ’

envelope in terms ofeffective stresses

ress

, cuMohr Coulombfailure envelope interms of total stresses

hear

st

Sh

or ’3a 1a3a 1a

Therefore one CU test would be sufficient to determine

3a 1a(d)fa

3a 1a

Therefore, one CU test would be sufficient to determine cu and ’= d) of sand or NC clay

Some practical applications of CU analysis forclaysclays1. Embankment constructed rapidly over a soft clay deposit

Soft clay

= in situ undrained shear strength

Some practical applications of CU analysis forclaysclays2. Rapid drawdown behind an earth dam

Core

= Undrained shear strength of clay corestrength of clay core

Some practical applications of CU analysis forclaysclays3. Rapid construction of an embankment on a natural slope

Note: Total stress parameters from CU test (ccu and cu) can be used for

= In situ undrained shear strength

p ( cu cu)stability problems where,

Soil have become fully consolidated and are at equilibrium withthe existing stress state; Then for some reason additionalstresses are applied quickly with no drainage occurring

Unconsolidated- Undrained test (UU Test)Data analysisData analysis

Initial specimen conditionSpecimen condition during shearing

C = 3

C = 3No drainage

3 + dNo drainageC 3drainage 3drainage

I i i l l f h l A HInitial volume of the sample = A0 × H0

Volume of the sample during shearing = A × H

Since the test is conducted under undrained condition,

A × H = A0 × H0A × H A0 × H0

A ×(H0 – H) = A0 × H0

AA 1

0

A ×(1 – H/H0) = A0 z1

Unconsolidated- Undrained test (UU Test)Step 1: Immediately after samplingStep 1: Immediately after sampling

0

00

Step 2: After application of hydrostatic cell pressure

= +C = 3

C = 3 u

’3 = 3 - uc

’ = - uNo drainage = +

u = B

C 3 uc 3 = 3 - ucdrainage

uc = B 3

Increase of pwp due toIncrease of cell pressure

Increase of pwp due toincrease of cell pressure Skempton’s pore water

pressure parameter, B

Note: If soil is fully saturated, then B = 1 (hence, uc = 3)

Unconsolidated- Undrained test (UU Test)

Step 3: During application of axial load

3 + dNo

’1 = 3 + d - uc ud3 d

3

No drainage ’3 = 3 - uc ud= +

uc ± ud

ud = ABd

Increase of pwp due toincrease of deviator stress

Increase of deviatorstress

Skempton’s pore water pressure parameter, A


Combining steps 2 and 3,

uc = B 3 ud = ABd

Total pore water pressure increment at any stage, u

u =uc + ud

u = B [3 + Ad]Skempton’s porewater pressure

tiu = B [3 + A(1 – 3]

equation

Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+

Step 1: Immediately after sampling0

Total, = Neutral, u Effective, +’V0 = ur

0 -ur

Step 2: After application of hydrostatic cell pressure

’h0 = ur

’ = + u = uStep 2: After application of hydrostatic cell pressureC

C-ur uc = -ur c

No drainage

VC = C + ur - C = ur

’h = ur(Sr = 100% ; B = 1)

Step 3: During application of axial load +

’V = C + + ur - c u C +

C

No drainage -ur c ± u ’h = C + ur - c u

Step 3: At failure ’Vf = C + f + ur - c uf = ’1fC + fNo

’hf = C + ur - c uf = ’3f

-ur c ± ufC

No drainage

Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+Total, = Neutral, u Effective, +


’hf = C + ur - c uf = ’3f

-ur c ± ufC

C fNo drainage

Mohr circle in terms of effective stresses do not depend on the cellpressure.

Therefore, we get only one Mohr circle in terms of effective stress fordifferent cell pressures

’’3 ’1f

Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+Total, = Neutral, u Effective, +


’hf = C + ur - c uf = ’3f

-ur c ± ufC

C fNo drainage

Mohr circles in terms of total stresses

Failure envelope, u = 0

uaub

cu

3b 1b3a 1af’3 ’1 or ’


Effect of degree of saturation on failure envelope

S < 100% S > 100%

or ’3b b3a a3c c or

Some practical applications of UU analysis forclaysclays1. Embankment constructed rapidly over a soft clay deposit

Soft clay

= in situ undrained shear strength

Some practical applications of UU analysis forclaysclays2. Large earth dam constructed rapidly with

no change in water content of soft clay

Core

= Undrained shear strength of clay corestrength of clay core

Some practical applications of UU analysis forclaysclays3. Footing placed rapidly on clay deposit

= In situ undrained shear strength

Note: UU test simulates the short term condition in the field.Thus c can be used to analyze the short termThus, cu can be used to analyze the short termbehavior of soils

Unconfined Compression Test (UC Test)

1 = VC +

3 = 0 3

Confining pressure is zero in the UC testConfining pressure is zero in the UC test

Unconfined Compression Test (UC Test)

1 = VC + f

ss,

ear s

tres

3 = 0

She

qu

Normal stress,

/2 /2τf = σ1/2 = qu/2 = cu

Various correlations for shear strengthFor NC clays, the undrained shear strength (cu) increases with theeffective overburden pressure, ’0

)(0037.011.0'0

PIcu

Skempton (1957)0

Plasticity Index as a %

For OC clays, the following relationship is approximately true

cc

8.0

'0

'0

)(OCRcc

edConsolidatNormally

u

idatedOverconsol

u

Ladd (1977)

For NC clays, the effective friction angle (’) is related to PI as follows

)log(234.0814.0' IPSin Kenny (1959)

Shear strength of partially saturated soilsIn the previous sections, we were discussing the shear strength ofsaturated soils. However, in most of the cases, we will encounterunsaturated soilsunsaturated soils

Pore airWater Pore water

pressure, uPore waterWater

Air Pore air pressure, ua

Solid Effective Solid

Pore water pressure, uw

Water

Solid stress, ’ SolidEffective stress, ’

Saturated soils Unsaturated soils

Pore water pressure can be negative in unsaturated soils

Shear strength of partially saturated soils

Bishop (1959) proposed shear strength equation for unsaturated soils asfollows

'tan)()(' waanf uuuc Where, n – ua = Net normal stressua – uw = Matric suctionua uw Matric suction= a parameter depending on the degree of saturation

( = 1 for fully saturated soils and 0 for dry soils)

Fredlund et al (1978) modified the above relationship as follows

bwaanf uuuc tan)('tan)('

WhereWhere, tanb = Rate of increase of shear strength with matric suction

Shear strength of partially saturated soils

bwaanf uuuc tan)('tan)('

Same as saturated soils Apparent cohesion due to matric suctiondue to matric suction

Therefore, strength of unsaturated soils is much higher than the strengthof saturated soils due to matric suctionof saturated soils due to matric suction

’

- ua

How it become possiblebuild a sand castlebuild a sand castle

buuuc tan)('tan)(' waanf uuuc tan)(tan)(

Same as saturated soils Apparent cohesion due to matric suction

’

Apparent - uacohesion

Shear Strength of Soil - Geotecnica e · PDF fileShear failure of soils Soils generally fail...

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