SHEAR STRENGTH IIYULVI ZAIKA

JURUSAN TEKNIK SIPILFAK.TEKNIK UNIV.BRAWIJAYA

Other laboratory tests include,Direct simple shear test, torsional ringshear test, plane strain triaxial test,laboratory vane shear test, laboratoryfall cone test

Determination of shear strength parameters ofsoils (c, φ or c’, φ’)

Laboratory tests onspecimens taken fromrepresentative undisturbedsamples

Field tests

Most common laboratory tests todetermine the shear strengthparameters are,

1.Direct shear test2.Triaxial shear test

1. Vane shear test2. Torvane3. Pocket penetrometer4. Fall cone5. Pressuremeter6. Static cone penetrometer7. Standard penetration test

Laboratory tests

Field conditions

zσvc

σvc

σhcσhc

Before construction

A representativesoil sample

zσvc + ∆ σ

σhcσhc

After and duringconstruction

σvc + ∆ σ

Laboratory testsSimulating field conditions inthe laboratory

Step 1Set the specimenin the apparatusand apply theinitial stresscondition

σvc

σvc

σhcσhc

Representativesoil sampletaken from thesite

0

00

0

Step 2Apply thecorresponding fieldstress conditions

σvc + ∆ σ

σhcσhc

σvc + ∆ σ

σvc

σvc

τ

τ

Direct shear testSchematic diagram of the direct shear apparatus

Direct shear test

Preparation of a sand specimen

Components of the shear box Preparation of a sand specimen

Porousplates

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

Direct shear test

Leveling the top surface ofspecimen

Preparation of a sand specimen

Specimen preparationcompleted

Pressure plate

Direct shear test

Test procedure

Porousplates

Pressure plate

Steel ball

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

P

Proving ring tomeasure shearforce

S

Direct shear test

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

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

PTest procedurePressure plate

Steel ball

Proving ring tomeasure shearforce

S

Porousplates

Direct shear test

Shear box

Loading frame toapply vertical load

Dial gauge tomeasure verticaldisplacement

Dial gauge to measurehorizontaldisplacement

Proving ring tomeasure shearforce

Direct shear testAnalysis of test results

sample theofsectioncrossofArea

(P)forceNormalstressNormal

sample theofsectioncrossofArea

(S)surfaceslidingat thedevelopedresistanceShearstressShear

Note: Cross-sectional area of the sample changes with the horizontaldisplacement

Direct shear tests on sands

Shea

r st

ress

,τ

Shear displacement

Dense sand/OC clay

τfLoose sand/ NCclayτf

Dense sand/OC Clay

Loose sand/NC Clay

Chan

ge in

hei

ght

of t

he s

ampl

e Expa

nsio

nCo

mpr

essi

on

Shear displacement

Stress-strain relationship

τf1

Normal stress = σ1

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

Shea

r st

ress

,τ

Shear displacement

τf2

Normal stress = σ2

τf3

Normal stress = σ3

Shea

r st

ress

at

failu

re, τ

f

Normal stress, σ

φMohr – Coulomb failure envelope

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

Sand is cohesionless hencec = 0

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

Therefore,φ’ = φ and c’ = c = 0

Direct shear tests on clays

Failure envelopes for clay from drained direct shear tests

Shea

r st

ress

at

failu

re, τ

f

Normal force, σ

φ’

Normally consolidated clay (c’ = 0)

In case of clay, horizontal displacement should be applied at a very slow rate toallow dissipation of pore water pressure (therefore, one test would take severaldays to finish)

Overconsolidated clay (c’ ≠ 0)

Interface tests on direct shear apparatusIn many foundation design problems and retaining wall problems, it is requiredto determine the angle of internal friction between soil and the structuralmaterial (concrete, steel or wood)

tan' af cWhere,ca = adhesion,δ = angle of internal friction

Foundation material

Soil

P

S

Foundation material

Soil

P

S

Triaxial Shear Test

Soil sample atfailure

Failure plane

Porousstone

imperviousmembrane

Piston (to apply deviatoric stress)

O-ring

pedestal

Perspexcell

Cell pressureBack pressure Pore pressure or

volume change

Water

Soilsample

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Sampling tubes

Sample extruder

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Edges of the sample arecarefully trimmed

Setting up the sample inthe triaxial cell

Triaxial Shear Test

Sample is covered with arubber membrane andsealed

Cell is completelyfilled with water

Specimen preparation (undisturbed sample)

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Proving ring tomeasure thedeviator load

Dial gauge tomeasure verticaldisplacement

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidatedsample

Unconsolidatedsample

Is the drainage valve open?

yes no

Drainedloading

Undrainedloading

Under all-around cell pressure c

cc

c

cStep 1

deviatoric stress( = q)

Shearing (loading)

Step 2

c c

c+ q

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidatedsample

Unconsolidatedsample

Under all-around cell pressure c

Step 1

Is the drainage valve open?

yes no

Drainedloading

Undrainedloading

Shearing (loading)

Step 2

CD test

CU test

UU test

Consolidated- drained test (CD Test)

Step 1: At the end of consolidationσVC

σhC

Total, σ = Neutral, u Effective, σ’+

0

Step 2: During axial stress increase

σ’VC = σVC

σ’hC = σhC

σVC + ∆ σ

σhC 0

σ’V = σVC + ∆ σ= σ’1

σ’h = σhC = σ’3

Drainage

Drainage

Step 3: At failure

σVC + ∆ σf

σhC 0

σ’Vf = σVC + ∆ σf = σ’1f

σ’hf = σhC = σ’3fDrainage

Deviator stress (q or ∆ σd) = σ1 – σ3

Consolidated- drained test (CD Test)

σ1 = σVC + ∆ σ

σ3 = σhC

Volu

me

chan

ge o

f th

esa

mpl

e

Expa

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Time

Volume change of sample during consolidation

Consolidated- drained test (CD Test)

Devi

ator

str

ess,

∆σ d

Axial strain

Dense sand orOC clay

(∆ σd)f

Dense sand orOC clay

Loose sand orNC clay

Volu

me

chan

ge o

fth

e sa

mpl

e

Expa

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Axial strain

Stress-strain relationship during shearing

Consolidated- drained test (CD Test)

Loose sand orNC Clay(∆ σd)f

CD tests How to determine strength parameters c and φ

Devi

ator

str

ess,

∆σ d

Axial strain

Shea

r st

ress

,τ

σ or σ’

φMohr – Coulombfailure envelope

(∆ σd)fa

Confining stress = σ3a(∆ σd)fb

Confining stress = σ3b

(∆ σd)fc

Confining stress = σ3c

σ3c σ1cσ3a σ1a

(∆ σd)fa

σ3b σ1b

(∆ σd)fb

σ1 = σ3 + (∆ σd)f

σ3

CD tests

Strength parameters c and φ obtained from CD tests

Since u = 0 in CDtests, σ = σ’

Therefore, c = c’and φ = φ’

cd and φd are usedto denote them

CD tests Failure envelopes

Shea

r st

ress

,τ

σ or σ’

φdMohr – Coulombfailure envelope

σ3a σ1a

(∆ σd)fa

For sand and NC Clay, cd = 0

Therefore, one CD test would be sufficient to determine φd ofsand or NC clay

CD tests Failure envelopes

For OC Clay, cd ≠ 0

τ

σ or σ’

φ

σ3 σ1(∆ σd)f

cσc

OC NC

Some practical applications of CD analysis forclays

τ τ = in situ drainedshear strength

Soft clay

1. Embankment constructed very slowly, in layers over a soft clay deposit

Some practical applications of CD analysis forclays

2. Earth dam with steady state seepage

τ = drained shear strength ofclay core

τ

Core

Some practical applications of CD analysis forclays

3. Excavation or natural slope in clay

τ = In situ drained shear strength

τ

Note: CD test simulates the long term condition in thefield. Thus, cd and φd should be used to evaluate thelong term behavior of soils

Consolidated- Undrained test (CU Test)

Step 1: At the end of consolidationσVC

σhC

Total, σ = Neutral, u Effective, σ’+

0

Step 2: During axial stress increase

σ’VC = σVC

σ’hC = σhC

σVC + ∆ σ

σhC ±∆u

Drainage

Step 3: At failure

σVC + ∆ σf

σhC

No drainage

Nodrainage ±∆uf

σ’V = σVC + ∆ σ± ∆u = σ’1

σ’h = σhC ± ∆u = σ’3

σ’Vf = σVC + ∆ σf ± ∆uf = σ’1f

σ’hf = σhC ± ∆uf = σ’3f

Volu

me

chan

ge o

f th

esa

mpl

e

Expa

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mpr

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Time

Volume change of sample during consolidation

Consolidated- Undrained test (CU Test)

Devi

ator

str

ess,

∆σ d

Axial strain

Dense sand orOC clay

(∆ σd)f

Dense sand orOC clay

Loose sand/NC Clay

∆u+

-

Axial strain

Stress-strain relationship during shearing

Consolidated- Undrained test (CU Test)

Loose sand orNC Clay(∆ σd)f

CU tests How to determine strength parameters c and φ

Devi

ator

str

ess,

∆σ d

Axial strain

Shea

r st

ress

,τ

σ or σ’

(∆ σd)fb

Confining stress = σ3b

σ3b σ1bσ3a σ1a(∆ σd)fa

φcuMohr – Coulomb failureenvelope in terms oftotal stresses

ccu

σ1 = σ3 + (∆ σd)f

σ3

Total stresses at failure(∆ σd)fa

Confining stress = σ3a

(∆ σd)fa

CU tests How to determine strength parameters c and φ

Shea

r st

ress

,τ

σ or σ’σ3b σ1bσ3a σ1a

(∆ σd)fa

φcu

Mohr – Coulomb failureenvelope in terms oftotal stresses

ccu σ’3b σ’1b

σ’3a σ’1a

Mohr – Coulomb failureenvelope in terms ofeffective stresses

φ’

C’ ufa

ufb

σ’1 = σ3 + (∆ σd)f - uf

σ’3 = σ3 - uf

Effective stresses at failure

uf

CU tests

Strength parameters c and φ obtained from CU tests

Shear strengthparameters in terms oftotal stresses are ccu andφcu

Shear strengthparameters in terms ofeffective stresses are c’and φ’

c’ = cd and φ’ = φd

CU tests Failure envelopes

For sand and NC Clay, ccu and c’ = 0

Therefore, one CU test would be sufficient to determine φcuand φ’(= φd) of sand or NC clay

Shea

r st

ress

,τ

σ or σ’

φcuMohr – Coulomb failureenvelope in terms oftotal stresses

σ3a σ1a

(∆ σd)fa

σ3a σ1a

φ’

Mohr – Coulomb failureenvelope in terms ofeffective stresses

Some practical applications of CU analysis forclays

τ τ = in situ undrainedshear strength

Soft clay

1. Embankment constructed rapidly over a soft clay deposit

Some practical applications of CU analysis forclays

2. Rapid drawdown behind an earth dam

τ = Undrained shear strengthof clay core

Core

τ

Some practical applications of CU analysis forclays

3. Rapid construction of an embankment on a natural slope

Note: Total stress parameters from CU test (ccu and φcu) can be used forstability problems where,Soil have become fully consolidated and are at equilibrium with theexisting stress state; Then for some reason additional stresses areapplied quickly with no drainage occurring

τ = In situ undrained shear strength

τ

Unconsolidated- Undrained test (UU Test)Data analysis

σC = σ3

σC = σ3

No drainage

Initial specimen condition

σ3 + ∆ σd

σ3

Nodrainage

Specimen conditionduring shearing

Initial 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 × H0

A ×(H0 – ∆H) = A0 × H0

A ×(1 – ∆H/H0) = A0 z

AA

10

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

0

0

= +

Step 2: After application of hydrostatic cell pressure

∆uc = B ∆ σ3

σC = σ3

σC = σ3 ∆uc

σ’3 = σ3 - ∆uc

σ’3 = σ3 - ∆uc

No drainage

Increase of pwp due toincrease of cell pressure

Increase of cell pressure

Skempton’s pore waterpressure parameter, B

Note: If soil is fully saturated, then B = 1 (hence, ∆uc = ∆ σ3)

Unconsolidated- Undrained test (UU Test)

Step 3: During application of axial load

σ3 + ∆ σd

σ3

Nodrainage

σ’1 = σ3 + ∆ σd - ∆uc ∆ud

σ’3 = σ3 - ∆uc ∆ud

∆ud = AB∆ σd

∆uc ± ∆ud

= +

Increase of pwp due toincrease of deviator stress

Increase of deviator stress

Skempton’s pore waterpressure parameter, A

Unconsolidated- Undrained test (UU Test)

Combining steps 2 and 3,

∆uc = B ∆ σ3 ∆ud = AB∆ σd

∆u = ∆uc + ∆ud

Total pore water pressure increment at any stage, ∆u

∆u = B [∆ σ3 + A∆ σd]Skempton’s porewater pressureequation

∆u = B [∆ σ3 + A(∆ σ1 – ∆ σ3]

Unconsolidated- Undrained test (UU Test)

Step 1: Immediately after sampling0

0

Total, σ = Neutral, u Effective, σ’+

-ur

Step 2: After application of hydrostatic cell pressure

σ’V0 = ur

σ’h0 = ur

σC

σC-ur + ∆uc = -ur + σc(Sr = 100% ; B = 1)

Step 3: During application of axial loadσC + ∆ σ

σC

No drainage

Nodrainage

-ur + σc ± ∆u

σ’VC = σC + ur - σC = ur

σ’h = ur

Step 3: At failure

σ’V = σC + ∆ σ+ ur - σc ∆u

σ’h = σC + ur - σc ∆u

σ’hf = σC + ur - σc ∆uf =σ’3f

σ’Vf = σC + ∆ σf + ur - σc ∆uf = σ’1f

-ur + σc ± ∆ufσC

σC + ∆ σfNodrainage

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

Step 3: At failure

σ’hf = σC + ur - σc ∆uf =σ’3f

σ’Vf = σC + ∆ σf + ur - σc ∆uf = σ’1f

-ur + σc ± ∆ufσC

σC + ∆ σfNodrainage

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

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

τ

σ’σ’3 σ’1∆ σf

σ3b σ1bσ3a σ1a∆ σfσ’3 σ’1

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

Step 3: At failure

σ’hf = σC + ur - σc ∆uf =σ’3f

σ’Vf = σC + ∆ σf + ur - σc ∆uf = σ’1f

-ur + σc ± ∆ufσC

σC + ∆ σfNodrainage

τ

σ or σ’

Mohr circles in terms of total stresses

uaub

Failure envelope, φu = 0

cu

σ3b σ1b

Unconsolidated- Undrained test (UU Test)

Effect of degree of saturation on failure envelope

σ3a σ1aσ3c σ1c

τ

σ or σ’

S < 100% S = 100%

Some practical applications of UU analysis forclays

τ τ = in situ undrainedshear strength

Soft clay

1. Embankment constructed rapidly over a soft clay deposit

Some practical applications of UU analysis forclays

2. Large earth dam constructed rapidly with nochange in water content of soft clay

Core

τ = Undrained shear strengthof clay core

τ

Some practical applications of UU analysis forclays3. Footing placed rapidly on clay deposit

τ = In situ undrained shear strength

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

Unconfined Compression Test (UC Test)

σ1 = σVC + ∆ σ

σ3 = 0

Confining pressure is zero in the UC test

Unconfined Compression Test (UC Test)

σ1 = σVC + ∆ σf

σ3 = 0

Shea

r st

ress

,τ

Normal stress, σ

qu

τf = σ1/2 = qu/2 = cu

Various correlations for shear strength

For NC clays, the undrained shear strength (cu) increases with the effectiveoverburden pressure, σ’0

)(0037.011.0'0

PIcu

Skempton (1957)

Plasticity Index as a %

For OC clays, the following relationship is approximately true

8.0'0

'0

)(OCRcc

edConsolidatNormally

u

idatedOverconsol

u

Ladd (1977)

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

)log(234.0814.0' IPSin Kenny (1959)

Shear strength of partially saturated soils

In the previous sections, we were discussing the shear strength of saturatedsoils. However, in most of the cases, we will encounter unsaturated soils

Solid

Water

Saturated soils

Pore waterpressure, u

Effectivestress, σ’ Solid

Unsaturated soils

Pore waterpressure, uw

Effectivestress, σ’

Water

Air Pore airpressure, ua

Pore water pressure can be negative in unsaturated soils

Shear strength of partially saturated soils

Bishop (1959) proposed shear strength equation for unsaturated soils as follows

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

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

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

bwaanf uuuc tan)('tan)('

Where,tanφb = 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 dueto matric suction

Therefore, strength of unsaturated soils is much higher than the strength ofsaturated soils due to matric suction

τ

σ - ua

φ’

τ

σ - ua

How it become possible builda sand castle

bwaanf uuuc tan)('tan)('

Same as saturated soils Apparent cohesion dueto matric suction

φ’

Apparentcohesion