57 · 2017. 8. 4. · compression, definition of failure, interlocking concept...

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

57


Module 4: Lecture 8 on Stress-strain relationship

and Shear strength of soils


Stress state, Mohr’s circle analysis and Pole, Principalstress space, Stress paths in p-q space;

Mohr-Coulomb failure criteria and its limitations,correlation with p-q space;

Stress-strain behavior; Isotropic compression andpressure dependency, confined compression, large stresscompression, Definition of failure, Interlocking conceptand its interpretations,

Triaxial behaviour, stress state and analysis of UC, UU, CU,CD, and other special tests, Drainage conditions; Stresspaths in triaxial and octahedral plane; Elastic modulusfrom triaxial tests.

Contents


The triaxial test: Introduction• Most widely used shear strength test and is suitable for all types of soil.• A cylindrical specimen, generally “L/D = 2” is used for the test, and stresses

are applied under conditions of axial symmetry.• Typical specimen diameters are 38mm and 100mm

Stress system in triaxial test

Equal all round

pressure

Axial stress


The triaxial test: Components

Loading ram

Perspex cell

Latex sheet

Soil sample

To pore pressure measuring device

Porous discs

Pressure supply to

cell


The triaxial test: Mechanism• Intermediate principal stress σ2 must be equal to major σ1 or

minor σ3 stress, so as to facilitate representation of stressstate in two dimensional Mohr’s circle.

• A cylindrical specimen is placed inside Perspex cell filledwith water.

• The specimen is covered with latex sheet so as to avoiddirect contact with water.

• The specimen is loaded initially by surrounding waterpressure so as to achieve isotropic loading conditions.

• A deviatoric stress is then applied gradually on the samplewith the help of Ram axially.

• A duct at the bottom of the sample allows water to passthrough the sample which is further monitored , orconversely, in some cases, no drainage is allowed.


The triaxial test: Mechanism

• Fine grained soil can stand the mould without anysupport

• But the coarse grained soils samples have to kept insome supporting mould until the application of negativepore pressure to the sample through drainage duct.

So,u = ue (negative)

σa = σr = 0

σ′a= σ′r = -ue

where, σa is the axial stress, σr is the radial stress


The triaxial test: Mechanism• If cell pressure increased to scp, this isotropic pressure is

taken entirely by the pore water. Thus pore pressureincreases, but no change occurs in effective stresses.

So,ui = σcp + ue (negative)

σa = σr = σcp

σ′a = σ′r = -uethus,

u – ue = ∆σcp

i.e. ∆u = ∆σcp

σa

σr


Drainage conditions : Combinations in triaxial test

drainage valve condition

Open Closed

Consolidated sample

Unconsolidated sample

drainage valve condition

Drained loading

Undrained loading

Under all-around cell pressure σc

Step 1

Shearing (loading)

Step 2

Open Closed

CD UU

CU


Drainage conditions : Combinations in triaxial test

Unconsolidated Undrainedtest (UU)

The drainage valve is initially opened toallow the pore pressure ui to dissipate tozero, and is kept open while thespecimen is taken to failure at asufficiently slow rate.

Consolidated Undrained test (CU)

Drainage valve initially opened to allowpore pressure ui to dissipate to zero, andthen closed so that specimen is taken tofailure without any further drainageApplying back pressure:

decreases cavitation, and reduction of voids.

Consolidated Drained test (CD)

Specimen is taken to failure with nodrainage permitted

Unconfined Compressive test (UC)

Specimen is taken to failure with noconfinement


Schematic of a Triaxial cell

Axial total stress σ1 = σ3+ P/A

Deviator stress σ1 - σ3= σd = P/A

Axial strain ε1 = ∆z/HoRadial strain εr = ∆r/ro

Volumetric strain εv = ε1 + 2 ε3

Deviatoric strain εd = 2/3( ε1 - ε3)

Axisymmetric condition, σ′2 = σ′3 or σ2 = σ3; ε2 = ε3p′ = (σ′1+ 2σ′3)/3 and p = (σ1+ 2σ3)/3 p′ = p- u

q = σ1- σ3;q′ = σ′1- σ′3= (σ1 - ∆u) –

(σ3 - ∆u) = σ1- σ3

Thus, q′ = q;Shear is unaffected by PWP.

Stresses and strains on a sample in the Triaxial compression test


Consolidated- drained test (CD Test)

Step 1: At the end of consolidation

σh

σ = u σ′+

0

Step 2: During axial stress increase

σ′V = σV

σ′hC = σhC

σV + ∆σ

σhC 0

σ′V = σV + ∆σ = σ′1

σ′h = σh = σ′3

Drainage

Drainage

Step 3: At failureσVC + ∆σf

σhC 0

σ′Vf = σV + ∆σf = σ′1f

σ′hf = σh = 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 the

sam

ple

Expa

nsio

nCo

mpr

essio

n

Time

Consolidated- drained test (CD Test) : Volume change of sample during consolidation


Devi

ator

stre

ss, ∆

σ d

Axial strain

Dense sand or OC clay

(∆σd)f


Loose sand /NC Clay

+-

Axial strain

CD Test :- Stress-strain relationship during shearing

Loose sand or NC Clay(∆σd)f

Volu

me

chan

ge o

f th

e sa

mpl

e


CD tests : How to determine strength parameters c and φDe

viat

or st

ress

,∆σ d

Axial strain

Shea

r str

ess,

τ

σ or σ’

φMohr – Coulomb failure envelope

(∆σd)faConfining stress = σ3a(∆σd)fb

Confining stress = σ3b

(∆σd)fc

Confining stress = σ3c

σ3c σ1cσ3a σ1a(∆σd)faσ3b σ1b

(∆σd)fb

σ1 = σ3 + (∆σd)f

σ3


CD testsStrength parameters c and φ obtained from CD tests

Since u = 0 in CD tests, σ = σ′

Therefore, c = c′ and φ = φ′

Parameters are denoted as cd and φd


CD tests : Failure envelopes

Shea

r str

ess,

τ

σ or σ’

φdMohr – Coulomb failure envelope

σ3a σ1a(∆σd)fa

For sand and NC Clay, cd = 0

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


CD tests : Failure envelopes

For OC Clay, cd ≠ 0

τ

σ or σ′

φ

σ3 σ1(∆σd)f

cσc

OC NC


Stress paths during CD Test

q = σ1-σ3

p, p′

ESP = TSP

31

p = (σ1+ 2σ3)/3

∆p = ∆σ1/3; ∆q = ∆σ1 ∆q/∆p = 3

∆σ1 = ∆σ′1 = ∆σ3 = ∆σ′3

∆σ3 = ∆σ′1∆u = 0

σ′1 = σ′3 + P/A σ3 = σ′3∆σ3 = 0 ∆u = 0

Consolidation phase

Shearing phase

Stage1: Isotropic consolidation phase

∆σ1 = ∆σ′1 = ∆σ3 = ∆σ′3; ∆σ1 > 0; ∆u = 0 (end of consolidation)∆p′ = ∆p = (∆σ1 + 2∆σ1)/3 = ∆σ1;∆q = ∆σ1 -∆σ3 = 0 ∆q/∆p′ =∆q/∆p = 0


Stress paths during CD TestStage 2: Shearing phase

∆σ1 = ∆σ′1 > 0 ;

∆σ3 = ∆σ′3 = 0 ; ∆u = 0;

∆p′ = ∆p = (∆σ1)/3 = ∆σ1 /3 ;

∆q = ∆σ1 -∆σ3 = 0; = ∆σ1 ; ∆q/∆p′ =∆q/∆p = 3


Consolidated- Undrained test (CU Test)

Step 1: At the end of consolidation

Total, σ = Neutral, u Effective, σ’+

0

Step 2: During axial stress increase

σ′VC = σVC

σ’hC = σhC

±∆u

Step 3: At failure

±∆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

σVC

σhC

Drainage

σVC + ∆σ

σhCNo drainage X

σVC + ∆σf

σhCNo drainage X


Volu

me

chan

ge o

f the

sa

mpl

e

Expa

nsio

nCo

mpr

essi

on

Time

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


Devi

ator

stre

ss, ∆

σ d

Axial strain


(∆σd)f


Loose sand /NC Clay

∆u

+-

Axial strain

CU Test :- Stress-strain relationship during shearing

Loose sand or NC Clay(∆σd)f

Pore water pressure varies with axial strain


CU tests :- How to determine strength parameters c and φD

evia

tor s

tress

,∆σ d

Axial strain

Shea

r stre

ss,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


CU tests: Strength parameters c and φ

Shea

r stre

ss,τ

σ or σ’σ3b σ1bσ3a σ1a

(∆σd)fa

φcu

Mohr – Coulomb failureenvelope in terms of totalstresses

ccu σ′3b σ’1bσ’3a σ’1a

Mohr – Coulomb failureenvelope in terms ofeffective stresses

φ′

c′ ufaufb

σ’1 = σ3 + (∆σd)f - uf

σ’3 = σ3 - uf

Effective stresses at failureuf


CU tests

Strength parameters cd and ϕd obtained from CD tests

Shear strength parameters in terms of total stresses are ccu and φcu

Shear strength parameters in terms of effective stresses are c′ and φ′

c′ = cd and φ′ = φd


Stress paths during CU Test

q = σ1-σ3

p, p′

TSP

31

p = (σ1+ 2σ3)/3

∆p = ∆σ1/3; ∆q = ∆σ1 ∆q/∆p = 3

∆σ1 = ∆σ′1 = ∆σ3 = ∆σ′3

∆σ3 = ∆σ′1∆u = 0

∆σ1 = ∆σ3 + P/A

∆σ3 = 0 ∆u ≠ 0

Con. phase

Shearing phase

Stage1: Isotropic consolidation phase

∆σ1 = ∆σ′1 = ∆σ3 = ∆σ′3; ∆σ1 > 0; ∆u = 0 (end of consolidation)∆p′ = ∆p = (∆σ1 + 2∆σ1)/3 = ∆σ1;∆q = ∆σ1 -∆σ3 = 0 ∆q/∆p′ =∆q/∆p = 0

∆σ′1 = ∆σ1 - ∆u

∆σ′3 = ∆σ3 - ∆u = -∆u

ESP ∆u


Stress paths during CU TestStage 2: Shearing phase

∆σ1 > 0; ∆σ3 = 0; ∆σ′1 = ∆σ1-∆u = 0; ∆σ′3 = -∆u

∆p = (∆σ1)/3 ; ∆q = ∆σ1, ∆q/∆p = 3 [For TSP]

∆p′ = ∆p - ∆u = (∆σ1)/3 - ∆u; [For ESP]

∆q = ∆σ1; ∆q/∆p′ =∆σ1/(∆σ1/3-∆u) = 3/[1-3(∆u/∆σ1)]


Stress conditions for the UU test The purpose of UU

test is todetermine the un-drained shearstrength of asaturated soil.

Quick test(Neither duringconsolidation andshearing stages,excess PWP isallowed to drain).


Mohr failure envelopes for UU tests

For 100% saturated clay

For partially saturated clay


Total stress path during UU Test

q = σ1-σ3

p

TSP

3

1

p = (σ1+ 2σ3)/3

∆p = ∆σ1/3; ∆q/∆p = 3

σ1 = σ3 + P/A

σ3∆u ≠ 0

∆σ1 = ∆σ3 ; ∆u ≠ 0∆p = ∆σ1, ∆q= 0; ∆q/ ∆p = 0

∆σ1 > 0 ; ∆σ3 = 0

∆p = ∆σ1/3 ; ∆q = ∆σ1

∆q/∆p = 3

Initial stage

Shearing phase


Unconfined compressive (UC) test

(After www.geocomp.com)

To determine the un-drained shearstrength of saturated clays quickly.

No radial stress (σ3 = 0)

Deviator load is increased rapidlyuntil the soil sample fails; Porewater can not drain from the soil;the soil sample is sheared atconstant volume.

σ1σ3 = 0


Stress conditions for the UC test


Total stress path during UC Test

q = σ1-σ3

p, p′

TSP

3

1

p = (σ1+ 2σ3)/3

∆p = ∆σ1/3; ∆q/∆p = 3

The effective stresspath is unknownsince PWP changesare not normallymeasured.

If ∆u is measured, itwould be negative.

Since σ3 = 0,σ′3 = σ3 - ∆u = - ∆u

∆u must be –vebecause as σ′3 cannot be –ve (soilscan not sustaintension). So σ′3 mustbe +ve.


Mohr Circles for UCSThe results of from UC tests can lead to:

Estimate the short-termbearing capacity offine-grained soils forfoundations.

Estimate the short-termstability of slopes.

Determine the stress-strain characteristicsunder fast (un-drainedloading conditions.

σ, σ′

τEffective stress Circle not determined UC test

Total stress Mohr Circle

45°

∆u

cu

Failure envelopeFailure plane


σ1 (kPa)

ε1 (-)

Typical variation of σ1 with ε1 (UCS Test)

Cu = 136/2 = 68 kPa


Consolidated undrained triaxial tests on Silty sand

Property Unit Silty sandSpecific gravity (Gs) -a 2.64

Particle size distribution

Sand (S) % 80

Silt (M) % 10

Clay (C) % 10

Classification (Unified soil classification system) -a SM

Compaction characteristics (standard Proctor)

Maximum dry unit weight (MDD) kN/m3 19.75

Optimum moisture content (OMC) % 10.5

Co-efficient of permeability (k) m/sec 4.0 x 10-7


Variation in deviator stress with axial strain

0

100

200

300

400

500

600

700

800

900

0 5 10 15 20 25Axial strain (%)

Dev

iato

r stre

ss (k

Pa)

σ' = 50 kPaσ' = 100 kPaσ' = 150 kPa


-50

0

50

100

150

200

0 5 10 15 20 25Axial strain (%)

Exce

ss p

ore

wat

er p

ress

ure

(kPa

)

σ' = 50 kPaσ' = 100 kPaσ' = 150 kPa

Variation in excess pore water pressure with axial strain


Status of silty sand sample after CU test


Variation in stress path at various effective stress

p ′ = (σ1′ + 2σ3 ′)/3p = (σ1 + 2σ3)/3

q = q′ = (σ1 - σ3)

Cambridge stress path is plotted between p or p′ and q. Where,

0

200

400

600

800

1000

0 200 400 600 800 1000

p, p' (kPa)

q (k

Pa)

TSP, σ' = 50 kPaTSP, σ' = 100 kPaTSP, σ' = 150 kPaESP, σ' = 50 kPaESP, σ' = 100 kPaESP, σ' = 150 kPaFailure envelope


Mohr circles for consolidated un-drained tests on silty sand

0

200

400

600

800

1000

1200

0 200 400 600 800 1000 1200

Normal stress (kPa)

Shea

r stre

ss (k

Pa)

Effective parameter (σ' = 50 kPa)Effective parameter (σ' = 150 kPa)Effective parameter (σ' = 100 kPa)Total parameter (σ' = 50 kPa)Total parameter (σ' = 100 kPa)Total parameter (σ' = 150 kPa)

Failure envelopes

c' = 2 kPaφ ' =35°

c = 7 kPaφ =32°


0

200

400

600

800

0 5 10 15 20 25

Axis Strain (%)

Dev

iato

r stre

ss (k

Pa)

Effective stress = 50 kPaEffective stress = 100 kPa

Variation in deviator stress with axial strain

Results of CU triaxial tests on Fine Sand

Max void ratio = 0.778 Min void ratio = 0.542

Void ratio after consolidation stage

i) e(σ′ = 50 kPa) = 0.723ii) e(σ′ = 100 kPa) = 0.741


Variation in excess pore water pressure with axial strain

-100

-50

0

50

100

150

200

250

0 5 10 15 20 25Axis strain (%)

Exc

ess

pore

wat

er p

ress

ure

(kPa

)



0

200

400

600

800

0 200 400 600 800

p (kPa)

q, q

' (kP

a)


0

200

400

600

800

0 200 400 600 800

p' (kPa)

q , q

' (kP

a)


Variation of TSP at various effective stresses

Variation of ESP at various effective stresses

At failure


Status of sample after termination of CU test

π/4 + 36°/2

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57 · 2017. 8. 4. · compression, definition of failure, interlocking concept...

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