Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
57
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Module 4: Lecture 8 on Stress-strain relationship
and Shear strength of soils
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
The triaxial test: Components
Loading ram
Perspex cell
Latex sheet
Soil sample
To pore pressure measuring device
Porous discs
Pressure supply to
cell
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Deviator stress (q or ∆σd) = σ1 – σ3
Consolidated- drained test (CD Test)
σ1 = σVC + ∆σ
σ3 = σhC
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Devi
ator
stre
ss, ∆
σ d
Axial strain
Dense sand or OC clay
(∆σd)f
Dense sand or OC clay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
CD tests : Failure envelopes
For OC Clay, cd ≠ 0
τ
σ or σ′
φ
σ3 σ1(∆σd)f
cσc
OC NC
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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)
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Devi
ator
stre
ss, ∆
σ d
Axial strain
Dense sand or OC clay
(∆σd)f
Dense sand or OC clay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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)]
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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).
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Mohr failure envelopes for UU tests
For 100% saturated clay
For partially saturated clay
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Stress conditions for the UC test
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
σ1 (kPa)
ε1 (-)
Typical variation of σ1 with ε1 (UCS Test)
Cu = 136/2 = 68 kPa
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
-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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Status of silty sand sample after CU test
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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°
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
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
)
Effective stress = 50 kPaEffective stress = 100 kPa
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
0
200
400
600
800
0 200 400 600 800
p (kPa)
q, q
' (kP
a)
Effective stress = 50 kPaEffective stress = 100 kPa
0
200
400
600
800
0 200 400 600 800
p' (kPa)
q , q
' (kP
a)
Effective stress = 50 kPaEffective stress = 100 kPa
Variation of TSP at various effective stresses
Variation of ESP at various effective stresses
At failure
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Status of sample after termination of CU test
π/4 + 36°/2
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