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Triaxial Shear Test
CEP 701 - Soil Engineering Laboratory
Strength of different
materials
Steel
Tensile
strength
Concrete
Compressive
strength
Soil
Shear
strength
Presence of pore water
Complex
behavior
Embankment
Strip footing
Shear failure of soils
Soils generally fail in shear
At failure, shear stress along the failure surface
(mobilized shear resistance) reaches the shear strength.
Failure surface
Mobilized shear
resistance
Retaining
wall
Shear failure of soils
Soils generally fail in shear
Retaining
wall
Shear failure of soils
At failure, shear stress along the failure surface
(mobilized shear resistance) reaches the shear strength.
Failure
surface
Mobilized
shear
resistance
Soils generally fail in shear
Shear failure
mechanism
The soil grains slide
over each other along
the failure surface.
No crushing of
individual grains.
failure surface
Shear failure mechanism
At failure, shear stress along the failure surface (τ)
reaches the shear strength (τf).
σ
τ
τ
σ
τ
τ
Mohr-Coulomb Failure Criterion
(in terms of total stresses)
τ
τf is the maximum shear stress the soil can take without
failure, under normal stress of σ.
σ
φστ tan+= cf
c
φ
failure envelope
Cohesio
n
Friction
angleτf
σ
Mohr-Coulomb Failure Criterion
(in terms of effective stresses)
τf is the maximum shear stress the soil can take without
failure, under normal effective stress of σ’.
τ
σ’
'tan'' φστ += cf
c’
φ’
failure envelope
Effective
cohesion Effective
friction angleτf
σ’
u−= σσ '
u = pore water
pressure
Mohr-Coulomb Failure Criterion
'tan'' φστ ff c +=
Shear strength consists of two
components: cohesive and frictional.
σ’f
τf
φ’
τ
σ'
c’ c’ cohesive component
σ’f tan φ’ frictional
component
c and φ are measures of shear strength.
Higher the values, higher the shear strength.
Mohr Circle of stress
Soil element
σ’
1
σ’
1
σ’
3
σ’
3
θ
σ’
τ
θ
σσσσ
σ
θ
σσ
τ
2
22
2
2
'
3
'
1
'
3
'
1'
'
3
'
1
Cos
Sin
−
+
+
=
−
=
Resolving forces in σ and τ directions,
2'
3
'
1
2'
3
'
1'2
22 






 −
=




 +
−+
σσσσ
στ
Mohr Circle of stress
2'
3
'
1
2'
3
'
1'2
22 






 −
=




 +
−+
σσσσ
στ
Soil element
σ’1
σ’1
σ’3
σ’3
θ
σ’
τ
Soil elementSoil element
σ’1
σ’1
σ’3
σ’3
σ’1
σ’1
σ’3
σ’3
θ
σ’
τ
θθ
σ’
τ
τ
σ’
2
'
3
'
1 σσ +
2
'
3
'
1 σσ −
'
3σ '
1σ
Mohr Circle of stress
2'
3
'
1
2'
3
'
1'2
22 






 −
=




 +
−+
σσσσ
στ
Soil element
σ’1
σ’1
σ’3
σ’3
θ
σ’
τ
Soil elementSoil element
σ’1
σ’1
σ’3
σ’3
σ’1
σ’1
σ’3
σ’3
θ
σ’
τ
θθ
σ’
τ
τ
σ’
2
'
3
'
1 σσ +
2
'
3
'
1 σσ −
'
3σ '
1σ
PD = Pole w.r.t. plane
θ
(σ’, τ)
Soil elements at different locations
Failure surface
Mohr Circles & Failure Envelope
X X
X ~ failure
Y
Y
Y ~ stable
τ
σ’
'tan'' φστ += cf
Mohr Circles & Failure Envelope
Y
σc
σc
σc
Initially, Mohr circle is a point
∆σ
σc+∆σ
∆σ
The soil element does not fail if
the Mohr circle is contained
within the envelope
GL
Mohr Circles & Failure Envelope
Y
σc
σc
σc
GL
As loading progresses, Mohr
circle becomes larger…
.. and finally failure occurs
when Mohr circle touches the
envelope
∆σ
σ’
2
'
3
'
1 σσ +
'
3σ '
1σ
PD = Pole w.r.t. plane
θ
(σ’, τf)
Orientation of Failure Plane
φ’
σ’1
σ’1
σ’3
σ’3
θ
σ’
τ
σ’1
σ’1
σ’3
σ’3
θθ
σ’
τ
Failure envelope
(90 –
θ)
Therefore,
90 – θ + φ’ = θ
θ = 45 + φ’/2
Mohr circles in terms of total & effective stresses
= X
σv’
σh’
X
u
u
+
σv’σh’
effective stresses
u
σvσh
X
σv
σh
total stresses
τ
σ or σ’
Failure envelopes in terms of total & effective
stresses
= X
σv’
σh’
X
u
u
+
σv’σh’
effective stresses
u
σvσh
X
σv
σh
total stresses
τ
σ or σ’
If X is on
failure
c
φ
Failure envelope in
terms of total stresses
φ’
c’
Failure envelope in terms
of effective stresses
Mohr Coulomb failure criterion with Mohr circle
of stress
X
σ’v = σ’1
σ’h = σ’3
X is on failure σ’1
σ’3
effective stresses
τ
σ
’
φ’ c’
Failure envelope in terms
of effective stresses
c’ Cotφ’ (σ’1+ σ’3)/
2
(σ’1 − σ’3)/
2





 −
=










 +
+
2
'
2
''
'
3
'
1
'
3
'
1 σσ
φ
σσ
φ SinCotc
Therefore,
Mohr Coulomb failure criterion with Mohr circle
of stress





 −
=










 +
+
2
'
2
''
'
3
'
1
'
3
'
1 σσ
φ
σσ
φ SinCotc
( ) ( ) ''2''
3
'
1
'
3
'
1 φφσσσσ CoscSin ++=−
( ) ( ) ''2'1'1 '
3
'
1 φφσφσ CoscSinSin ++=−
( )
( ) ( )'1
'
'2
'1
'1'
3
'
1
φ
φ
φ
φ
σσ
Sin
Cos
c
Sin
Sin
−
+
−
+
=






++





+=
2
'
45'2
2
'
452'
3
'
1
φφ
σσ TancTan
Other laboratory tests include,
Direct simple shear test, torsional
ring shear test, plane strain triaxial
test, laboratory vane shear test,
laboratory fall cone test
Determination of shear strength parameters of
soils (c, φ or c’, φ’)
Laboratory tests on
specimens taken from
representative undisturbed
samples
Field tests
Most common laboratory tests
to determine the shear strength
parameters are,
1.Direct shear test
2.Triaxial shear test
1. Vane shear test
2. Torvane
3. Pocket penetrometer
4. Fall cone
5. Pressuremeter
6. Static cone penetrometer
7. Standard penetration test
Laboratory tests
Field conditions
z
σvc
σvc
σhcσhc
Before construction
A representative
soil sample
z
σvc + ∆σ
σhcσhc
After and during
construction
σvc + ∆σ
Laboratory tests
Simulating field conditions
in the laboratory
Step 1
Set the specimen in
the apparatus and
apply the initial
stress condition
σvc
σvc
σhc
σhc
Representative
soil sample
taken from the
site
0
00
0
Step 2
Apply the
corresponding field
stress conditions
σvc + ∆σ
σhc
σhc
σvc + ∆σTraxial test
σvc
σvc
τ
τ
Direct shear test
Triaxial Shear Test
Soil sample
at failure
Failure plane
Porous
stone
impervious
membrane
Piston (to apply deviatoric stress)
O-ring
pedestal
Perspex
cell
Cell pressure
Back pressure Pore pressure or
volume change
Water
Soil
sample
Triaxial Shear Test
Specimen preparation (undisturbed sample)
Sampling tubes
Sample extruder
Triaxial Shear Test
Specimen preparation (undisturbed sample)
Edges of the sample
are carefully trimmed
Setting up the sample
in the triaxial cell
Triaxial Shear Test
Sample is covered
with a rubber
membrane and sealed
Cell is completely
filled with water
Specimen preparation (undisturbed sample)
Triaxial Shear Test
Specimen preparation (undisturbed sample)
Proving ring to
measure the
deviator load
Dial gauge to
measure vertical
displacement
Types of Triaxial Tests
Is the drainage valve open?
yes no
Consolidated
sample
Unconsolidated
sample
Is the drainage valve open?
yes no
Drained
loading
Undrained
loading
Under all-around cell pressure σc
σc
σc
σc
σcStep 1
deviatoric stress
(∆σ = q)
Shearing (loading)
Step 2
σc σc
σc+ q
Types of Triaxial Tests
Is the drainage valve open?
yes no
Consolidated
sample
Unconsolidated
sample
Under all-around cell pressure σc
Step 1
Is the drainage valve open?
yes no
Drained
loading
Undrained
loading
Shearing (loading)
Step 2
CD test
CU test
UU test
Consolidated- drained test (CD Test)
Step 1: At the end of consolidation
σVC
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’3f
Drainage
Deviator stress (q or ∆σd) = σ1 – σ3
Consolidated- drained test (CD Test)
σ1 = σVC + ∆σ
σ3 = σhC
Volumechangeofthe
sample
ExpansionCompression
Time
Volume change of sample during consolidation
Consolidated- drained test (CD Test)
Deviator
stress,∆σd
Axial strain
Dense sand
or OC clay
(∆σd)f
Dense sand
or OC clay
Loose sand
or NC clay
Volumechange
ofthesample
ExpansionCompression
Axial strain
Stress-strain relationship during shearing
Consolidated- drained test (CD Test)
Loose sand
or NC Clay(∆σd)f
CD tests How to determine strength parameters c and φ
Deviator
stress,∆σd
Axial strain
Shear
stress,τ
σ or
σ’
φ
Mohr – Coulomb
failure envelope
(∆σd)f
a
Confining stress = σ3a(∆σd)f
b
Confining stress = σ3b
(∆σd)f
c
Confining stress = σ3c
σ3c σ1c
σ3a σ1a
(∆σd)f
σ3b σ1b
(∆σd)fb
σ1 = σ3 +
(∆σd)f
σ3
CD tests
Strength parameters c and φ obtained from CD tests
Since u = 0 in CD
tests, σ = σ’
Therefore, c = c’
and φ = φ’
cd and φd are used
to denote them
CD tests Failure envelopes
Shear
stress,τ
σ or
σ’
φd
Mohr – Coulomb
failure envelope
σ3a σ1a
(∆σd)f
a
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
c
σc
OC NC
Some practical applications of CD analysis for
clays
τ τ = in situ drained
shear strength
Soft clay
1. Embankment constructed very slowly, in layers over a soft clay
deposit
Some practical applications of CD analysis for
clays
2. Earth dam with steady state seepage
τ = drained shear
strength of clay core
τ
Core
Some practical applications of CD analysis for
clays
3. Excavation or natural slope in clay
τ = In situ drained shear strength
τ
Note: CD test simulates the long term condition in the field.
Thus, cd and φd should be used to evaluate the long
term behavior of soils
Consolidated- Undrained test (CU Test)
Step 1: At the end of consolidation
σVC
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
No
drainage ±∆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

Volumechangeofthe
sample
ExpansionCompression
Time
Volume change of sample during consolidation
Consolidated- Undrained test (CU Test)
Deviator
stress,∆σd
Axial strain
Dense sand
or OC clay
(∆σd)f
Dense sand
or OC clay
Loose
sand /NC
Clay
∆u
+-
Axial strain
Stress-strain relationship during shearing
Consolidated- Undrained test (CU Test)
Loose sand
or NC Clay(∆σd)f
CU tests How to determine strength parameters c and φ
Deviator
stress,∆σd
Axial strain
Shear
stress,τ
σ or
σ’
(∆σd)f
b Confining stress = σ3b
σ3b σ1bσ3a σ1a
(∆σd)fa
φcuMohr – Coulomb
failure envelope in
terms of total stresses
ccu
σ1 = σ3 +
(∆σd)f
σ3
Total stresses at failure
(∆σd)f
a
Confining stress = σ3a
(∆σd)fa
CU tests How to determine strength parameters c and φ
Shear
stress,τ
σ or
σ’
σ3b σ1bσ3a σ1a
(∆σd)fa
φcu
Mohr – Coulomb
failure envelope in
terms of total stresses
ccu
σ’3b σ’1b
σ’3a σ’1a
Mohr – Coulomb failure
envelope in terms of
effective 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 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
CU tests Failure envelopes
For sand and NC Clay, ccu and c’ = 0
Therefore, one CU test would be sufficient to determine
φcu and φ (’ = φd) of sand or NC clay
Shear
stress,τ
σ or
σ’
φcu
Mohr – Coulomb
failure envelope in
terms of total stresses
σ3a σ1a
(∆σd)f
a
σ3a σ1a
φ’
Mohr – Coulomb failure
envelope in terms of
effective stresses
Some practical applications of CU analysis for
clays
τ τ = in situ
undrained shear
strength
Soft clay
1. Embankment constructed rapidly over a soft clay deposit
Some practical applications of CU analysis for
clays
2. Rapid drawdown behind an earth dam
τ = Undrained shear
strength of clay core
Core
τ
Some practical applications of CU analysis for
clays
3. Rapid construction of an embankment on a natural slope
Note: Total stress parameters from CU test (ccu and φcu) can be used for
stability problems where,
Soil have become fully consolidated and are at equilibrium with
the existing stress state; Then for some reason additional
stresses are applied 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
No
drainage
Specimen condition
during 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
A
A
ε−
=
1
0
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 to
increase of cell pressure
Increase of cell pressure
Skempton’s pore water
pressure 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
No
drainage
σ’1 = σ3 + ∆σd - ∆uc ∆ud
σ’3 = σ3 - ∆uc ∆ud
∆ud = A∆σd
∆uc ± ∆ud
= +
Increase of pwp due to
increase of deviator stress
Increase of deviator
stress
Skempton’s pore water
pressure parameter, A
Unconsolidated- Undrained test (UU Test)
Combining steps 2 and 3,
∆uc = B ∆σ3 ∆ud = A∆σd
∆u = ∆uc + ∆ud
Total pore water pressure increment at any stage, ∆u
∆u = B ∆σ3 + A∆σd
Skempton’s pore
water pressure
equation
∆u = B ∆σ3 + A(∆σ1 – ∆σ3)
Unconsolidated- Undrained test (UU Test)
Step 1: Immediately after sampling
0
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 load
σC + ∆σ
σC
No
drainage
No
drainage
-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 ± ∆uf
σC
σC + ∆σf
No
drainage
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 ± ∆uf
σC
σC + ∆σf
No
drainage
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 for
different cell pressures
τ
σ’
σ’3 σ’1∆σ
σ3b σ1bσ3a σ1a∆σf
σ’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 ± ∆uf
σC
σC + ∆σf
No
drainage
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 σ1a
σ3c σ1c
τ
σ or
σ’
S < 100% S > 100%
Some practical applications of UU analysis for
clays
τ τ = in situ
undrained shear
strength
Soft clay
1. Embankment constructed rapidly over a soft clay deposit
Some practical applications of UU analysis for
clays
2. Large earth dam constructed rapidly with
no change in water content of soft clay
Core
τ = Undrained shear
strength of clay core
τ
Some practical applications of UU analysis for
clays
3. Footing placed rapidly on clay deposit
τ = In situ undrained shear strength
Note: UU test simulates the short term condition in the field.
Thus, cu can be used to analyze the short term
behavior 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
Shearstress,τ
Normal stress, σ
qu
τf = σ1/2 = qu/2 = cu
Triaxial shear test

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Triaxial shear test

  • 1. Triaxial Shear Test CEP 701 - Soil Engineering Laboratory
  • 3. Embankment Strip footing Shear failure of soils Soils generally fail in shear At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. Failure surface Mobilized shear resistance
  • 4. Retaining wall Shear failure of soils Soils generally fail in shear
  • 5. Retaining wall Shear failure of soils At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. Failure surface Mobilized shear resistance Soils generally fail in shear
  • 6. Shear failure mechanism The soil grains slide over each other along the failure surface. No crushing of individual grains. failure surface
  • 7. Shear failure mechanism At failure, shear stress along the failure surface (τ) reaches the shear strength (τf). σ τ τ σ τ τ
  • 8. Mohr-Coulomb Failure Criterion (in terms of total stresses) τ τf is the maximum shear stress the soil can take without failure, under normal stress of σ. σ φστ tan+= cf c φ failure envelope Cohesio n Friction angleτf σ
  • 9. Mohr-Coulomb Failure Criterion (in terms of effective stresses) τf is the maximum shear stress the soil can take without failure, under normal effective stress of σ’. τ σ’ 'tan'' φστ += cf c’ φ’ failure envelope Effective cohesion Effective friction angleτf σ’ u−= σσ ' u = pore water pressure
  • 10. Mohr-Coulomb Failure Criterion 'tan'' φστ ff c += Shear strength consists of two components: cohesive and frictional. σ’f τf φ’ τ σ' c’ c’ cohesive component σ’f tan φ’ frictional component c and φ are measures of shear strength. Higher the values, higher the shear strength.
  • 11. Mohr Circle of stress Soil element σ’ 1 σ’ 1 σ’ 3 σ’ 3 θ σ’ τ θ σσσσ σ θ σσ τ 2 22 2 2 ' 3 ' 1 ' 3 ' 1' ' 3 ' 1 Cos Sin − + + = − = Resolving forces in σ and τ directions, 2' 3 ' 1 2' 3 ' 1'2 22         − =      + −+ σσσσ στ
  • 12. Mohr Circle of stress 2' 3 ' 1 2' 3 ' 1'2 22         − =      + −+ σσσσ στ Soil element σ’1 σ’1 σ’3 σ’3 θ σ’ τ Soil elementSoil element σ’1 σ’1 σ’3 σ’3 σ’1 σ’1 σ’3 σ’3 θ σ’ τ θθ σ’ τ τ σ’ 2 ' 3 ' 1 σσ + 2 ' 3 ' 1 σσ − ' 3σ ' 1σ
  • 13. Mohr Circle of stress 2' 3 ' 1 2' 3 ' 1'2 22         − =      + −+ σσσσ στ Soil element σ’1 σ’1 σ’3 σ’3 θ σ’ τ Soil elementSoil element σ’1 σ’1 σ’3 σ’3 σ’1 σ’1 σ’3 σ’3 θ σ’ τ θθ σ’ τ τ σ’ 2 ' 3 ' 1 σσ + 2 ' 3 ' 1 σσ − ' 3σ ' 1σ PD = Pole w.r.t. plane θ (σ’, τ)
  • 14. Soil elements at different locations Failure surface Mohr Circles & Failure Envelope X X X ~ failure Y Y Y ~ stable τ σ’ 'tan'' φστ += cf
  • 15. Mohr Circles & Failure Envelope Y σc σc σc Initially, Mohr circle is a point ∆σ σc+∆σ ∆σ The soil element does not fail if the Mohr circle is contained within the envelope GL
  • 16. Mohr Circles & Failure Envelope Y σc σc σc GL As loading progresses, Mohr circle becomes larger… .. and finally failure occurs when Mohr circle touches the envelope ∆σ
  • 17. σ’ 2 ' 3 ' 1 σσ + ' 3σ ' 1σ PD = Pole w.r.t. plane θ (σ’, τf) Orientation of Failure Plane φ’ σ’1 σ’1 σ’3 σ’3 θ σ’ τ σ’1 σ’1 σ’3 σ’3 θθ σ’ τ Failure envelope (90 – θ) Therefore, 90 – θ + φ’ = θ θ = 45 + φ’/2
  • 18. Mohr circles in terms of total & effective stresses = X σv’ σh’ X u u + σv’σh’ effective stresses u σvσh X σv σh total stresses τ σ or σ’
  • 19. Failure envelopes in terms of total & effective stresses = X σv’ σh’ X u u + σv’σh’ effective stresses u σvσh X σv σh total stresses τ σ or σ’ If X is on failure c φ Failure envelope in terms of total stresses φ’ c’ Failure envelope in terms of effective stresses
  • 20. Mohr Coulomb failure criterion with Mohr circle of stress X σ’v = σ’1 σ’h = σ’3 X is on failure σ’1 σ’3 effective stresses τ σ ’ φ’ c’ Failure envelope in terms of effective stresses c’ Cotφ’ (σ’1+ σ’3)/ 2 (σ’1 − σ’3)/ 2       − =            + + 2 ' 2 '' ' 3 ' 1 ' 3 ' 1 σσ φ σσ φ SinCotc Therefore,
  • 21. Mohr Coulomb failure criterion with Mohr circle of stress       − =            + + 2 ' 2 '' ' 3 ' 1 ' 3 ' 1 σσ φ σσ φ SinCotc ( ) ( ) ''2'' 3 ' 1 ' 3 ' 1 φφσσσσ CoscSin ++=− ( ) ( ) ''2'1'1 ' 3 ' 1 φφσφσ CoscSinSin ++=− ( ) ( ) ( )'1 ' '2 '1 '1' 3 ' 1 φ φ φ φ σσ Sin Cos c Sin Sin − + − + =       ++      += 2 ' 45'2 2 ' 452' 3 ' 1 φφ σσ TancTan
  • 22. Other laboratory tests include, Direct simple shear test, torsional ring shear test, plane strain triaxial test, laboratory vane shear test, laboratory fall cone test Determination of shear strength parameters of soils (c, φ or c’, φ’) Laboratory tests on specimens taken from representative undisturbed samples Field tests Most common laboratory tests to determine the shear strength parameters are, 1.Direct shear test 2.Triaxial shear test 1. Vane shear test 2. Torvane 3. Pocket penetrometer 4. Fall cone 5. Pressuremeter 6. Static cone penetrometer 7. Standard penetration test
  • 23. Laboratory tests Field conditions z σvc σvc σhcσhc Before construction A representative soil sample z σvc + ∆σ σhcσhc After and during construction σvc + ∆σ
  • 24. Laboratory tests Simulating field conditions in the laboratory Step 1 Set the specimen in the apparatus and apply the initial stress condition σvc σvc σhc σhc Representative soil sample taken from the site 0 00 0 Step 2 Apply the corresponding field stress conditions σvc + ∆σ σhc σhc σvc + ∆σTraxial test σvc σvc τ τ Direct shear test
  • 25. Triaxial Shear Test Soil sample at failure Failure plane Porous stone impervious membrane Piston (to apply deviatoric stress) O-ring pedestal Perspex cell Cell pressure Back pressure Pore pressure or volume change Water Soil sample
  • 26.
  • 27. Triaxial Shear Test Specimen preparation (undisturbed sample) Sampling tubes Sample extruder
  • 28. Triaxial Shear Test Specimen preparation (undisturbed sample) Edges of the sample are carefully trimmed Setting up the sample in the triaxial cell
  • 29. Triaxial Shear Test Sample is covered with a rubber membrane and sealed Cell is completely filled with water Specimen preparation (undisturbed sample)
  • 30. Triaxial Shear Test Specimen preparation (undisturbed sample) Proving ring to measure the deviator load Dial gauge to measure vertical displacement
  • 31. Types of Triaxial Tests Is the drainage valve open? yes no Consolidated sample Unconsolidated sample Is the drainage valve open? yes no Drained loading Undrained loading Under all-around cell pressure σc σc σc σc σcStep 1 deviatoric stress (∆σ = q) Shearing (loading) Step 2 σc σc σc+ q
  • 32. Types of Triaxial Tests Is the drainage valve open? yes no Consolidated sample Unconsolidated sample Under all-around cell pressure σc Step 1 Is the drainage valve open? yes no Drained loading Undrained loading Shearing (loading) Step 2 CD test CU test UU test
  • 33. Consolidated- drained test (CD Test) Step 1: At the end of consolidation σVC 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’3f Drainage
  • 34. Deviator stress (q or ∆σd) = σ1 – σ3 Consolidated- drained test (CD Test) σ1 = σVC + ∆σ σ3 = σhC
  • 35. Volumechangeofthe sample ExpansionCompression Time Volume change of sample during consolidation Consolidated- drained test (CD Test)
  • 36. Deviator stress,∆σd Axial strain Dense sand or OC clay (∆σd)f Dense sand or OC clay Loose sand or NC clay Volumechange ofthesample ExpansionCompression Axial strain Stress-strain relationship during shearing Consolidated- drained test (CD Test) Loose sand or NC Clay(∆σd)f
  • 37. CD tests How to determine strength parameters c and φ Deviator stress,∆σd Axial strain Shear stress,τ σ or σ’ φ Mohr – Coulomb failure envelope (∆σd)f a Confining stress = σ3a(∆σd)f b Confining stress = σ3b (∆σd)f c Confining stress = σ3c σ3c σ1c σ3a σ1a (∆σd)f σ3b σ1b (∆σd)fb σ1 = σ3 + (∆σd)f σ3
  • 38. CD tests Strength parameters c and φ obtained from CD tests Since u = 0 in CD tests, σ = σ’ Therefore, c = c’ and φ = φ’ cd and φd are used to denote them
  • 39. CD tests Failure envelopes Shear stress,τ σ or σ’ φd Mohr – Coulomb failure envelope σ3a σ1a (∆σd)f a For sand and NC Clay, cd = 0 Therefore, one CD test would be sufficient to determine φd of sand or NC clay
  • 40. CD tests Failure envelopes For OC Clay, cd ≠ 0 τ σ or σ’ φ σ3 σ1 (∆σd)f c σc OC NC
  • 41. Some practical applications of CD analysis for clays τ τ = in situ drained shear strength Soft clay 1. Embankment constructed very slowly, in layers over a soft clay deposit
  • 42. Some practical applications of CD analysis for clays 2. Earth dam with steady state seepage τ = drained shear strength of clay core τ Core
  • 43. Some practical applications of CD analysis for clays 3. Excavation or natural slope in clay τ = In situ drained shear strength τ Note: CD test simulates the long term condition in the field. Thus, cd and φd should be used to evaluate the long term behavior of soils
  • 44. Consolidated- Undrained test (CU Test) Step 1: At the end of consolidation σVC 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 No drainage ±∆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 
  • 45. Volumechangeofthe sample ExpansionCompression Time Volume change of sample during consolidation Consolidated- Undrained test (CU Test)
  • 46. Deviator stress,∆σd Axial strain Dense sand or OC clay (∆σd)f Dense sand or OC clay Loose sand /NC Clay ∆u +- Axial strain Stress-strain relationship during shearing Consolidated- Undrained test (CU Test) Loose sand or NC Clay(∆σd)f
  • 47. CU tests How to determine strength parameters c and φ Deviator stress,∆σd Axial strain Shear stress,τ σ or σ’ (∆σd)f b Confining stress = σ3b σ3b σ1bσ3a σ1a (∆σd)fa φcuMohr – Coulomb failure envelope in terms of total stresses ccu σ1 = σ3 + (∆σd)f σ3 Total stresses at failure (∆σd)f a Confining stress = σ3a
  • 48. (∆σd)fa CU tests How to determine strength parameters c and φ Shear stress,τ σ or σ’ σ3b σ1bσ3a σ1a (∆σd)fa φcu Mohr – Coulomb failure envelope in terms of total stresses ccu σ’3b σ’1b σ’3a σ’1a Mohr – Coulomb failure envelope in terms of effective stresses φ’ C’ ufa ufb σ’1 = σ3 + (∆σd)f - uf σ’3 = σ3 - uf Effective stresses at failure uf
  • 49. CU tests Strength parameters c and φ 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
  • 50. CU tests Failure envelopes For sand and NC Clay, ccu and c’ = 0 Therefore, one CU test would be sufficient to determine φcu and φ (’ = φd) of sand or NC clay Shear stress,τ σ or σ’ φcu Mohr – Coulomb failure envelope in terms of total stresses σ3a σ1a (∆σd)f a σ3a σ1a φ’ Mohr – Coulomb failure envelope in terms of effective stresses
  • 51. Some practical applications of CU analysis for clays τ τ = in situ undrained shear strength Soft clay 1. Embankment constructed rapidly over a soft clay deposit
  • 52. Some practical applications of CU analysis for clays 2. Rapid drawdown behind an earth dam τ = Undrained shear strength of clay core Core τ
  • 53. Some practical applications of CU analysis for clays 3. Rapid construction of an embankment on a natural slope Note: Total stress parameters from CU test (ccu and φcu) can be used for stability problems where, Soil have become fully consolidated and are at equilibrium with the existing stress state; Then for some reason additional stresses are applied quickly with no drainage occurring τ = In situ undrained shear strength τ
  • 54. Unconsolidated- Undrained test (UU Test) Data analysis σC = σ3 σC = σ3 No drainage Initial specimen condition σ3 + ∆σd σ3 No drainage Specimen condition during 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 A A ε− = 1 0
  • 55. 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 to increase of cell pressure Increase of cell pressure Skempton’s pore water pressure parameter, B Note: If soil is fully saturated, then B = 1 (hence, ∆uc = ∆σ3)
  • 56. Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load σ3 + ∆σd σ3 No drainage σ’1 = σ3 + ∆σd - ∆uc ∆ud σ’3 = σ3 - ∆uc ∆ud ∆ud = A∆σd ∆uc ± ∆ud = + Increase of pwp due to increase of deviator stress Increase of deviator stress Skempton’s pore water pressure parameter, A
  • 57. Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3, ∆uc = B ∆σ3 ∆ud = A∆σd ∆u = ∆uc + ∆ud Total pore water pressure increment at any stage, ∆u ∆u = B ∆σ3 + A∆σd Skempton’s pore water pressure equation ∆u = B ∆σ3 + A(∆σ1 – ∆σ3)
  • 58. Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling 0 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 load σC + ∆σ σC No drainage No drainage -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 ± ∆uf σC σC + ∆σf No drainage
  • 59. 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 ± ∆uf σC σC + ∆σf No drainage 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 for different cell pressures τ σ’ σ’3 σ’1∆σ
  • 60. σ3b σ1bσ3a σ1a∆σf σ’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 ± ∆uf σC σC + ∆σf No drainage τ σ or σ’ Mohr circles in terms of total stresses uaub Failure envelope, φu = 0 cu
  • 61. σ3b σ1b Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope σ3a σ1a σ3c σ1c τ σ or σ’ S < 100% S > 100%
  • 62. Some practical applications of UU analysis for clays τ τ = in situ undrained shear strength Soft clay 1. Embankment constructed rapidly over a soft clay deposit
  • 63. Some practical applications of UU analysis for clays 2. Large earth dam constructed rapidly with no change in water content of soft clay Core τ = Undrained shear strength of clay core τ
  • 64. Some practical applications of UU analysis for clays 3. Footing placed rapidly on clay deposit τ = In situ undrained shear strength Note: UU test simulates the short term condition in the field. Thus, cu can be used to analyze the short term behavior of soils
  • 65. Unconfined Compression Test (UC Test) σ1 = σVC + ∆σ σ3 = 0 Confining pressure is zero in the UC test
  • 66. Unconfined Compression Test (UC Test) σ1 = σVC + ∆σf σ3 = 0 Shearstress,τ Normal stress, σ qu τf = σ1/2 = qu/2 = cu
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