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
nsio
nCo
mpr
essi
on
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
nsio
nCo
mpr
essi
on
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
nsio
nCo
mpr
essi
on
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
Top Related