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Shear Strength of SoilsShear Strength of Soils
CEL 610 – Foundation Engineering
Strength of different materials
Steel Concrete Soil
Tensilestrength
Compressivestrength
Shearstrengthstrength strength strength
Presence of pore waterComplexbehaviorbehavior
Shear failure of soilsSoils generally fail in shear
EmbankmentEmbankment
Strip footing
Failure surface
Mobilized shear resistance
At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.
Shear failure of soilsSoils generally fail in shear
Retaining wall
Shear failure of soilsSoils generally fail in shear
Retaining wall
Mobilized shear resistance
Failure surface
resistance
At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.
Shear failure mechanismShear failure mechanism
failure surface
The soil grains slideover each other alongthe failure surface.
No crushing ofNo crushing ofindividual grains.
Shear failure mechanismShear failure mechanism
At failure, shear stress along the failure surface ()reaches the shear strength (f).g ( f)
Mohr-Coulomb Failure Criterion( )(in terms of total stresses)
tan cf
Cohesion Friction angle
cf
is the maximum shear stress the soil can take without
f is the maximum shear stress the soil can take without failure, under normal stress of .
Mohr-Coulomb Failure Criterion( )(in terms of effective stresses)
'tan'' cf'
’u '
u = pore waterEffective cohesion Effective
f i ti l
u pore water pressure
’
c’friction anglef
’
is the maximum shear stress the soil can take without
’’
f is the maximum shear stress the soil can take without failure, under normal effective stress of ’.
Mohr-Coulomb Failure CriterionShear strength consists of two components: cohesive and frictional.p
'tan'' c tan ff c f
’f tan ’ frictional ’
c’ c’
f tan component
’f '
c and are measures of shear strengthc and are measures of shear strength.
Higher the values higher the shear strengthHigher the values, higher the shear strength.
Mohr Circle of stress’1
’3’3
’
Soil element
’1
Resolving forces in and directions
2'3
'1 Sin
Resolving forces in and directions,
22
2
22
'3
'1
'3
'1' C
Sin
2'3
'1
2'3
'1'2
22
2
223131 Cos 22
Mohr Circle of stress’1
’
’1’1
’’
Soil element
’3’3
Soil elementSoil element
’3’3 ’3’3
Soil element
’1
Soil elementSoil element
’1’1
22
2'3
'1
2'3
'1'2
22
2
'3
'1
22
’
2
'3
'1 '
3'1
Mohr Circle of stress’1
’
’1’1
’’
Soil element
’3’3
Soil elementSoil element
’3’3 ’3’3
Soil element
’1
Soil elementSoil element
’1’1
22
’, 2'
3'1
2'3
'1'2
22
2
'3
'1
22
’
2
'3
'1 '
3'1
PD = Pole w.r.t. plane
Mohr Circles & Failure Envelopep
Failure surface
'tan'' cf f
X XYY
Soil elements at different locations
’
X ~ failure
Y ~ stable
X ~ failure
Mohr Circles & Failure EnvelopepThe soil element does not fail if the Mohr circle is containedthe Mohr circle is contained within the envelope
GL
GL
Y
c
cc
Initially, Mohr circle is a pointc+
Mohr Circles & Failure EnvelopepAs loading progresses, Mohr circle becomes larger…
GLGL
Y
c
c
c
.. and finally failure occurs when Mohr circle touches the envelopeenvelope
Orientation of Failure Plane’1’1 Failure envelope
’, f’3’3
’
’3’3
’
’1
’1
–
’'3
'1 '
3'1
’
2
PD = Pole w.r.t. plane
Therefore,45 + ’/2
– ’ = 45 + /2
Mohr circles in terms of total & effective stresses
v’ uv
= Xh’
X u+Xh +
effective stressestotal stresses
v’h’ vh or ’vhu
vh
Failure envelopes in terms of total & effectivestressesstresses
v’ uv
= Xh’
X u+Xh +
If X is on failure
Failure envelope interms of total stresses
’
Failure envelope in termsof effective stresses
effective stressestotal stresses
v’h’ vh or ’cc’vh
uvh
Mohr Coulomb failure criterion with Mohr circle of stressof stress
’v = ’1 Failure envelope in termsof effective stresses
X’h = ’3
effective stresses
’ c’’ ’
X is on failure ’1’3 ’ c
c’ Cot’ ’’c Cot
''''
Therefore,
2'
2''
'3
'1
'3
'1 SinCotc
Mohr Coulomb failure criterion with Mohr circle of stressof stress
''''3
'1
'3
'1 SinCotc
22
SinCotc
''2''''' CS ''2'3131 CoscSin
''2'1'1 '' CoscSinSin 211 31 CoscSinSin
''2'1'' CoscSin
'1
2'131
Sin
cSin
'45'2'452'' TancTan
2
4522
4531 TancTan
Determination of shear strength parameters ofsoils (c or c’ ’soils (c, or c
Laboratory tests on Field testsLaboratory tests onspecimens taken fromrepresentative undisturbed
Field tests
samples
M t l b t t t 1 V h t tMost common laboratory teststo determine the shear strengthparameters are,
1. Vane shear test2. Torvane3. Pocket penetrometer
1.Direct shear test2.Triaxial shear test
4. Fall cone5. Pressuremeter6. Static cone penetrometer
Other laboratory tests include,Direct simple shear test, torsionali h t t l t i t i i l
p7. Standard penetration test
ring shear test, plane strain triaxialtest, laboratory vane shear test,laboratory fall cone test
Laboratory tests
Field conditions
A representative
z vc
A representative soil sample z
vc +
hchc hchc
hc hc
+ vc
Aft d d i
vc +
Before construction After and during construction
Laboratory testsvc +
Simulating field conditions in the laboratory
hchc
vc0 vc +
hchc00vc
vc0
Step 1Representativesoil sample St 2
vc
Set the specimen inthe apparatus and
l th i iti l
soil sampletaken from thesite
Step 2
Apply theapply the initialstress condition
corresponding fieldstress conditions
Direct shear testS h ti di f th di t h tSchematic diagram of the direct shear apparatus
Direct shear testDi t h t t i t it bl f lid t d d i d t t
Preparation of a sand specimen
Direct shear test is most suitable for consolidated drained testsspecially on granular soils (e.g.: sand) or stiff clays
Preparation of a sand specimen
PorousPorous plates
Components of the shear box Preparation of a sand specimen
Direct shear testPreparation of a sand specimen Pressure plate
L li h fLeveling the top surface of specimen
Specimen preparation completed
Direct shear testSt l b llTest procedurePressure plate
Steel ballP
Porous plates
S
Proving ring tto measure shear force
Step 1: Apply a vertical load to the specimen and wait for consolidation
Direct shear testSt l b llPTest procedurePressure plate
Steel ball
Porous plates
S
Proving ring tto measure shear force
Step 1: Apply a vertical load to the specimen and wait for consolidation
Step 2: Lower box is subjected to a horizontal displacement at a constant rate
Direct shear test
Shear box
Dial gauge tomeasure verticaldisplacementdisplacement
Proving ringto measure
Loading frame to Dial gauge to
shear force
gapply vertical load
g gmeasure horizontaldisplacement
Direct shear testfAnalysis of test results
sampletheofsectioncrossofArea(P) force Normal stress Normal
sampletheofsection crossofArea
(S)flidihd l diShsample theofsection cross of Area
(S) surfaceslidingat thedevelopedresistanceShear stressShear
Note: Cross-sectional area of the sample changes with the horizontal displacementd sp ace e t
Direct shear tests on sandsStress-strain relationship
ess,
Dense sand/ OC l
p
ear s
tre OC clay
fLoose sand/
She
NC clayf
Shear displacement
ion
Dense sand/OC Clay hei
ght
mpl
e
Expa
nsi
Loose sand/NC Clayhang
e in
the
sam
ssio
n Shear displacementy
Ch
of
Com
pres
Direct shear tests on sandsHow to determine strength parameters c and How to determine strength parameters c and
s,
Normal stress = 1ear s
tres
s
Normal stress = 2
Normal stress = 3
f1
Normal stress 1
She f2
f3
Shear displacement
re,
fs
at fa
ilu
Mohr – Coulomb failure envelope
ar s
tres
s
Shea
Normal stress,
Direct shear tests on sandsSome important facts on strength parameters c and of sandSome important facts on strength parameters c and of sand
Direct shear tests areSand is cohesionlesshence c = 0
Direct shear tests aredrained and pore waterpressures aredissipated hence u = 0dissipated, hence u = 0
Therefore,
’ = and c’ = c = 0
Direct shear tests on clays
In case of clay, horizontal displacement should be applied at a veryslow rate to allow dissipation of pore water pressure (therefore, onetest would take several days to finish)
Failure envelopes for clay from drained direct shear tests
test would take several days to finish)
e, f Overconsolidated clay (c’ ≠ 0)
at fa
ilure
Normally consolidated clay (c’ = 0)
y ( )
rstr
ess
a
’
Shea
r
Normal force,
Interface tests on direct shear apparatusI f d ti d i bl d t i i ll bl itIn many foundation design problems and retaining wall problems, itis required to determine the angle of internal friction between soiland the structural material (concrete, steel or wood)
PP
Soil SSoil S
Foundation materialFoundation material
tan' af c Where,
dh iaf ca = adhesion,
= angle of internal friction
Triaxial Shear TestPiston (to apply deviatoric stress)
Failure planeO-ring
Soil sample
impervious membrane
SoilSoil sample at failure Porous
stonePerspex
Soil sample
cellWater
pedestal
Cell pressureBack pressure Pore pressure or
volume change
Triaxial Shear TestSpecimen preparation (undisturbed sample)
Sampling tubes
Sample extruder
Triaxial Shear TestSpecimen preparation (undisturbed sample)
Edges of the sample are carefully trimmed
Setting up the sample in the triaxial celly
Triaxial Shear TestSpecimen preparation (undisturbed sample)
Sample is coveredwith a rubber Cell is completelywith a rubbermembrane and sealed
p yfilled with water
Triaxial Shear TestSpecimen preparation (undisturbed sample)
Proving ring toProving ring tomeasure thedeviator load
Dial gauge tomeasure verticalmeasure verticaldisplacement
Types of Triaxial Tests deviatoric stress ( = q)cStep 1( q)
Step 2
ccc c
c+ q
Under all-around cell pressure c
c
Shearing (loading)
c q
Is the drainage valve open? Is the drainage valve open?
yes no
Consolidated U lid t d
yes no
D i d U d i dConsolidatedsample
Unconsolidatedsample
Drained loading
Undrainedloading
Types of Triaxial Tests
Under all aro nd cell press re
Step 1
Sh i (l di )
Step 2
Under all-around cell pressure c Shearing (loading)
Is the drainage valve open?
yes no
Is the drainage valve open?
yes no
Consolidatedsample
Unconsolidatedsample
yes no
Drained l di
Undrainedloadingsample sample loading loading
CD test
CU test
UU test
CU test
Consolidated- drained test (CD Test)Total, = Neutral u Effective ’+
Step 1: At the end of consolidationVC
Total, = Neutral, u Effective, +
’VC = VCVC
hC 0
VC VC
’hC =Drainage 0
Step 2: During axial stress increase
hC
’
Drainage
VC +
’V = VC + =’1
’ ’hC 0 ’h = hC = ’3Drainage
Step 3: At failureStep 3: At failureVC + f ’Vf = VC + f = ’1f
hC 0 ’hf = hC = ’3fDrainage
Consolidated- drained test (CD Test)
1 = VC + 1 VC +
3 = hC
Deviator stress (q or d) = 1 – 3
Volume change of sample during consolidationConsolidated- drained test (CD Test)
g p g
of th
e
xpan
sion
chan
ge o Ex
Time
Volu
me
csa
mpl
e
ssio
n
V s
Com
pres
Stress-strain relationship during shearingConsolidated- drained test (CD Test)
s, d Dense sand
or OC clay
p g g
r str
ess or OC clay
d)fLoose sand
Dev
iato or NC Clayd)f
D
Axial strain
ion
Dense sand or OC clayha
nge
mpl
e
Expa
nsi
y
L dolum
e ch
the
sam
ssio
n Axial strain
Loose sand or NC clay
Vo of
Com
pres
CD tests How to determine strength parameters c and
d d)fc = + ( )re
ss,
d
C fi i t
d)fc
Confining stress = 3c
1 = 3 + (d)fvi
ator
str
)Confining stress = 3ad)fb
Confining stress = 3b
3
Dev
Axial strain
d)fa
ess,
Mohr – Coulomb failure envelope
ear s
tre failure envelope
She
or ’ 3c 1c3a 1a
(d)fa
3b 1b
(d)fb
CD testsSt th t d bt i d f CD t tStrength parameters c and obtained from CD tests
Since u = 0 in CD tests = ’
Therefore, c = c’ and = ’ tests, =
cd and d are used to denote them
CD tests Failure envelopes
For sand and NC Clay, cd = 0s, d
M h C l b
stre
ss Mohr – Coulomb failure envelope
Shea
r S
or ’3a 1a
( )(d)fa
Therefore, one CD test would be sufficient to determine dTherefore, one CD test would be sufficient to determine d of sand or NC clay
CD tests Failure envelopes
For OC Clay, cd ≠ 0
OC NC
OC NC
c or ’3 1
(d)f
cc(d)f
Some practical applications of CD analysis forclaysclays1. Embankment constructed very slowly, in layers over a soft clay
deposit
Soft clay
= in situ drained shear strength
Some practical applications of CD analysis forclaysclays2. Earth dam with steady state seepage
Core
= drained shear strength of clay corestrength of clay core
Some practical applications of CD analysis forclaysclays3. Excavation or natural slope in clay
= In situ drained shear strength
Note: CD test simulates the long term condition in the field.Thus, cd and d should be used to evaluate the longterm behavior of soils
Consolidated- Undrained test (CU Test)Total, = Neutral u Effective ’+
Step 1: At the end of consolidationVC
Total, = Neutral, u Effective, +
’VC = VCVC
hC 0
VC VC
’hC =Drainage 0
Step 2: During axial stress increase
hC
Drainage
’ = + ± u = ’VC +
No
V = VC + ± u = 1
’ ± ’hC ±u
Step 3: At failure
drainage ’h = hC ± u = ’3
Step 3: At failureVC + f
N
’Vf = VC + f ± uf = ’1f
hCNo drainage ±uf ’hf = hC ± uf = ’3f
Volume change of sample during consolidationConsolidated- Undrained test (CU Test)
g p g
of th
e
xpan
sion
chan
ge o Ex
Time
Volu
me
csa
mpl
e
ssio
n
V s
Com
pres
Stress-strain relationship during shearingConsolidated- Undrained test (CU Test)
s, d Dense sand
or OC clay
p g g
r str
ess or OC clay
d)fLoose sand
Dev
iato or NC Clayd)f
D
Axial strain
+
Loose sand /NC Clay
+
Dense sand
u
-
Axial strain
Dense sand or OC clay
CU tests How to determine strength parameters c and
d d)fb 1 = 3 + (d)fre
ss,
d
d)fb
Confining stress = 3b
1 3 + (d)f
Confining stress =
viat
or s
tr
3
)
Confining stress = 3a
Dev
Axial strain
Total stresses at failured)fa
stre
ss, cuMohr – Coulomb
failure envelope interms of total stresses
hear
sS
or ’
ccu3b 1b3a 1a
(d)fa
CU tests How to determine strength parameters c and ’1 = 3 + (d)f - uf
Mohr – Coulomb failureenvelope in terms of
’ = 3 - ufuf
ss,
Mohr – Coulombf il l i
peffective stresses
’Effective stresses at failure
ar s
tres cu
failure envelope interms of total stresses
Shea
or ’
ccu ’3b ’1b’ ’
C’ ufa
ufb
(d)fa
3b 1b3a 1a(d)fa
’3a ’1a
CU testsSt th t d bt i d f CD t tStrength parameters c and obtained from CD tests
Shear strengthShear strengthparameters in terms
Shear strengthparameters in termsof effective stressesp
of total stresses areccu and cu
are c’ and ’
c’ = cd and ’ = dd d
CU tests Failure envelopes
F d d NC Cl d ’ 0For sand and NC Clay, ccu and c’ = 0
Mohr – Coulomb failurel i t f
Mohr – Coulomb ’
envelope in terms ofeffective stresses
ress
, cuMohr Coulombfailure envelope interms of total stresses
hear
st
Sh
or ’3a 1a3a 1a
Therefore one CU test would be sufficient to determine
3a 1a(d)fa
3a 1a
Therefore, one CU test would be sufficient to determine cu and ’= d) of sand or NC clay
Some practical applications of CU analysis forclaysclays1. Embankment constructed rapidly over a soft clay deposit
Soft clay
= in situ undrained shear strength
Some practical applications of CU analysis forclaysclays2. Rapid drawdown behind an earth dam
Core
= Undrained shear strength of clay corestrength of clay core
Some practical applications of CU analysis forclaysclays3. Rapid construction of an embankment on a natural slope
Note: Total stress parameters from CU test (ccu and cu) can be used for
= In situ undrained shear strength
p ( cu cu)stability problems where,
Soil have become fully consolidated and are at equilibrium withthe existing stress state; Then for some reason additionalstresses are applied quickly with no drainage occurring
Unconsolidated- Undrained test (UU Test)Data analysisData analysis
Initial specimen conditionSpecimen condition during shearing
C = 3
C = 3No drainage
3 + dNo drainageC 3drainage 3drainage
I i i l l f h l A HInitial 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 × H0A × H A0 × H0
A ×(H0 – H) = A0 × H0
AA 1
0
A ×(1 – H/H0) = A0 z1
Unconsolidated- Undrained test (UU Test)Step 1: Immediately after samplingStep 1: Immediately after sampling
0
00
Step 2: After application of hydrostatic cell pressure
= +C = 3
C = 3 u
’3 = 3 - uc
’ = - uNo drainage = +
u = B
C 3 uc 3 = 3 - ucdrainage
uc = B 3
Increase of pwp due toIncrease of cell pressure
Increase of pwp due toincrease of cell pressure Skempton’s pore water
pressure parameter, B
Note: If soil is fully saturated, then B = 1 (hence, uc = 3)
Unconsolidated- Undrained test (UU Test)
Step 3: During application of axial load
3 + dNo
’1 = 3 + d - uc ud3 d
3
No drainage ’3 = 3 - uc ud= +
uc ± ud
ud = ABd
Increase of pwp due toincrease of deviator stress
Increase of deviatorstress
Skempton’s pore water pressure parameter, A
Unconsolidated- Undrained test (UU Test)
Combining steps 2 and 3,
uc = B 3 ud = ABd
Total pore water pressure increment at any stage, u
u =uc + ud
u = B [3 + Ad]Skempton’s porewater pressure
tiu = B [3 + A(1 – 3]
equation
Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+
Step 1: Immediately after sampling0
Total, = Neutral, u Effective, +’V0 = ur
0 -ur
Step 2: After application of hydrostatic cell pressure
’h0 = ur
’ = + u = uStep 2: After application of hydrostatic cell pressureC
C-ur uc = -ur c
No drainage
VC = C + ur - C = ur
’h = ur(Sr = 100% ; B = 1)
Step 3: During application of axial load +
’V = C + + ur - c u C +
C
No drainage -ur c ± u ’h = C + ur - c u
Step 3: At failure ’Vf = C + f + ur - c uf = ’1fC + fNo
’hf = C + ur - c uf = ’3f
-ur c ± ufC
No drainage
Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+Total, = Neutral, u Effective, +
Step 3: At failure ’Vf = C + f + ur - c uf = ’1fC + fNo
’hf = C + ur - c uf = ’3f
-ur c ± ufC
C fNo drainage
Mohr circle in terms of effective stresses do not depend on the cellpressure.
Therefore, we get only one Mohr circle in terms of effective stress fordifferent cell pressures
’’3 ’1f
Unconsolidated- Undrained test (UU Test)Total, = Neutral u Effective ’+Total, = Neutral, u Effective, +
Step 3: At failure ’Vf = C + f + ur - c uf = ’1fC + fNo
’hf = C + ur - c uf = ’3f
-ur c ± ufC
C fNo drainage
Mohr circles in terms of total stresses
Failure envelope, u = 0
uaub
cu
3b 1b3a 1af’3 ’1 or ’
Unconsolidated- Undrained test (UU Test)
Effect of degree of saturation on failure envelope
S < 100% S > 100%
or ’3b b3a a3c c or
Some practical applications of UU analysis forclaysclays1. Embankment constructed rapidly over a soft clay deposit
Soft clay
= in situ undrained shear strength
Some practical applications of UU analysis forclaysclays2. Large earth dam constructed rapidly with
no change in water content of soft clay
Core
= Undrained shear strength of clay corestrength of clay core
Some practical applications of UU analysis forclaysclays3. Footing placed rapidly on clay deposit
= In situ undrained shear strength
Note: UU test simulates the short term condition in the field.Thus c can be used to analyze the short termThus, cu can be used to analyze the short termbehavior of soils
Unconfined Compression Test (UC Test)
1 = VC +
3 = 0 3
Confining pressure is zero in the UC testConfining pressure is zero in the UC test
Unconfined Compression Test (UC Test)
1 = VC + f
ss,
ear s
tres
3 = 0
She
qu
Normal stress,
/2 /2τf = σ1/2 = qu/2 = cu
Various correlations for shear strengthFor NC clays, the undrained shear strength (cu) increases with theeffective overburden pressure, ’0
)(0037.011.0'0
PIcu
Skempton (1957)0
Plasticity Index as a %
For OC clays, the following relationship is approximately true
cc
8.0
'0
'0
)(OCRcc
edConsolidatNormally
u
idatedOverconsol
u
Ladd (1977)
For NC clays, the effective friction angle (’) is related to PI as follows
)log(234.0814.0' IPSin Kenny (1959)
Shear strength of partially saturated soilsIn the previous sections, we were discussing the shear strength ofsaturated soils. However, in most of the cases, we will encounterunsaturated soilsunsaturated soils
Pore airWater Pore water
pressure, uPore waterWater
Air Pore air pressure, ua
Solid Effective Solid
Pore water pressure, uw
Water
Solid stress, ’ SolidEffective stress, ’
Saturated soils Unsaturated soils
Pore water pressure can be negative in unsaturated soils
Shear strength of partially saturated soils
Bishop (1959) proposed shear strength equation for unsaturated soils asfollows
'tan)()(' waanf uuuc Where, n – ua = Net normal stressua – uw = Matric suctionua uw Matric suction= a parameter depending on the degree of saturation
( = 1 for fully saturated soils and 0 for dry soils)
Fredlund et al (1978) modified the above relationship as follows
bwaanf uuuc tan)('tan)('
WhereWhere, tanb = 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 due to matric suctiondue to matric suction
Therefore, strength of unsaturated soils is much higher than the strengthof saturated soils due to matric suctionof saturated soils due to matric suction
’
- ua
How it become possiblebuild a sand castlebuild a sand castle
buuuc tan)('tan)(' waanf uuuc tan)(tan)(
Same as saturated soils Apparent cohesion due to matric suction
’
Apparent - uacohesion
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