CEL 610 – Founda Foundation tion Enginee Engineerin ring g
Strength of different materials
Steel
Tensile
Concrete
Compressive
Complex
Soil
Shear
Presence of pore water
Strength of different materials
Steel
Tensile
Concrete
Compressive
Complex
Soil
Shear
Presence of pore water
Shear failure of soils Soils generally fail in shear Embankment Strip footing
Failure surface Mobilized shear resistance
At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. s trength.
Shear failure of soils Soils generally fail in shear
Retaining wall
Shear failure of soils Soils generally fail in shear
Retaining wall
Mobilized shear resistance Failure surface
At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength.
failure surface
The soil grains slide over each other along the failure surface. ru individual grains.
At failure, shear stress along the failure surface () reaches the shear stren th .
Mohr-Coulomb Failure Criterion in terms of total stresses
f
c tan Friction angle
Cohesion
f
c
f
failure, under normal stress of .
Mohr-Coulomb Failure Criterion in terms of effective stresses
f
c' ' tan '
’ Effective cohesion
c’
f
f
= pressure
Effective r c on ang e
’ failure, under normal effective stress of ’.
u
Mohr-Coulomb Failure Criterion Shear strength consists of two com onents: cohesive and frictional.
f
f ’
’ c’
’
c’
’f
'
'
' f frictional component
'
. ,
.
Mohr Circle of stress ’1
’ ’3
’3
Soil element
’1 ,
1' 3' 2 '
' ' 2
' ' 2
os
' ' ' ' 3 2 ' 1 3 1
Mohr Circle of stress ’1
’ ’3
’3
Soil element
’1
1 ' 2
' 1
' 3
'
1' 3'
' 3
2
3'
1' 3' 2
1'
’
Mohr Circle of stress ’1
’ ’3
’3
Soil element
’1
’,
1 ' 2
' 1
' 3
'
1' 3'
' 3
2
3'
1' 3' 2
PD = Pole w.r.t. plane
1'
’
Mohr Circles & Failure Envelo e
Failure surface
Y X
c ' ' tan
'
Y X
’ Soil elements at different locations
Y ~ stable
Mohr Circles & Failure Envelo e The soil element does not fail if within the envelope
c Y
c
c Initially, Mohr circle is a point
c+
Mohr Circles & Failure Envelo e As loading progresses, Mohr circle becomes larger…
c Y
c c .. and finally failure occurs when Mohr circle touches the
Orientation of Failure Plane ’1
Failure envelope ’
’3
’3
’,
f
– ’1
’
3'
1' 3' 2
PD = Pole w.r.t. plane
Therefore, ’ –
’=
1'
’
Mohr circles in terms of total & effective stresses
v X
v’
h
=
X
u h’
effective stresses
’
+
X
u
total stresses
’
u
or ’
Failure envelopes in terms of total & effective
v X
v’
h
=
X
u h’
+
Failure envelope in terms of effective stresses
If X is on failure
’ effective stresses
c’ c
’
X
u
Failure envelope in terms of total stresses
total stresses
’
u
or ’
Mohr Coulomb failure criterion with Mohr circle
’v = ’1 X
Failure envelope in terms of effective stresses
’h = ’3 effective stresses
’
X is on failure ’
’
c’
’3
’
’
’1 ’
Therefore,
c' Cot '
1 3 2
Sin '
1 3 2
’
’
Mohr Coulomb failure criterion with Mohr circle
'
'
1' 3' ' 1
' 1
' 1
1' 3' 2
' 3
2
1' 3' in '2c' os ' ' 3
' ' 3
'
1' 3'
'
1 Sin ' 1 Sin ' 2
'
' 2
'
'
Cos '
1 Sin ' '
' 2
Determination of shear strength parameters of , ’ ’
specimens taken from representative undisturbed samples os common a ora ory es s to determine the shear strength parameters are, 1.Direct shear test 2.Triaxial shear test Other laboratory tests include, Direct simple shear test, torsional r ng s ear es , p ane s ra n r ax a test, laboratory vane shear test, laboratory fall cone test
. 2. 3. 4. 5. 6. 7.
ane s ear es Torvane Pocket penetrometer Fall cone Pressuremeter Static cone enetrometer Standard penetration test
Laboratory tests Field conditions
A re resentative soil sample
z
vc hc
vc
Before construction
vc + hc
hc vc
er an ur ng construction
z
vc +
Laboratory tests Simulating field conditions in the laboratory 0
vc 0
0
hc
0
hc
hc vc +
vc
hc vc
Representative
Step 1
taken from the site
Set the specimen in the apparatus and app y e n a stress condition
vc ep Apply the corresponding field stress conditions
Direct shear test c ema c
agram o
e
rec s ear appara us
Direct shear test rec s ear es s mos su a e or conso a e specially on granular soils (e.g.: sand) or stiff clays
ra ne
es s
plates
Components of the shear box
Preparation of a sand specimen
Direct shear test Preparation of a sand specimen
eve ng t e top sur ace of specimen
Pressure plate
Specimen preparation completed
Direct shear test Test procedure
P
ee
a
Pressure plate Porous plates S
Proving ring o measure shear force
Step 1: Apply a vertical load to the specimen and wait for consolidation
Direct shear test Test procedure
P
ee
a
Pressure plate Porous plates S
Proving ring o 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 to measure vertical
Proving ring to measure shear force
Loadin frame to apply vertical load
Dial au e to measure horizontal displacement
Direct shear test Analysis of test results
Normal stress
Shear stress
Normal force (P)
ear res stance eve ope at t e s
ng sur ace
Area of cross section of the sample
Note: Cross-sectional area of the sample changes with the horizontal dis lacement
Direct shear tests on sands Stress-strain relationshi , s s r t s r a e h S
Dense sand/ c ay
f f
Loose sand/ NC clay
Shear displacement t h g i e e l h p n i a e s g e n h a t f C o
n o s n a p x E n o i s s e r p m o C
Dense sand/OC Clay Shear displacement
Loose sand/NC Cla
Direct shear tests on sands
, s e r t s r a h S
Normal stress =
3
Normal stress =
2
= f2
f1
f3
f
, e r u l i a f t a s e r t s r e h S
Shear displacement
Mohr – Coulomb failure envelope
Normal stress,
Direct shear tests on sands
Sand is cohesionless hence c = 0
drained and pore water pressures are , Therefore, ’=
and c’ = c = 0
Direct shear tests on clays In case of clay, horizontal displacement should be applied at a very slow rate to allow dissipation of pore water pressure (therefore, one
Failure envelopes for clay from drained direct shear tests f
Overconsolidated clay (c’ ≠ 0)
, r u l i a f t s s e r t s r a e h S
Normally consolidated clay (c’ = 0)
’
Normal force,
Interface tests on direct shear apparatus n many oun a on es gn pro ems an re a n ng wa pro ems, is required to determine the angle of internal friction between soil and the structural material (concrete, steel or wood) P
Soil
S
Foundation material
c ' tan
Where, ca = a
es on,
= angle of internal friction
Triaxial Shear Test Piston (to apply deviatoric stress)
Failure plane
O-ring impervious membrane
sample
at failure Perspex cell
Porous stone Water
Cell pressure Back pressure
Pore pressure or pedestal
volume change
Triaxial Shear Test Specimen preparation (undisturbed sample)
Sampling tubes
Sample extruder
Triaxial Shear Test Specimen preparation (undisturbed sample)
Edges of the sample are carefull trimmed
Setting up the sample in the triaxial cell
Triaxial Shear Test Specimen preparation (undisturbed sample)
Sample
is
covered
membrane and sealed
Cell is com letel filled with water
Triaxial Shear Test Specimen preparation (undisturbed sample) Provin rin to measure the deviator load Dial gauge to measure vertical displacement
Types of Triaxial Tests c
Step 1
deviatoric stress =
Step 2
c
c
c
c
+ c Under all-around cell pressure
c
Is the drainage valve open? yes
sample
no
nconsolidated sample
Shearing (loading) Is the drainage valve open? yes
ra ne loading
no
n ra ne loading
Types of Triaxial Tests Step 2
Step 1 -
u
u
c
Is the drainage valve open? yes
Consolidated sam le
ear ng oa ng
Is the drainage valve open?
no
Unconsolidated
Drained oa ng
CD test
UU test
Undrained
Consolidated- drained test (CD Test) ,
, ’
Step 1: At the end of consolidation ’
hC
= ’hC =
0
hC
Step 2: During axial stress increase VC +
VC +
=
VC +
=
’1 hC
Drainage
Drainage
V
0
’Vf =
f
hC
h
0
VC
=
+
’hf =
hC =
f =
hC
3
’1f
= ’3f
Consolidated- drained test (CD Test)
=
3=
Deviator stress (q or
d)
=
1 –
hC
3
Consolidated- drained test (CD Test) Volume chan e of sam le durin consolidation
e h t f e g n a h e l e m p u m l o a
n o i s n a p E
Time n o i s e r p m o C
Consolidated- drained test (CD Test) Stre res sss-s stra rain in re rela lati tio ons nsh hi du duri rin n sh she ear arin in d
, s e r t s r o t a i v e
Dense sand d)f d)f
Loose sand or NC Clay
Axial strain e g e n l a p c a e s m e u h t l f V o
n o s n a p x E n o i s s e r p m o C
Dense sand or OC cla Axial strain
oose san or NC clay
CD tests
How to determine strength parameters c and 1
, s s e t s r o t a i e D
Confining stress =
d f
3c
on n ng s ress =
3b
Confining stress =
d)fb
3
3a
3
d fa
Axial strain
, s s r t s r a h S
Mohr Mo hr – Co Coul ulom omb b
or ’ 3a
3b
(
3c
d)fa
1a
(
1b d)fb
1c
CD tests reng
parame ers c an
Since u = 0 in CD , = ’
o
a ne
ro m
es s
Therefo re, c = c’ Therefore, and = ’ cd and d are used to denote them
CD tests Failure envelopes For sand and NC Clay, cd = 0
, s e r t s r a e h
d
o r – ou om failure envelope
or ’ 3a
1a d fa
Therefore one CD test would be sufficient to determine of sand or NC clay
CD tests Failure envelopes For OC Clay, cd ≠ 0
or ’ 3
1
c
Some practical applications of CD analysis for 1. Embankment constructed very slowly, in layers over a soft clay deposit
Soft clay
= in situ drained shear strength
Some practical applications of CD analysis for 2. Earth dam with steady state seepage
Core
= drained shear
Some practical applications of CD analysis for 3. 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 long term behavior of soils
Consolidated- Undrained test (CU Test) ,
, ’
Step 1: At the end of consolidation ’
hC
= ’hC =
0
hC
Step 2: During axial stress increase ’V
VC +
No drainage
hC
VC +
o drainage
± u
h
’Vf =
f
hC
’1
VC
± uf
=
u=
hC
+
f ±
’hf =
hC ±
VC
3
uf = ’1f
uf = ’3f
Consolidated- Undrained test (CU Test) Volume chan e of sam le durin consolidation
e h t f e g n a h e l e m p u m l o a
n o i s n a p E
Time n o i s e r p m o C
Consolidated- Undrained test (CU Test) Stress-strain relationshi durin shearin d
, s e r t s r o t a i v e
Dense sand d)f d)f
Loose sand or NC Clay
Axial strain
+ Loose sand /NC Clay
u
Axial strain
or OC clay
CU tests
How to determine strength parameters c and =
, s s e t s r o t a i e D
, s s e r t r a e h S ccu
Confining stress =
3b 3a
3 d fa
Total stresses at failure Axial strain Mohr – Coulomb failure envelope in terms of total stresses
cu
or ’ 3a
3b
1a
(
d)fa
1b
CU tests
How to determine strength parameters c and ’1 = 3 + ( d)f - uf
uf
Mohr – Coulomb failure envelo e in terms of effective stresses
, s e r t s r e h S C’
’ cu
ufa
’3b 3a
3a
3 - uf
Effective stresses at failure
Mohr – Coulomb a ure enve ope n terms of total stresses
ccu
’ =
3b
(
1a ) d fa
ufb
’1b 1a
or ’ 1b
CU tests reng
parame ers c an
Shear strength parameters in terms of total stresses are ccu and cu
o
a ne
rom
es s
parameters in terms of effective stresses are c’ and ’
c’ = c and ’ =
CU tests Failure envelopes or san an
ay, ccu an c =
Mohr – Coulomb failure enve ope n erms o effective stresses
, s s e r t s r a e
Mohr – Coulomb failure envelope in terms of total stresses
’
S
or ’ 3a
( , cu and ’ =
cu
d)
d)fa
of sand or NC clay
Some practical applications of CU analysis for 1. Embankment constructed rapidly over a soft clay deposit
Soft clay
= in situ undrained shear strength
Some practical applications of CU analysis for 2. Rapid drawdown behind an earth dam
Core
= Undrained shear
Some practical applications of CU analysis for 3. Rapid construction of an embankment on a natural slope
= In situ undrained shear strength Note: Total stress arameters from CU test c stability problems where,
and
can be used for
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
Unconsolidated- Undrained test (UU Test)
Initial specimen condition C=
Specimen condition during shearing
3
No
=
No
3+
d 3
Initial volume of the sample = A 0 × H0 Volume of the sample during shearing = A × H Since the test is conducted under undrained condition, = A ×(H0 – H) = A0 × H0 A ×(1 – H/H0) = A0
A
A0
z
Unconsolidated- Undrained test (UU Test) 0
Step 2: After application of hydrostatic cell pressure ’3 = = C
No
3
=
’3
c
c
3-
uc 3
c
3 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 =
3)
Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load ’1 =
+ drainage
3
=
’3 =
+ uc ±
d
- uc 3-
ud
uc ud
ud
ud = AB Increase of pwp due to increase of deviator stress
3+
d Increase stress
Skempton’s pore water pressure parameter, A
of
deviator
Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3,
uc = B
ud = AB
3
d
Total pore water pressure increment at any stage, u
u = uc + ud u=B [ u=B [
3+ 3+
A A(
d] 1 –
3]
Skempton’s pore water pressure equa on
Unconsolidated- Undrained test (UU Test) ,
, ’ ’V0 = ur
Step 1: Immediately after sampling
0
’h0 = ur
-ur
0
’VC No drainage
-ur C
uc = -ur
-
C
r
c
’h = ur
(Sr = 100% ; B = 1)
’V =
C+
+ ur -
u
c
C C
-ur
c
C+
’h =
± u
Step 3: At failure
drainage
r
C
Step 3: During application of axial load
No drainage
C
’Vf =
C
+
f +
C + ur
ur -
-
c
c
u
uf = ’1f
f C
-ur
c
± uf
’hf = C + ur = ’3f
c
uf
Unconsolidated- Undrained test (UU Test) ,
, ’
Step 3: At failure
’Vf =
+ drainage
C
-ur
c
C
+
f +
ur -
c
’hf = C + ur = ’3f
± uf
uf = ’1f c
uf
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
’3
f
’1
’
Unconsolidated- Undrained test (UU Test) ,
, ’
Step 3: At failure
’Vf =
+ drainage
C
-ur
c
C
+
f +
ur -
uf = ’1f
c
’hf = C + ur = ’3f
± uf
c
uf
Mohr circles in terms of total stresses Failure envelope,
u
=0
cu ub
’3a 3b 3
f
ua
’1b 1a 1
or ’
Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope
S < 100%
3c
3b
S > 100%
c
3a
b
a
’
Some practical applications of UU analysis for 1. Embankment constructed rapidly over a soft clay deposit
Soft clay
= in situ undrained shear strength
Some practical applications of UU analysis for 2. Large earth dam constructed rapidly with no change in water content of soft clay
Core
= Undrained shear
Some practical applications of UU analysis for 3. Footing placed rapidly on clay deposit
= In situ undrained shear strength
Note: UU test simulates the short term condition in the field. , u behavior of soils
Unconfined Compression Test (UC Test) 1=
VC
+
=0
Unconfined Compression Test (UC Test) 1=
VC
+
f
3=
, s e r t s r a h S
0 qu
Normal stress,
f = σ1/2
τ
= qu/2 = cu
Various correlations for shear strength For NC clays, the undrained shear strength (c u) increases with the effective overburden pressure, ’0
cu
'
0.11 0.0037( PI )
Skempton (1957)
Plasticity Index as a % For OC clays, the following relationship is approximately true u ' 0
u ' 0
Overconsolidated Normally Consolidat ed
(OCR)0.8
For NC clays, the effective friction angle (’) is related to
Sin ' 0.814 0.234 log( IP)
Kenny (1959)
Ladd (1977)
PI
as follows
Shear strength of partially saturated soils In the previous sections, we were discussing the shear strength of saturated soils. However, in most of the cases, we will encounter
Water
Pore water pressure, u
Air
pressure, ua ore wa er pressure, uw
Effective stress, ’
Saturated soils
Effective stress, ’ 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 as follows
f
c'( n ua )
(ua
u w )tan '
Where, n –
ua
u –u
= Net normal stress = 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
f
c'( n ua ) tan '(ua u w ) tan
b
, tan
b=
Rate of increase of shear strength with matric suction
Shear strength of partially saturated soils
f
c'( n ua ) tan '(ua u w ) tan
Same as saturated soils
b
Apparent cohesion
Therefore, strength of unsaturated soils is much higher than the strength
’
- ua
How
f
it
'
become
n
a
Same as saturated soils
possible
'
a
b
w
Apparent cohesion due to matric suction
’ Apparent cohesion
- ua
