Triaxial shear test of soils

It's required for finding c and phi of the soil which is further used to find bearing capacity of soils

1. Triaxial Shear Test
By:
Amardeep Singh

2. Strength of different
materials
Steel
Tensile
strength
Concrete
Compressive
strength
Soil
Shear
strength
Presence of pore water
Complex
behavior

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 (t)
reaches the shear strength (t
f
).
s
t
t
s
t
t

8. Mohr-Coulomb Failure Criterion
(in terms of total stresses)
t
t
f
is the maximum shear stress the soil can take without
failure, under normal stress of s.
s
fst tan+=c
f
c
f
failure envelope
Cohesion
Friction angle
t
f
s

9. Mohr-Coulomb Failure Criterion
(in terms of effective stresses)
t
f
is the maximum shear stress the soil can take without
failure, under normal effective stress of s’.
t
s’
'tan'' fst +=c
f
c’
f’
failure envelope
Effective
cohesion
Effective
friction anglet
f
s’
u-=ss
'
u = pore water
pressure

10. Mohr-Coulomb Failure Criterion
'tan'' fst
ffc+=
Shear strength consists of two
components: cohesive and frictional.
s’
f
t
f
f’
t
s'
c’ c’
cohesive component
s’
f
tan f’
frictional
component
c and f are measures of shear strength.
Higher the values, higher the shear strength.

11. Mohr Circle of stress
Soil element
s’
1
s’
1
s’
3
s’
3
q
s’
t
q
ssss
s
q
ss
t
2
22
2
2
'
3
'
1
'
3
'
1'
'
3
'
1
Cos
Sin
-
+
+
=
-
=
Resolving forces in s and t directions,
2
'
3
'
1
2
'
3
'
1'2
22
÷
÷
ø
ö
ç
ç
è
æ -
=
÷
÷
ø
ö
ç
ç
è
æ +
-+
ssss
st

12. Mohr Circle of stress
2
'
3
'
1
2
'
3
'
1'2
22
÷
÷
ø
ö
ç
ç
è
æ -
=
÷
÷
ø
ö
ç
ç
è
æ +
-+
ssss
st
Soil element
s’
1
s’
1
s’
3
s’
3
q
s’
t
Soil elementSoil element
s’
1
s’
1
s’
3
s’
3
s’
1
s’
1
s’
3
s’
3
q
s’
t
qq
s’
t
t
s’
2
'
3
'
1ss+
2
'
3
'
1ss-
'
3s
'
1
s

13. Mohr Circle of stress
2
'
3
'
1
2
'
3
'
1'2
22
÷
÷
ø
ö
ç
ç
è
æ -
=
÷
÷
ø
ö
ç
ç
è
æ +
-+
ssss
st
Soil element
s’
1
s’
1
s’
3
s’
3
q
s’
t
Soil elementSoil element
s’
1
s’
1
s’
3
s’
3
s’
1
s’
1
s’
3
s’
3
q
s’
t
qq
s’
t
t
s’
2
'
3
'
1ss+
2
'
3
'
1ss-
'
3s
'
1
s
P
D
= Pole w.r.t. plane
q
(s’, t)

14. Soil elements at different locations
Failure surface
Mohr Circles & Failure Envelope
X X
X~ failure
Y
Y
Y~ stable
t
s’
'tan'' fst +=c
f

15. Mohr Circles & Failure Envelope
Y
s
c
s
c
s
c
Initially, Mohr circle is a point
Ds
s
c
+Ds
Ds
The soil element does not fail if
the Mohr circle is contained within
the envelope
GL

16. Mohr Circles & Failure Envelope
Y
s
c
s
c
s
c
GL
As loading progresses, Mohr
circle becomes larger…
.. and finally failure occurs
when Mohr circle touches the
envelope
Ds

17. s’
2
'
3
'
1ss+
'
3s
'
1
s
P
D
= Pole w.r.t. plane
q
(s’, t
f
)
Orientation of Failure Plane
f’
s’
1
s’
1
s’
3
s’
3
q
s’
t
s’
1
s’
1
s’
3
s’
3
qq
s’
t
Failure envelope
(90 –
q)
Therefore,
90 – q + f’ = q
q = 45 + f’/2

18. Mohr circles in terms of total & effective stresses
=
X
s
v
’
s
h
’
X
u
u
+
s
v
’s
h
’
effective stresses
u
s
v
s
h
X
s
v
s
h
total stresses
t
s or s’

19. Failure envelopes in terms of total & effective
stresses
=
X
s
v
’
s
h
’
X
u
u
+
s
v
’s
h
’
effective stresses
u
s
v
s
h
X
s
v
s
h
total stresses
t
s or s’
If X is on
failure
c
f
Failure envelope in
terms of total stresses
f’
c’
Failure envelope in terms
of effective stresses

20. Mohr Coulomb failure criterion with Mohr circle
of stress
X
s’
v
= s’
1
s’
h
= s’
3
X is on failure s’
1
s’
3
effective stresses
t
s
’
f’c’
Failure envelope in terms
of effective stresses
c’ Cotf’(s’
1
+ s’
3
)/
2
(s’
1
- s’
3
)/
2
÷
÷
ø
ö
ç
ç
è
æ-
=ú
û
ù
ê
ë
é
÷
÷
ø
ö
ç
ç
è
æ+
+
2
'
2
''
'
3
'
1
'
3
'
1
ss
f
ss
f SinCotc
Therefore,

21. Mohr Coulomb failure criterion with Mohr circle
of stress
÷
÷
ø
ö
ç
ç
è
æ-
=ú
û
ù
ê
ë
é
÷
÷
ø
ö
ç
ç
è
æ+
+
2
'
2
''
'
3
'
1
'
3
'
1 ss
f
ss
f SinCotc
( )( ) ''2'
'
3
'
1
'
3
'
1
ffssss CoscSin++=-
( ) ( ) ''2'1'1
'
3
'
1
ffsfs CoscSinSin ++=-
( )
( ) ( )'1
'
'2
'1
'1
'
3
'
1
f
f
f
f
ss
Sin
Cos
c
Sin
Sin
-
+
-
+
=
÷
ø
ö
ç
è
æ
++÷
ø
ö
ç
è
æ
+=
2
'
45'2
2
'
45
2'
3
'
1
ff
ss 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, f or c’, f’)
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
s
vc
s
vc
s
hcs
hc
Before construction
A representative
soil sample
z
s
vc
+ Ds
s
hcs
hc
After and during
construction
s
vc
+ Ds

24. Laboratory tests
Simulating field conditions
in the laboratory
Step 1
Set the specimen in
the apparatus and
apply the initial
stress condition
s
vc
s
vc
s
hc
s
hc
Representative
soil sample
taken from the
site
0
00
0
Step 2
Apply the
corresponding field
stress conditions
s
vc + Ds
s
hc
s
hc
s
vc
+ Ds
T
ra
x
ia
l te
s
t
s
vc
s
vc
t
t
D
i
r
e
c
t

s
h
e
a
r

t
e
s
t

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

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 s
c
s
c
s
c
s
c
s
c
Step 1
deviatoric stress
(Ds = q)
Shearing (loading)
Step 2
s
c
s
c
s
c
+ q

32. Types of Triaxial Tests
Is the drainage valve open?
yes no
Consolidated
sample
Unconsolidated
sample
Under all-around cell pressure s
c
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
s
VC
s
hC
Total, s =
Neutral, u Effective, s’+
0
Step 2: During axial stress increase
s’
VC
=

s
VC
s’
hC
=

s
hC
s
VC
+ Ds
s
hC 0
s’
V
=

s
VC
+

Ds = s’
1
s’
h
=

s
hC
= s’
3
Drainage
Drainage
Step 3: At failure
s
VC
+ Ds
f
s
hC 0
s’
Vf
=

s
VC
+

Ds
f
= s’
1f
s’
hf
=

s
hC
= s’
3f
Drainage

34. Deviator stress (q or Ds
d
) = s
1
– s
3
Consolidated- drained test (CD Test)
s
1
= s
VC
+ Ds
s
3
= s
hC

35. V
o
l
u
m
e

c
h
a
n
g
e

o
f

t
h
e

s
a
m
p
l
e
E
x
p
a
n
s
i
o
n
C
o
m
p
r
e
s
s
i
o
n
Time
Volume change of sample during consolidation
Consolidated- drained test (CD Test)

36. D
e
v
i
a
t
o
r

s
t
r
e
s
s
,

D
s
d
Axial strain
Dense sand
or OC clay
(Ds
d
)
f
Dense sand
or OC clay
Loose sand
or NC clay
V
o
l
u
m
e

c
h
a
n
g
e

o
f

t
h
e

s
a
m
p
l
e
E
x
p
a
n
s
i
o
n
C
o
m
p
r
e
s
s
i
o
n Axial strain
Stress-strain relationship during shearing
Consolidated- drained test (CD Test)
Loose sand
or NC Clay(Ds
d
)
f

37. CD tests How to determine strength parameters c and f
D
e
v
i
a
t
o
r

s
t
r
e
s
s
,

D
s
d
Axial strain
S
h
e
a
r

s
t
r
e
s
s
,

t
s or
s’
f
Mohr – Coulomb
failure envelope
(Ds
d
)
f
a
Confining stress = s
3a(Ds
d
)
f
b
Confining stress = s
3b
(Ds
d
)
f
c
Confining stress = s
3c
s
3c s
1c
s
3a s
1a
(Ds
d
)
f
a
s
3b s
1b
(Ds
d
)
fb
s
1
= s
3
+
(Ds
d
)
f

s
3

38. CD tests
Strength parameters c and f obtained from CD tests
Since u = 0 in CD
tests, s = s’
Therefore, c = c’
and f = f’
c
d and f
d are used
to denote them

39. CD tests Failure envelopes
S
h
e
a
r

s
t
r
e
s
s
,

t
s or
s’
f
d
Mohr – Coulomb
failure envelope
s
3a s
1a
(Ds
d
)
f
a
For sand and NC Clay, c
d
= 0
Therefore, one CD test would be sufficient to determine f
d
of sand or NC clay

40. CD tests Failure envelopes
For OC Clay, c
d
≠ 0
t
s or
s’
f
s
3 s
1
(Ds
d)
f
c
s
c
OC NC

41. Some practical applications of CD analysis for
clays
t
t = 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
t = drained shear
strength of clay core
t
Core

43. Some practical applications of CD analysis for
clays
3. Excavation or natural slope in clay
t = In situ drained shear strength
t
Note: CD test simulates the long term condition in the field.
Thus, c
d
and f
d
should be used to evaluate the long
term behavior of soils

44. Consolidated- Undrained test (CU Test)
Step 1: At the end of consolidation
s
VC
s
hC
Total, s =
Neutral, u Effective, s’+
0
Step 2: During axial stress increase
s’
VC
=

s
VC
s’
hC
=

s
hC
s
VC
+ Ds
s
hC ±D
u
Drainage
Step 3: At failure
s
VC
+ Ds
f
s
hC
No
drainage
No
drainage
±Du
f
s’
V
=

s
VC
+

Ds ± Du
= s’
1
s’
h
=

s
hC
± Du

= s’
3
s’
Vf
=

s
VC
+

Ds
f
± Du
f

= s’
1f
s’
hf
=

s
hC
± Du
f

= s’
3f


45. V
o
l
u
m
e

c
h
a
n
g
e

o
f

t
h
e

s
a
m
p
l
e
E
x
p
a
n
s
i
o
n
C
o
m
p
r
e
s
s
i
o
n
Time
Volume change of sample during consolidation
Consolidated- Undrained test (CU Test)

46. D
e
v
i
a
t
o
r

s
t
r
e
s
s
,

D
s
d
Axial strain
Dense sand
or OC clay
(Ds
d
)
f
Dense sand
or OC clay
Loose
sand /NC
Clay
D
u
+
-
Axial strain
Stress-strain relationship during shearing
Consolidated- Undrained test (CU Test)
Loose sand
or NC Clay(Ds
d
)
f

47. CU tests How to determine strength parameters c and f
D
e
v
i
a
t
o
r

s
t
r
e
s
s
,

D
s
d
Axial strain
S
h
e
a
r

s
t
r
e
s
s
,

t
s or
s’
(Ds
d
)
f
b
Confining stress = s
3b
s
3b s
1b
s
3a s
1a
(Ds
d
)
fa
f
cuMohr – Coulomb
failure envelope in
terms of total stresses
c
cu
s
1
= s
3
+
(Ds
d
)
f

s
3
Total stresses at failure
(Ds
d
)
f
a
Confining stress = s
3a

48. (Ds
d
)
fa
CU tests How to determine strength parameters c and f
S
h
e
a
r

s
t
r
e
s
s
,

t
s or
s’
s
3b s
1b
s
3a s
1a
(Ds
d
)
fa
f
cu
Mohr – Coulomb
failure envelope in
terms of total stresses
c
cu
s’
3b s’
1b
s’
3a s’
1a
Mohr – Coulomb failure
envelope in terms of
effective stresses
f’
C
’ u
fa
u
fb
s’
1
= s
3
+ (Ds
d
)
f
-

u
f

s’
3
= s
3
-

u
f

Effective stresses at failure
u
f

49. CU tests
Strength parameters c and f obtained from CD tests
Shear strength
parameters in terms
of total stresses are
c
cu
and f
cu
Shear strength
parameters in terms
of effective stresses
are c’ and f’
c’ = c
d
and f’ = f
d

50. CU tests Failure envelopes
For sand and NC Clay, c
cu
and c’ = 0
Therefore, one CU test would be sufficient to determine
f
cu
and f’(= f
d
) of sand or NC clay
S
h
e
a
r

s
t
r
e
s
s
,

t
s or
s’
f
cu
Mohr – Coulomb
failure envelope in
terms of total stresses
s
3a s
1a
(Ds
d
)
f
a
s
3a s
1a
f
’
Mohr – Coulomb failure
envelope in terms of
effective stresses

51. Some practical applications of CU analysis for
clays
t
t = 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
t = Undrained shear
strength of clay core
Core
t

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 (c
cu
and f
cu
) 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
t = In situ undrained shear strength
t

54. Unconsolidated- Undrained test (UU Test)
Data analysis
s
C
= s
3
s
C
= s
3
No
drainage
Initial specimen condition
s
3
+ Ds
d
s
3
No
drainage
Specimen condition
during shearing
Initial volume of the sample = A
0
× H
0
Volume of the sample during shearing = A × H
Since the test is conducted under undrained condition,
A × H = A
0
× H
0
A ×(H
0
– DH) = A
0
× H
0
A ×(1

– DH/H
0
) = A
0
z
A
A
e-
=
1
0

55. Unconsolidated- Undrained test (UU Test)
Step 1: Immediately after sampling
0
0
= +
Step 2: After application of hydrostatic cell pressure
Du
c
=

B Ds
3
s
C
= s
3
s
C
= s
3 Du
c
s’
3
=

s
3
- Du
c
s’
3
=

s
3
- Du
c
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, Du
c
= Ds
3
)

56. Unconsolidated- Undrained test (UU Test)
Step 3: During application of axial load
s
3
+ Ds
d
s
3
No
drainage
s’
1
=

s
3
+

Ds
d
- Du
c
Du
d

s’
3
=

s
3
- Du
c
Du
d
Du
d
=

ADs
d
Du
c
±

Du
d
= +
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,
Du
c
=

B Ds
3
Du
d
=

ADs
d
Du

=

Du
c
+ Du
d
Total pore water pressure increment at any stage, Du
Du

=

B Ds
3
+ ADs
d
Skempton’s pore
water pressure
equation
Du

=

B Ds
3
+ A(Ds
1
– Ds
3
)

58. Unconsolidated- Undrained test (UU Test)
Step 1: Immediately after sampling
0
0
Total, s =
Neutral, u Effective, s’+
-u
r
Step 2: After application of hydrostatic cell pressure
s’
V0
=

u
r
s’
h0
=

u
r
s
C
s
C
-u
r
+ Du
c
=
-u
r
+ s
c
(S
r
= 100%

;

B = 1)
Step 3: During application of axial load
s
C
+ Ds
s
C
No
drainage
No
drainage
-u
r
+ s
c
±
Du
s’
VC
=

s
C
+ u
r
-

s
C
= u
r
s’
h
=

u
r
Step 3: At failure
s’
V
=

s
C
+

Ds + u
r
- s
c

Du
s’
h
=

s
C
+

u
r
- s
c
Du
s’
hf
=

s
C
+

u
r
- s
c
Du
f
= s’
3f

s’
Vf
=

s
C
+

Ds
f
+ u
r
- s
c
Du
f
= s’
1f


-u
r
+ s
c
± Du
f
s
C
s
C
+ Ds
f
No
drainage

59. Unconsolidated- Undrained test (UU Test)
Total, s =
Neutral, u Effective, s’+
Step 3: At failure
s’
hf
=

s
C
+

u
r
- s
c
Du
f
= s’
3f

s’
Vf
=

s
C
+

Ds
f
+ u
r
- s
c
Du
f
= s’
1f


-u
r
+ s
c
± Du
f
s
C
s
C
+ Ds
f
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
t
s’
s’
3
s’
1Ds
f

60. s
3b
s
1bs
3a
s
1aDs
f
s’
3 s’
1
Unconsolidated- Undrained test (UU Test)
Total, s =
Neutral, u Effective, s’+
Step 3: At failure
s’
hf
=

s
C
+

u
r
- s
c
Du
f
= s’
3f

s’
Vf
=

s
C
+

Ds
f
+ u
r
- s
c
Du
f
= s’
1f


-u
r
+ s
c
± Du
f
s
C
s
C
+ Ds
f
No
drainage
t
s or
s’
Mohr circles in terms of total stresses
u
au
b
Failure envelope, f
u
= 0
c
u

61. s
3b s
1b
Unconsolidated- Undrained test (UU Test)
Effect of degree of saturation on failure envelope
s
3a s
1a
s
3c
s
1c
t
s or
s’
S < 100% S > 100%

62. Some practical applications of UU analysis for
clays
t
t = 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
t = Undrained shear
strength of clay core
t

64. Some practical applications of UU analysis for
clays
3. Footing placed rapidly on clay deposit
t = In situ undrained shear strength
Note: UU test simulates the short term condition in the field.
Thus, c
u
can be used to analyze the short term
behavior of soils

65. Unconfined Compression Test (UC Test)
s
1
= s
VC
+ Ds
s
3
= 0
Confining pressure is zero in the UC test

66. Unconfined Compression Test (UC Test)
s
1
= s
VC
+
Dsf
s
3
= 0
S
h
e
a
r

s
t
r
e
s
s
,

t
Normal stress, s
q
u
τ
f
= σ
1
/2 = q
u
/2 = c
u

67. Example of Triaxial Graph :