shear strength of soil

1. Shear strength of
soil
UNIT- 5

2.  The strength of a material is the greatest
stress it can sustain
 The safety of any geotechnical structure is
dependent on the strength of the soil
 If the soil fails, the structure founded on it can
collapse
Shear Failure in Soils

3. Shear Strength in Soils
 The shear strength of a soil is its resistance to
shearing stresses.
 It is a measure of the soil resistance to deformation by
continuous displacement of its individual soil particles
 Shear strength in soils depends primarily on
interactions between particles
 Shear failure occurs when the stresses between the
particles are such that they slide or roll past each other

4. o Soil derives its shear strength from two sources:
o Cohesion between particles (stress independent
component)
o Cementation between sand grains
o Electrostatic attraction between clay particles
o Frictional resistance between particles (stress dependent
component)

5. Embankment
Strip footing
SHEAR FAILURE OF SOILS
Soils generally fail in shear
Failure surface
Mobilized shear resistance
At failure, shear stress along the
failure surface (mobilized shear
resistance) reaches the shear strength.

6. 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

7. Shear failure mechanism
failure surface
The soil grains slide over each
other along the failure surface.
No crushing of individual
grains.

8. Mohr-Coulomb Failure Criteria:
 This theory states that a material fails because of a
critical combination of normal stress and shear stress,
and not from their either maximum normal or shear
stress alone.
 The relationship between normal stress and shear is
given as
s  ctan
s - shear strength
C - cohesion
 - angle of internal
friction

9. Mohr-Coulomb Failure Criterion
(in terms of effective stresses)
 f is the maximum shear stress the soil can take without
failure, under normal effective stress of .
 f  c' ' tan
'
’
c
’
Effective
cohesion Effective
friction anglef
 '
  - u
u = pore water
pressure

10. MOHR-COULOMB FAILURE CRITERION
tan' f  c''f
’f
’
Shear strength consists of two components:
cohesive and frictional.
f
'
c’
’f tan ’
c’
frictional
component
c and  are measures of shear strength. Higher the values, higher the shear
strength.

11. MOHR CIRCLE OF STRESS
Soil element
’
1
’
Cos2
 - 
'
22
'
311

’
1
Resolving forces in  and  directions,
’
3
’
3


  Sin2
-
2
' '
3
'
'
3
'
1
2
-'

3 
'
1
2'
3
'
12 '
22 




 


  -

 

12. 22 





Soil element
’


Soil element
Mohr Circle of stress
’1
2
'
 -3 
'
1
2'
3
'
1
2
'
- 




 
Soil element
’1
’3
’3


’

’
2
 '
 '
1 
3
2
 '
- '
1 
3
'
3
'
1

13. 3
'
3
2
'
-
22 





 -





  
Soil element
’


Soil element
Mohr Circle of stress
’1
' '

1
2'
1
Soil element
’1
’3



’
’3
’
’, 
 '
- 
'
1 3
2

 '
 '
  '
 '
3 1 3 1
2
PD = Pole w.r.t. plane

14. Soil elements at different locations
Failure surface
Mohr Circle & Failure Envelope
X X
Y ~ stable
X ~ failure
Y Y

’
 f  c''tan'

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

16. ’
2
PD = Pole w.r.t. plane
 '
 '
1 
3
'
3
'
1

’, f
ORIENTATION OF FAILURE PLANE
’


’1
 –

’
’1
’3
’3

’

Failure envelope
Therefore,
 –   ’ = 
  45 + ’/2

17. Mohr circles in terms of total & effective stresses
= +
Failure envelope in terms

v’
X
u
u
v’ hh’
effective stresses
u
v
v
total stresses

or

If X is on
failure
c’ c

Failure envelope in
terms of total stresses
’
of effective stresses
XX

18. Mohr Coulomb failure criterion with Mohr circle of stress
’v = ’1
of effective stresses
’h = ’3
X is on failure ’1’3
effective stresses
  - 


’ c’
Failure envelope in terms
c’
Cot
  





Sin' 

 '
- '


c'Cot'



 '
 '
22
1 31 3
Therefore,
X

19. 


Sin' 

 '
- '

Sin'

c'Cot'

 -  


 '
 '
22
1 31 3
2c'Cos'
'
3
'
1
' '
1 3 
 1- Sin'  1 Sin' 2c'Cos'
'
3
'
1
1 Sin'
1- Sin' 1- Sin'
2c'
Cos'
  '
3 '1 

2 
'
45   2c'Tan45


2 
'



  Tan
2'
3 '1

20. Most common laboratory tests to
determine the shear strength
parameters are,
1.Direct shear test
2.Triaxial shear test
3. Direct simple shear test
Determination of shear strength parameters of soils
(c, or c’ ’
Laboratory
specimens
tests on
taken from
representative
samples
undisturbed
Field tests
1. Vane shear test
2. Torvane
3. Pocket penetrometer
4. Fall cone
5. Pressuremeter
6. Static cone penetrometer
7. Standard penetration test

21. LABORATORY TESTSField conditions
z
A representative
soil sample
vc
vc
hchc
Before construction
z
vc + 
hc
vc + 
hc
After and during
construction

22. LABORATORY TESTS
Simulating field conditions
in the laboratory
Step 1
Set the specimen in
the apparatus and
initialapply the
stress condition
vc
vc
hchc
Representative
soil sample
taken from the
site
0
00
0
Step 2 Apply
corresponding the
field
stress conditions
vc + 
hchc
vc + 
vc
vc



23. Direct shear test
Schematic diagram of the direct shear apparatus

24. DIRECT SHEAR TEST
Preparation of a sand specimen
Porous plates
Components of the shear box Preparation of a sand specimen
Direct shear test is most suitable for consolidated drained tests
specially on granular soils (e.g.: sand) or stiff clays

25. Leveling the top surface
of specimen
Preparation of a sand specimen
Specimen preparation
completed
Pressure plate

26. 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
PTest procedure
Pressure plate
Porous plates
S
Steel ball
Proving ring
to measure
shear force

27. DIRECT SHEAR TEST
Shear box
Loading frame to
apply vertical load
Dial gauge to
measure vertical
displacement
Dial gauge to
measure horizontal
displacement
Proving ring
to measure
shear force

28. Analysis of test results
Area of cross section of the sample
Normalforce (P)
 Normalstress 
  Shear stress 
Shear resistance developed at the sliding surface (S)
Area of cross section of the sample
Note: Cross-sectional area of the sample changes with the horizontal
displacement

29. Shear
stress,
Shear displacement
Dense sand/
OC clay
f
Loose sand/
NC clayf
Dense sand/OC Clay
DIRECT SHEAR TESTS ON SANDS
STRESS-STRAIN RELATIONSHIP
Shear displacement
Loose sand/NC Clay
Changeinheight
ofthesample
ExpansionCompression

30. f1
Normal stress =
 1

How to determine strength parameters c and 
Normal stress = 3
Shear
stress,
Shear displacement
f2
f3
Shearstressat
failure,f
Normal stress,
Mohr – Coulomb failure envelope


31. DIRECT SHEAR TESTS ON CLAYS
Failure envelopes for clay from drained direct shear tests
Shearstressat
failure,f
Normal force,

Normally consolidated clay (c’ = 0)

In case of clay, horizontal displacement should be applied at a very
slow rate to allow dissipation of pore water pressure (therefore, one
test would take several days to finish)
Overconsolidated clay (c’ ≠ 0)

32. INTERIRECT SHEAR APPARATUS
In many foundation design problems and retaining wall problems, it
is required to determine the angle of internal friction between
soil and the structural material (concrete, steel or wood)
 f  ca  '
tan
Where,
c = adhesion,a
 = angle of internal friction
Soil
Foundation material
Foundation material
Soil
P
S

33. Due to the smaller thickness of the sample, rapid
drainage can be achieved
Can be used to determine interface strength parameters
Clay samples can be oriented along the plane of
weakness or an identified failure plane
Disadvantages of direct shear apparatus
Failure occurs along a predetermined failure plane
Area of the sliding surface changes as the test progresses
Non-uniform distribution of shear stress along the failure surface
Advantages of direct shear apparatus

34. 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

35. Specimen preparation (undisturbed sample)
Sampling tubes
Sample extruder

36. Specimen preparation (undisturbed sample)
Sampling tubes
Sample extruder

37. Cell is completely
filled with water
Specimen preparation (undisturbed sample)
Sample is covered
with a
membrane an
rubber
d sealed

38. TRIAXIAL SHEAR TESTSpecimen preparation (undisturbed sample)
Proving ring to
measure the
deviator load
Dial gauge to
measure vertical
displacement
In some tests

39. 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
 c

c

c

c

cStep 1
deviatoric stress
( = q)
Shearing (loading)
Step 2

c

c
 c+
q

40. TYPES OF TRIAXIAL TESTS
Is the drainage valve open?
yes no
Consolidated
sample
Unconsolidated
sample
Under all-around cell pressure
 c
Step 1
Is the drainage valve open?
yes no
Drained
loading
Undrained
loading
Shearing (loading)
Step 2
CD test
CU test
UU test

41. Deviator
stress,d
Axial strain
Dense sand or OC
clay
Axial strain
Loose sand or NC
clay
Stress-strain relationship during shearing
Dense sand or OC clay
d)f
Volumechange
ofthesample
ExpansionCompression
Consolidated- drained test (CD Test)
Loose sand
or NC Clay
d)f

42. CD TESTS
Deviator
stress,d
a
Axial strain

Mohr – Coulomb failure envelope
Shear
stress,
or



d)f
Confining stress =
 3a
How to determine strength parameters c and 
 d)f

 )
d f
b
Confining stress =  3c
Confining stress = 
3b
c

3c

1c
 3a
 3b

1a
(

1b
( d fb
 =
 +
1 3
(
d)f
3

43. CD TESTS
Failure envelopes
Shear
stress,
or

d
Mohr – Coulomb
failure envelope

3a

1a(
d)fa
For sand and NC Clay, cd = 0
Therefore, one CD test would be sufficient to determine d
of sand or NC clay

44. CD TESTS
Failure envelopes
For OC Clay, cd ≠ 0

or



3

1(
d)f
c

c
OC NC

45. Some practical applications of CD analysis for
clays
  = in situ drained
shear strength
Soft clay
1. Embankment constructed very slowly, in layers over a soft clay
deposit

46. Some practical applications of CD analysis for
clays
2. Earth dam with steady state seepage
• 
• Core
•  = drained shear strength of
clay core

47. Some practical applications of CD analysis for
clays
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

48. Volumechangeofthe
sample
ExpansionCompression
Time
Volume change of sample during consolidation
Consolidated- Undrained test (CU Test)

49. Deviator
stress,d
Axial strain
Loose sand /NC Clay
Axial strain
Dense sand or OC
clay
Stress-strain relationship during shearing
Dense sand or OC clay
d)f
u
+-
Consolidated- Undrained test (CU Test)
Loose sand
or NC Clay
d)f

50. CU TESTS
How to determine strength parameters c and 
Deviator
stress,d
Axial strain
Shear
stress,
or


d)fb

3b

1b

3a

1a( d fa
Mohr – Coulomb
failure envelope in
terms of total stresses
ccu
Confining stress = 
3b
Confining stress =
 3a
cu
 1 =  3
+ (d)f

3
Total stresses at failure
d)f
a

51. ( d)fa
CU TESTS
How to determine strength parameters c and 
Shear
stress,
or


3b

1b

3a

11ba( 1da fa
cu
Mohr – Coulomb
failure envelope
in terms of total
stresses
ccu  3
b


3a

Mohr – Coulomb failure
envelope in terms
of effective stresses

C’ ufa
ufb
 1 =  3 + (d)f -
uf
 =  3
- u

f
Effective stresses at failure
uf

52. CU TESTS
Failure envelopes
For sand and NC Clay, ccu and c’ = 0
Therefore, one
CU test
would be sufficient to determine
cu and  = d) of sand or NC clay
Shear
stress,
or

cuMohr – Coulomb
failure envelope in
terms of total stresses

3a
 1a
 1a(
d)fa

3a
’
Mohr – Coulomb failure
envelope in terms of
effective stresses

53. Some practical applications of CU analysis for
clays
  = in situ
undrained shear
strength
Soft clay
1. Embankment constructed rapidly over a soft clay deposit

54. Some practical applications of CU analysis for
clays
2. Rapid drawdown behind an earth dam
• 
• Core
• = Undrained shear strength of clay
core

55. 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 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

 = In situ undrained shear
strength

56. Unconsolidated- Undrained test (UU Test)
Step 1: Immediately after sampling
0
0
= +
Step 2: After application of hydrostatic cell pressure
uc =B
 3
 C =
 3
 C =
 3
uc
 3
=  3
- uc  =

3 3
- 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 = 
3)

57. Unconsolidated- Undrained test (UU Test)
Step 3: During application of axial load
 3 +
 d

3
No
drainage
 1 =  3 +
 d - 
uc
ud
 =  -
u u
3 3 c d

uc ± ud
ud =
AB d
= +
Increase of pwp due to
increase of deviator stress
Increase
stress
of deviator
Skempton’s pore water
pressure parameter, A

58. UNCONSOLIDATED- UNDRAINED TEST (UU TEST)
Combining steps 2 and 3,
uc = B  3 ud = AB d
Total pore water pressure increment at any stage, u
u= uc + ud
u= B [ 3 +
A d]
Skempton’s pore
water pressure
equation
u= B [ 3 + A( 1 –
 3]

59. Some practical applications of UU analysis for
clays
  = in situ
undrained shear
strength
Soft clay
1. Embankment constructed rapidly over a soft clay deposit

60. Some practical applications of UU analysis for
clays
2. Large earth dam constructed rapidly with
no change in water content of soft clay
• 
• Core
• = Undrained shear strength of clay
core

61. Some practical applications of UU analysis for
clays
3. Footing placed rapidly on clay deposit
 = 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

62. UNCONFINED COMPRESSION TEST (UC
TEST)
 1 =  VC
+ 
 3 =
0
Confining pressure is zero in the UC test

63.  1 =
 VC +
 f
3 = 0
qu
Normal stress, 
Shearstress,
Note: Theoritically qu = cu , However in the actual case qu
< cu due to premature failure of the sample

64. STRESS INVARIANTS (P AND Q)p (or s) = ( 1 +
 3)/2
q (or t) = ( 1 -
 3)/2

(1 - 3)/2
3 1
( 1 +
 3)/2
c
p and q can be used to illustrate the variation of the stress
state of a soil specimen during a laboratory triaxial test

65. GL

c
c
STRESS INVARIANTS (P AND Q)

c
p (or s) = ( 1 + 
3)/2
q (or t) = ( 1 - 
3)/2
orq
or
p

66. Mohr Coulomb failure envelope in terms of stress invariants
p (or s) = ( 1 +
 3)/2
q (or t) = ( 1 - 
3)/2



Sin' 

 '
- '


c'Cot'



 '
 '
22
1 31 3

-


Sin'c'Cos'
22
'
3
'
1
' '
31


3 1
( 1 +
 3)/2
( 1 -
 3)/2
c
q  pSin'c'Cos'

67. p (or s) = ( 1 +
 3)/2
q (or t) = ( 1 - 
3)/2orq
or
p
Mohr Coulomb failure envelope in terms of stress invariants

c cos
Therefore, sin = tan  = sin-1(tan


68. STRESS PATH FOR CD TRIAXIAL TEST

3
p, p’ (or s, s’) = ( 1 + 
3)/2
= ( 1 +
 3S tep3)1
/2
q (or t) = ( 1 -  3)/
2
orq
or
In CD tests pore water pressure is equal to zero. Therefore, total and
effective stresses are equal

3
p, p’ (or s, s’) = 
3
q (or t) = 0
Step 2  3 +
 d

3
p, p’ (or s, s’) =  3 + q (or t) = d/


450

69. STRESS PATH FOR CU TRIAXIAL TEST

3 =
p (or s) = ( 1 +  3)/2
p’ (or s’) = ( 1 +  3)/2
- u
q (or t) = ( 1 -  3)/
2
In CU tests pore water pressure develops during shearing
Step 1 
3

3
p, p’ (or s, s’) = 
3
q (or t) = 0
p (or s) =  3 +


450
ud
Step 2  3 +
 d

3
ud
q
  
or p,p’q (or t) = d/

70. DIRECT SIMPLE SHEAR TEST
Direct shear test
 = 80 mm
Soil specimenPorous
stones
Spiral wire
in rubber
membrane
Direct simple shear test

71. Strength
envelope of
soil:
Strength
envelope or
failure envelope is
the combination of
normal and shear
stress at which
the soil fails.

72. Dilatancy :
Dilatancy is the volume change observed in granular materials when they are
subjected to shear deformations.This effect was first described scientifically
by Osborne Reynolds in 1885/1886 and is also known as Reynolds dilatancy.
Unlike most other solid materials, the tendency of a compacted granular material
is to dilate (expand in volume) as it is sheared. This occurs because the grains in a
compacted state are interlocking and therefore do not have the freedom to move
around one another.

73. Critical Void Ratio:
An increase or decrease of volume means change in the void ratio of
soil. In a soil mass, the ratio of the volume of the void space to the
volume of the solid particles is defined as the critical void ratio.

74. Soil liquefaction :
Soil liquefaction describe
s a phenomenon whereby
a saturated or partially
saturated soil
substantially loses
strength and stiffness in
response to an applied
stress, usually earthquake
shaking or other sudden
change in stress
condition, causing it to
behave like a liquid.