1. Soil Shear Strength

2. Strength of a material the greatest stress it can sustain
If the stress exceeds the strength, failure occurs
Strength analysis can be performed for tensile, compressive or shear stress.
Tensile strength
A36  248 MPA

3. Strength of a material
The structural engineer define compressive strength for various material and design structural members accordingly.
The soil and rock do not fail in compression, but in shear only

4. Importance
The safety of any geotechnical structure is dependent on the shear strength of soil.
If the soil fails, a structure founded on or in it can collapse, endangering lives and causing economic damage.
North Klang Valley Expressway (Nov 2003)

5. Example
Massive slope failure at Bukit Antarabangsa on 4th December 2008
There were three people dead and 15 injured. 2,000 residents in the vicinity have been evacuated

6. Block A at Highland Towers collapsed on 11 December 1993
48 dead bodies with 30bags of human flesh

7. Question about shear strength of soil
What is meant by the shear strength of soils?
What factor affect the shear strength?
What are the assumptions in the Mohr-Coulomb failure criteria?
What are the differences between drained and undrained shear strength?
Under what conditions should the drained and undrained shear strength parameters be used?
What laboratory and field tests are used to determine shear strength?
How do I know what laboratory test to specify for a project?

8. Application
When the ground surface is inclined, gravity produces large geostatic stresses in the soil or rock.
If the stresses exceed the shear strength, a landslide occurs

9. Application
Load from a structure are transferred to the ground through structural foundations
This produces compressive and shear stress in the nearby soil.
If the shear stress exceed the shear strength, thus producing a shear failure.
This is known as bearing capacity failure and can cause the structure to collapse

10. Application
The weight of soil behind a retaining wall produces shear stresses
Its shear strength resists some of this stress and the wall resists the rest
Thus, the load carried by the wall depends on the shear strength of the retained soil.

11. Application
Tunnels in soil or weak rock normally require linings of steel or concrete for support
Such linings must resist pressures exerted by the surrounding ground, thus keeping the tunnel from collapsing.

12. Application
Wheel loads from vehicles spread through the pavement and into the ground below.
These loads produces shear stresses that could causes a shear failure.

13. Engineering Material Soil SHEAR FAILURE IN SOILS
Shear strength is controlled by their molecular structure.
Failure of these materials requires breaking the molecular bonds that hold the material together
Steel – very strong molecular bonds *» high shear strength
Plastic – weaker bonds *» lower shear strength
Soil
Shear strength primarily depends on interactions between the particles, not their internal strength.
These interaction can be divided into frictional strength and cohesive strength

14. FRICTIONAL STRENGTH
Similar to sliding friction N V
Forces resist sliding
 = coefficient of friction N
Geotechnical engineers prefer to describe using effective friction angle (’) instead of  , where:
’ = tan-1
So, Shear strength, f = (-u) tan ’= ’tan’

15. Factors influence friction angle ()
FRICTIONAL STRENGTH
Factors influence friction angle ()
Mineralogy
Shape
Gradation
Void ratio
Organic Material
Mineralogy – Soil includes many different minerals. The friction angle in Sands of pure quartz is typically . However, the presence of mica given smaller ’.
Shape – angular particles is much higher than rounded ones
Gradation – Well graded soil have more interlocking between particles than poorly graded
Void ratio – decreasing the void ratio (eg. by compaction) increase interlocking and give higher ’.
Organic material – the presence of organic material will decrease the friction angle.

16. COHESIVE STRENGTH
Some soils have shear strength even when the effective stress, ’ is zero.
If a soil has both frictional and cohesive strength, the equation becomes:
f = c’ + ’tan’
Where:
f = Shear strength
c’ = Effective cohesion
’= Effective stress on shear surface
’= Effective friction angle

17. Factors influence cohesive strength (c)
Cementation
Electrostatic
Negative pore water pressures
Negative excess pore water pressures due to dilation
Cementation – Chemical bonding due to the presence of cementing agents (CaCO3, Fe2O3)
Electrostatic – can hold particles together, forces very small
Negative pore water pressures – presence in soils above ground water table. Example, moist unsaturated sands can stand in vertical cut
Negative excess pore water pressures– some soil tend to dilate or expand when sheared. Dilation draws water into voids. When the rate of shearing is rapid than the rate at which water can flow, large negative excess pore water pressures can develop in the soil

18. Mohr-Coulomb Failure Criterion 
friction angle
failure envelope f 
cohesion c 
f is the maximum shear stress the soil can take without failure, under normal stress of .

19. Mohr-Coulomb Failure Criterion
Shear strength consists of two components: cohesive and frictional.    c
frictional component
cohesive component f f
f tan  c

20. c and  are measures or parameters of soil shear strength.
Higher the values, higher the shear strength.
c’ = 0
 = 0   c      c 
overconsolidated clays
Sand
Inorganic silt
Normally consolidated clays

21. FRICTIONAL STRENGTH EXAMPLE:
Compute the shear strength at point A along surface when the groundwater table is at level B, then compute the new shear strength if it rose to level C. Given the unit weight of soil is 18.8 kN/m3 above water table and 19.3 kN/m3 below
c = 0
 = 30
4.3 m C
3.7 m B
7 m A

22. Mohr Circles & Failure Envelope Y X X  Y
Soil elements at
different locations X
~ failure Y
~ stable

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

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

25. How to determine the shear strength of soil ?

26. How to determine the shear strength of soil ?

27. Determination of the shear strength of soil
Laboratory Test
Field Test
Vane shear test
Standard Penetration test (SPT)
Mackintosh/JKR probe test
Dutch cone penetrometer
Piezocone
Direct shear test
Triaxial Compression test
Unconfined compression test
Vane shear test

28. Direct shear test

29. Direct shear test
(Max 1000 kN/m2)

30. Direct shear test
Can be performed using two test methods:
i) stress controlled
ii) strain controlled
Stress control – the shear force is applied in equal increments (Example: reading the displacement gauge every increment 0.01 kN/m2 of shear force)
Strain controlled - a constant shear rate is applied
(Example: reading the shear force (proving ring) for every 5 mm displacement)

31. Shear stress and change in the height of specimen versus strain()

32. Example of the results of direct shear test for a dry sand

33. Test on saturated sand and clay
the container will be filled with water to saturate the specimens.
A drained test will be performed and the rate of loading must be slow enough to avoid build up of pore water pressure
The rate of loading for clay must be much slower than sand due to its permeability properties

34. Advantage and disadvantage of direct shear test
Simple and quick test
Most economical for dry and saturated sand
Very suitable to measure the angle of friction between soil and foundation (concrete, steel, rock, wood)
1. The specimen is forced to fail along the plane of split box not to the weakest plane
2. Can not measure the pore water pressure during the test

35. Example
A dry sand sample is subjected to a normal stress ’ = 20 kN/m2 in a direct shear test. Calculate the shear force at failure if the soil sample is 10 cm x 10cm in plane and 2.5 cm in height. The strength parameters of the sand are c’=0 and ’ = 38.
Solution:
At failure, the applied shear stress = shear strength of soil
So,

36. Triaxial shear test
Most reliable methods for determining the shear strength parameters (c’, ’)
provides stress-strain behaviour
uniform stress condition
more flexibility in loading path
A soil specimen – 38mm in diameter and 76 mm long
- 50mm in diameter and 100 mm long
- 100mm in diameter and 200 mm long
Compression medium – water or air

37. Triaxial Apparatus

38. Triaxial Apparatus

39. Types of Triaxial Tests
Depending on whether drainage is allowed or not during
initial isotropic cell pressure application, and
shearing,
there are three special types of triaxial tests that have practical significances. They are:
Consolidated Drained (CD) test
Consolidated Undrained (CU) test
Unconsolidated Undrained (UU) test

40. Types of Triaxial Tests
deviatoric stress ()
Consolidation stage
Under all-around cell pressure 3
Shearing (loading)
Is the drainage valve open?
Is the drainage valve open?
yes no
yes no
Consolidated sample
Unconsolidated sample
Drained loading
Undrained loading

41. For unconsolidated undrained test, in terms of total stresses, u = 0
For normally consolidated clays, c’ = 0 & c = 0.
Granular soils have no cohesion.
c = 0 & c’= 0

42. CD, CU and UU Triaxial Tests
Consolidated Drained (CD) Test
no excess pore pressure throughout the test
very slow shearing to avoid build-up of pore pressure
Can be days!
 not desirable
gives c’ and ’
Use c’ and ’ for analysing fully drained
situations (e.g., long term stability,
very slow loading)

43. CD, CU and UU Triaxial Tests
Consolidated Undrained (CU) Test
pore pressure develops during shear
Measure  ’
gives c’ and ’
faster than CD (preferred way to find c’ and ’)

44. Consolidated Drained Test
Loose sand and normally consolidated clay
Dense sand and over consolidated clay
Deviator stress
Deviator stress

45. Consolidated Drained Test

46. Consolidated Drained Test

47. Example
Try to understand :-
Example 10.2
Example 10.3
Example 10.4

48. Example of CD test
Find the friction angle, ’ for sand sample in CD test if the confining pressure, 3 is 19 kN/m2 and deviator stress,  = 48 kN/m2.
Answers
34
36
38
40

49. Example of CD test
Solution 1

50. Example of CD test
Solution 2
For sand, c=0 

51. Consolidated undrained test
Loose sand and normally consolidated clay
Deviator stress
Dense sand and over consolidated clay

52. Consolidated undrained test

53. Consolidated undrained test

54. Unconsolidated undrained test

55. Unconfined Compression Test
No confining pressure, 3

56. Unconfined Compression Test
The shear strength can be calculated using this equation
qu = unconfined compression strength

57. Unconfined Compression Test

58. Sensitivity and Thixotropy of clay
Sensitivity is the reducing of the strength due to remolding even without changing the moisture content.
The degree of sensitivity can be calculated using this equation
Type of clay St
Most clays
1 to 8
Flocculent marine clay
10 to 80

59. Sensitivity and Thixotropy of clay
Thixotropy is the increasing of the strength with time after remolding and is kept in undisturbed state.
But most soils are partially thixotropy

60. Vane Shear Test Apparatus
Degree of rotation
Torque
Spring 1
Spring 2
Spring 3
Spring 4
Cu=T / [πd2(h/2 + d/6)]
Cu      Undrained shear strength of the soil T        Maximum torque at failure h        height of the vane d        diameter of the vane

61. Vane Shear Test At Site
To estimate the undrained shear strength of the soil
The blade is rotated at a specified rate that should not exceed 1degree every 10sec).

62. Vane Shear Test At Site Device that measures the required Torque
The amount of rotation

63. Vane Shear Test At Site

64. SPT Test At Site
This is a dynamic test as described in BS1377 (Part 9) and is a measure of the density of the soil.
The test incorporates a small diameter tube with a cutting shoe known as the 'split barrel sampler' of about 650mm length, 50mm external diameter and 35mm internal diameter.
The sampler is forced into the soil dynamically using blows from a 63.5kg hammer dropped through 760mm. The sampler is forced 150mm into the soil then the number of blows required to lower the sampler each 75mm up to a depth of 300mm is recorded. This is known as the "N" value.
For coarse gravels the split barrel is replaced by a 60 degree cone.

65. SPT Test At Site

66. Dutch cone penetrometer Test

67. Dutch cone penetrometer Test

69. Cone Penetration Testing (CPT)
Real-Time readings in computer screen
Penetration at 2 cm/s
Sand
Clay
Buried Crust

70. Characteristics of the failure plane
Shear stress, f
Strength Envelope D
180-2θ θ  c 2θ θ A B 3 C
Normal stress,  1

71. Characteristics of the failure plane
(180-2θ) + 90 +  = 180
Therefore, θ = 45 + (/2)
Sin  = DC/AC
Large Triangle

72. Characteristics of the failure plane
Normal stress and shear stress on the failure plane can be calculated using this equations.
Normal stress, n =
Shear stress, f =

73. 1- 3 Relation at FailureX
soil element at failure X 3 1 3 1

74. Stress Point v h X  t stress point stress point v h  s (v-h)/2

75. Stress Path During loading…  t  s
Stress path is the locus of stress points   t s
Stress path
Stress path is a convenient way to keep track of the
progress in loading with respect to failure envelope.

76. Failure Envelopes  t  s During loading (shearing)…. failure c 
tan-1 (sin )
stress path
c cos 
During loading (shearing)….

77. Shear force at failure (N)
Example 3.1
Direct shear tests were performed on a dry, sandy soil. The size of the specimen was 50 mm x 50 mm x 20 mm. Test results were as given in the table.
Test no.
Normal force (N)
Shear force at failure (N) 1 90 54 2
135
82.35 3
315
189.5 4
450
270.5
Find the shear stress parameters.

78. Example 3.1 Solution f = shear force / area of specimen
Test no.
Normal force (N)
Normal stress,  (kN/m2)
Shear force at failure (N)
Shear stress at failure, f (kN/m2) 1 90 36 54
21.6 2
135
82.35
32.9 3
315
126
189.5
75.8 4
450
180
270.5
108.2
= normal force / area of specimen
= normal force x 10-3 kN/ 50 x 50 x 10-6 m2
f = shear force / area of specimen
f = shear force x 10-3 kN/ 50 x 50 x 10-6 m2

79. Example 3.1
Solution
The shear stress parameters
c = 0 dan  = 31

80. Example 3.2
For a normally consolidated clay, these are the results of a drained triaxial test:
Chamber confining pressure = 112 kN/m2
Deviator stress at failure = 175 kN/m2
(a) Find the angle of friction, .
(b) Determine the angle θ that the failure plane makes with
the major principal plane.
Find the normal stress, ’, and the shear stress, f on the failure plane.
Determine the effective normal stress on the plane of maximum shear stress.

81. Example 3.2
Solution
For a normally consolidated soil, the failure envelope equation is f = ’ tan  ( c = 0)
For triaxial test, the effective major and minor principle stress at failure are:
3 = 112 kN/m2
1- 3 = 175 kN/m2, so, 1 = = 287 kN/m2

82. Example 3.2 Solution (a) (b) θ 3  1 B 2θ O A Shear Stress (kN/m2)
3 = 112
1 = 287
Normal Stress (kN/m2)
Shear Stress (kN/m2) A O B  2θ
(a)
(b)

83. Example 3.2
(c) ’ (on the failure plane )
Shear stress, f
(d) The maximum shear stress will occur on the plane with θ = 45.

84. Example 3.3 Solution = 210 – 70 = 140 kN/m2
The equation of the effective stress failure envelope for normally consolidated clayey soil is f = ’ tan 30. A drained triaxial test was conducted with the same soil at a chamber confining pressure of 70 kN/m2. Calculate the deviator stress at failure.
Solution
Deviator stress = 1’ - 3’
= 210 – 70 = 140 kN/m2