GEOTECHNICAL ENGINEERING (201003) Assistant Professor
Rajiv Kumar VISHWAKARMA INSTITUTE OF INFORMATION TECHNOLOGY
PUNE 1

UNIT 4SHEAR STRENGTH OF SOIL :

UNIT 4SHEAR STRENGTH OF SOIL 2

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Strength of different materials 3

INTRODUCTION :

INTRODUCTION Soil failure usually occurs in the form of “shearing” along internal surface within the soil.
Thus, structural strength is primarily a function of shear strength.
Shear strength is a soils’ ability to resist sliding along internal surfaces within the soil mass. 4

Mass Wasting: Shear Failure :

Mass Wasting: Shear Failure 5

Shear Failure: Earth Dam :

Shear Failure: Earth Dam 6

Shear Failure Under Foundation Load :

Shear Failure Under Foundation Load 7

Shear failure of soils :

Shear failure of soils Soils generally fail in shear At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. 8

Shear failure of soils :

Shear failure of soils Soils generally fail in shear 9

Shear failure of soils :

Retaining wall Shear failure of soils At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. Soils generally fail in shear 10

Shear failure mechanism :

Shear failure mechanism The soil grains slide over each other along the failure surface. No crushing of individual grains. 11

Shear failure mechanism :

Shear failure mechanism At failure, shear stress along the failure surface () reaches the shear strength (f). 12

What is Shear Strength? :

What is Shear Strength? Thus shear strength is “The capacity of a material to resist the internal and external forces which slide past each other” .
Shear strength in soils is the resistance to movement between particles due to physical bonds from:
Particle interlocking
Atoms sharing electrons at surface contact points
Chemical bonds (cementation) such as crystallized calcium carbonate 13

Influencing Factors on Shear Strength :

Influencing Factors on Shear Strength The shearing strength, is affected by:
soil composition: mineralogy, grain size and grain size distribution, shape of particles, pore fluid type and content, ions on grain and in pore fluid.
Initial state: State can be describe by terms such as: loose, dense, over consolidated, normally consolidated, stiff, soft, etc.
Structure: Refers to the arrangement of particles within the soil mass; the manner in which the particles are packed or distributed. Features such as layers, voids, pockets, cementation, etc, are part of the structure. 14

Shear Strength of Soil :

Shear Strength of Soil In reality, a complete shear strength formulation would account for all previously stated factors.
Soil behavior is quite complex due to the possible variables stated.
Laboratory tests commonly used:
Direct Shear Test
Unconfined Compression Testing ….etc 15

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Mohr presented in 1900 a theory of rupture of materials, that was the result of a combination of both normal and shear stresses. The shear stress at failure is thus a function of normal stress and the Mohr circle is tangential to the functional relationship given by Coulomb 18

The Mohr-Coulomb Failure Criterion: :

The Mohr-Coulomb Failure Criterion: 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” 19

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Mohr-Coulomb Failure Criterion(in terms of total stresses) :

Mohr-Coulomb Failure Criterion(in terms of total stresses) f is the maximum shear stress the soil can take without failure, under normal stress of . 21

Mohr-Coulomb Failure Criterion(in terms of effective stresses) :

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 ’. u = pore water pressure 22

Mohr-Coulomb Failure Criterion :

Mohr-Coulomb Failure Criterion Shear strength consists of two components: cohesive and frictional. 23

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Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress. t s s1 s3 c The Mohr-Coulomb failure locus is tangent to the Mohr circles at failure 24

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Mohr Circles t s s1 s3 f 2a (ta, sa) From the Mohr Circle geometry 25

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The Mohr circle construction enables the stresses acting in different directions at a point on a plane to be determined, provided that the stress acting normal to the plane is a principal stress.
The construction is useful in Soil Mechanics because many practical situations may be approximated as plane strain.
The sign convention is different to that used in Structural analysis because it is conventional to take compressive stresses positive
Sign convention: Compressive normal stresses positive
Anti-clockwise shear stresses positive (from inside element)
Angles measured clockwise are positive Mohr Circles 26

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Mohr-Coulomb criterion (Principal stresses) t s f s1 s3 c c cot f p R Failure occurs if a Mohr circle touches the failure criterion. Then
R = sin f ( p + c cot f ) 27

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Mohr-Coulomb criterion (Principal stresses) t s f s1 s3 c c cot f p R Failure occurs if a Mohr circle touches the failure criterion. Then
R = sin f ( p + c cot f ) 28

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Mohr-Coulomb criterion (Principal stresses) t s f s1 s3 c c cot f p R Failure occurs if a Mohr circle touches the failure criterion. Then
R = sin f ( p + c cot f ) 29

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Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses 30

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Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses 31

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Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses where 32

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t s s1 s3 u u Effective and total stress Mohr circles For any point in the soil a total and an effective stress Mohr circle can be drawn. These are the same size with The two circles are displaced horizontally by the pore pressure, u. 33

Mohr-Coulomb failure criterion :

Mohr-Coulomb failure criterion The limiting shear stress (soil strength) is given by
t = c + sn tan f
where c = cohesion (apparent)
f = friction angle t sn 34

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Other laboratory tests include,
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’) 39

Direct shear test :

NEED AND SCOPE
In many engineering problems such as
design of foundation,
retaining walls,
slab bridges,
pipes,
sheet piling,
The value of the angle of internal friction and cohesion of the soil involved are required for the design.
Direct shear test is used to predict these parameters quickly. Direct shear test 40

Direct shear test :

This test is performed to determine the consolidated - drained shear strength of a sandy to silty soil.
The shear strength is one of the most important engineering properties of a soil, because it is required whenever a structure is dependent on the soil’s shearing resistance.
The shear strength is needed for engineering situations such as determining the stability of slopes or cuts, finding the bearing capacity for foundations, and calculating the pressure exerted by a soil on a retaining wall. Direct shear test 41

PROCEDURE
Check the inner dimension of the soil container.
Put the parts of the soil container together.
Calculate the volume of the container. Weigh the container.
Place the soil in smooth layers (approximately 10 mm thick). If a dense sample is desired tamp the soil.
Weigh the soil container, the difference of these two is the weight of the soil. Calculate the density of the soil.
Make the surface of the soil plane.
Put the upper grating on stone and loading block on top of soil. 43

Direct shear test :

Direct shear test 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 44

Direct shear test :

Direct shear test Preparation of a sand specimen 45

Direct shear test :

Direct shear test Test procedure 46

Direct shear test :

Direct shear test Step 2: Lower box is subjected to a horizontal displacement at a constant rate 47

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PROCEDURE
Measure the thickness of soil specimen.
Apply the desired normal load.
Remove the shear pin.
Attach the dial gauge which measures the change of volume.
Record the initial reading of the dial gauge and calibration values.
Before proceeding to test check all adjustments to see that there is no connection between two parts except sand/soil.
Start the motor. Take the reading of the shear force and record the reading.
Take volume change readings till failure.
Add 5 kg normal stress 0.5 kg/cm2 and continue the experiment till failure
Record carefully all the readings. Set the dial gauges zero, before starting the experiment 48

Direct shear test :

Direct shear test 49

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Direct shear test Analysis of test results Note: Cross-sectional area of the sample changes with the horizontal displacement 50

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Direct shear tests on sands Stress-strain relationship 51

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Direct shear tests on sands How to determine strength parameters c and f 52

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Direct shear tests on sands Sand is cohesionless hence c = 0 Direct shear tests are drained and pore water pressures are dissipated, hence u = 0 Therefore,
f’ = f and c’ = c = 0 53

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Direct shear tests on clays Failure envelopes for clay from drained direct shear tests 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) 54

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Interface tests on direct 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) 55

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Advantages of direct shear apparatus 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 56

Triaxial Shear Test :

Triaxial Shear Test 57

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Triaxial Shear Test :

Triaxial Shear Test Specimen preparation (undisturbed sample) 59

Triaxial Shear Test :

Triaxial Shear Test Specimen preparation (undisturbed sample) 60

Triaxial Shear Test :

Triaxial Shear Test Specimen preparation (undisturbed sample) 61

Triaxial Shear Test :

Triaxial Shear Test Specimen preparation (undisturbed sample) 62

Types of Triaxial Tests :

Types of Triaxial Tests Is the drainage valve open? Is the drainage valve open? 63

Types of Triaxial Tests :

Types of Triaxial Tests 64

Consolidated- drained test (CD Test) :

Consolidated- drained test (CD Test) Step 1: At the end of consolidation Step 2: During axial stress increase Step 3: At failure 65

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Deviator stress (q or Dsd) = s1 – s3 Consolidated- drained test (CD Test) 66

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Volume change of sample during consolidation Consolidated- drained test (CD Test) 67

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Stress-strain relationship during shearing Consolidated- drained test (CD Test) 68

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CD tests How to determine strength parameters c and f 69

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CD tests Since u = 0 in CD tests, s = s’ Therefore, c = c’ and f = f’ cd and fd are used to denote them 70

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CD tests Failure envelopes For sand and NC Clay, cd = 0 Therefore, one CD test would be sufficient to determine fd of sand or NC clay 71

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CD tests Failure envelopes For OC Clay, cd ≠ 0 72

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Some practical applications of CD analysis for clays t = in situ drained shear strength 1. Embankment constructed very slowly, in layers over a soft clay deposit 73

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Some practical applications of CD analysis for clays 2. Earth dam with steady state seepage 74

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Some practical applications of CD analysis for clays 3. Excavation or natural slope in clay t = In situ drained shear strength Note: CD test simulates the long term condition in the field. Thus, cd and fd should be used to evaluate the long term behavior of soils 75

Consolidated- Undrained test (CU Test) :

Consolidated- Undrained test (CU Test) Step 1: At the end of consolidation Step 2: During axial stress increase Step 3: At failure 76

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Volume change of sample during consolidation Consolidated- Undrained test (CU Test) 77

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Stress-strain relationship during shearing Consolidated- Undrained test (CU Test) 78

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CU tests How to determine strength parameters c and f 79

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CU tests How to determine strength parameters c and f fcu Mohr – Coulomb failure envelope in terms of total stresses ccu Effective stresses at failure uf 80

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CU tests Shear strength parameters in terms of total stresses are ccu and fcu Shear strength parameters in terms of effective stresses are c’ and f’ c’ = cd and f’ = fd 81

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CU tests Failure envelopes For sand and NC Clay, ccu and c’ = 0 Therefore, one CU test would be sufficient to determine fcu and f’(= fd) of sand or NC clay 82

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Some practical applications of CU analysis for clays t = in situ undrained shear strength 1. Embankment constructed rapidly over a soft clay deposit 83

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Some practical applications of CU analysis for clays 2. Rapid drawdown behind an earth dam t = Undrained shear strength of clay core Core 84

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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 fcu) 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 85

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Data analysis Initial 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 × H0 A ×(H0 – DH) = A0 × H0 A ×(1 – DH/H0) = A0 86

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling = + Step 2: After application of hydrostatic cell pressure Duc = B Ds3 Note: If soil is fully saturated, then B = 1 (hence, Duc = Ds3) 87

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load Dud = ADsd = + 88

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3, Du = Duc + Dud Total pore water pressure increment at any stage, Du Du = B Ds3 + ADsd 89

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling Step 2: After application of hydrostatic cell pressure Step 3: During application of axial load Step 3: At failure 90

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) 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 91

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Mohr circles in terms of total stresses 92

Unconsolidated- Undrained test (UU Test) :

Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope S < 100% S > 100% 93

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Some practical applications of UU analysis for clays t = in situ undrained shear strength 1. Embankment constructed rapidly over a soft clay deposit 94

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Some practical applications of UU analysis for clays 2. Large earth dam constructed rapidly with no change in water content of soft clay 95

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Some practical applications of UU analysis for clays 3. Footing placed rapidly on clay deposit Note: UU test simulates the short term condition in the field. Thus, cu can be used to analyze the short term behavior of soils 96

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Unconfined Compression Test (UC Test) s1 = sVC + Ds s3 = 0 Confining pressure is zero in the UC test 97

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Unconfined Compression Test (UC Test) τf = σ1/2 = qu/2 = cu 98

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Specimens are subjected to (approximately) uniform stresses and strains
The complete stress-strain-strength behaviour can be investigated
Drained and undrained tests can be performed
Pore water pressures can be measured in undrained tests, allowing effective stresses to be determined
Different combinations of cell pressure and axial stress can be applied Advantages of the triaxial test 99

UNCONFINED COMPRESSION TEST :

UNCONFINED COMPRESSION TEST PURPOSE
The purpose of this test is to determine the unconfined compressive strength of a cohesive soil sample.
We will measure this with the unconfined compression test, which is an unconsolidated undrained (UU or Q-type) test where the lateral confining pressure is equal to zero (atmospheric pressure). 100

TEST PROCEDURE 1. Extrude the soil sample from Shelby tube sampler. Cut a soil specimen so that the ratio (L/d) is approximately between 2 and 2.5. Where L and d are the length and diameter of soil specimen, respectively.
2. Measure the exact diameter of the top of the specimen at three locations 120° apart, and then make the same measurements on the bottom of the specimen. Average the measurements and record the average as the diameter on the data sheet.
3. Measure the exact length of the specimen at three locations 120° apart, and then average the measurements and record the average as the length on the data sheet. 104

TEST PROCEDURE :

TEST PROCEDURE (4) Weigh the sample and record the mass on the data sheet.
(5) Calculate the deformation (DL) corresponding to 15% strain (e).
(6) Carefully place the specimen in the compression device and center it on the bottom plate. Adjust the device so that the upper plate just makes contact with the specimen and set the load and deformation dials to zero. 105

TEST PROCEDURE :

TEST PROCEDURE (7) Apply the load so that the device produces an axial strain at a rate of 0.5% to 2.0% per minute, and then record the load and deformation dial readings on the data sheet at every 20 to 50 divisions on deformation the dial.
(8) Keep applying the load until (1) the load (load dial) decreases on the specimen significantly, (2) the load holds constant for at least four deformation dial readings, or (3) the deformation is significantly past the 15% strain that was determined in step 5. 106

TEST PROCEDURE :

TEST PROCEDURE (9) Draw a sketch to depict the sample failure.
(10) Remove the sample from the compression device and obtain a sample for water content determination. Determine the water content as in Experiment 1. 107

Analysis :

Analysis Calibrate the dial gage reading and proving dial gage.
Compute the sample C.S.A= A0
Compute the strain e=
Compute the corrected area A’ =
Compute the sample stress sc =
Compute the water content = 108

Analysis :

Analysis Plot the stress versus strain. Show qu as the peak stress (or at 15% strain) of the test. Be sure that the strain is plotted on the abscissa. See example data.
Draw Mohr’s circle using qu from the last step and show the undrained shear strength, su = c (or cohesion) = qu/2. See the example data. 109

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112

VANE SHEAR TEST :

VANE SHEAR TEST OBJECTIVE: To find shear strength of a given soil specimen.
NEED AND SCOPE:
The structural strength of soil is basically a problem of shear strength.
Vane shear test is a useful method of measuring the shear strength of clay. It is a cheaper and quicker method. 113

VANE SHEAR TEST :

VANE SHEAR TEST The test can also be conducted in the laboratory.
The laboratory vane shear test for the measurement of shear strength of cohesive soils, is useful for soils of low shear strength (less than 0.3 kg/cm2) for which triaxial or unconfined tests can not be performed. The test gives the undrained strength of the soil.
The undisturbed and remoulded strength obtained are useful for evaluating the sensitivity of soil. 114

VANE SHEAR TEST :

VANE SHEAR TEST EQUIPMENT:
Vane shear apparatus.
Specimen.
Specimen container.
Callipers GENERAL REMARKS:
This test is useful when the soil is soft and its water content is nearer to liquid limit. 115

VANE SHEAR TEST :

VANE SHEAR TEST PROCEDURE:
1.Prepare two or three specimens of the soil sample of dimensions of at least 37.5 mm diameter and 75 mm length in specimen.(L/D ratio 2 or 3).
2.Mount the specimen container with the specimen on the base of the vane shear apparatus. If the specimen container is closed at one end, it should be provided with a hole of about 1 mm diameter at the bottom.
3.Gently lower the shear vanes into the specimen to their full length without disturbing the soil specimen. The top of the vanes should be atleast 10 mm below the top of the specimen. Note the readings of the angle of twist. 116

VANE SHEAR TEST :

VANE SHEAR TEST PROCEDURE:
4.Rotate the vanes at an uniform rate say 0.1º/s by suitable operating the torque application handle until the specimen fails.
5.Note the final reading of the angle of twist.
6.Find the value of blade height in cm.
7.Find the value of blade width in cm. 117

VANE SHEAR TEST :

VANE SHEAR TEST CALCULATIONS: 118

VANE SHEAR TEST :

VANE SHEAR TEST CALCULATIONS: 119

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Vane shear test This is one of the most versatile and widely used devices used for investigating undrained shear strength (Cu) and sensitivity of soft clays Rate of rotation : 60 – 120 per minute
Test can be conducted at 0.5 m vertical intervals 120

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Surface area of the cylinder = 2prh= pdh 121

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Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected Ms – Shaft shear resistance along the circumference T = Ms + Me + Me = Ms + 2Me 122

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Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected T = Ms + Me + Me = Ms + 2Me Me – Assuming a triangular distribution of shear strength Can you derive this ??? 123

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Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected T = Ms + Me + Me = Ms + 2Me Me – Assuming a parabolic distribution of shear strength Can you derive this ??? 124

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Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected After the initial test, vane can be rapidly rotated through several revolutions until the clay become remoulded 125

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Some important facts on vane shear test Insertion of vane into soft clays and silts disrupts the natural soil structure around the vane causing reduction of shear strength The above reduction is partially regained after some time Cu as determined by vane shear test may be a function of the rate of angular rotation of the vane 126

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Correction for the strength parameters obtained from vane shear test Bjerrum (1974) has shown that as the plasticity of soils increases, Cu obtained by vane shear tests may give unsafe results for foundation design. Therefore, he proposed the following correction. Cu(design) = lCu(vane shear) Where, l = correction factor = 1.7 – 0.54 log (PI)
PI = Plasticity Index 127

QUIZ :

QUIZ . The equation τ = C + σ tan φ is given by
a) Rankine
b) Coulomb
c) Mohr
d) Atterberg
Depending upon the properties of a material, the failure envelope may
a) be either straight or curved
b) pass through the origin of stress
c) intersect the shear stress axis
d) all the above 128

QUIZ :

QUIZ Angle between the directions of the failure and the major principal plane, is equal to
A) 90° + effective angle of shearing resistance
B) 90° + half of the angle of shearing resistance
C) 45° - half of the angle of shearing resistance
D) 45° + half of the angle of shearing resistance
The vane shear test is used for the in-situ determination of the undrained strength of the intact fully saturated
A) sands
B) clays 129

QUIZ :

QUIZ The shear box is separated in two halves in case of
a) Triaxial test
b) Unconfined Compression test
c) Direct shear test
d) All of above
The shearing strength of a cohesion-less soil depends upon
a) dry density
b) rate of loading
c) confining pressure
d) nature of loading 130

QUIZ :

QUIZ The shear strength of granular soil depends upon
a) Particle shape
b) Orientation
c) Initial void ratio
d) All of above
The pore pressure is measured accurately in case of
a) triaxial test
b) direct shear test
c) Vane shear test
d) None of above 131

QUIZ :

QUIZ The undrained shear strength parameter is….
Drainage could not be controlled in case of…
The most suitable shear test for Sensitive Clays is……………..
The most complex shear strength test is……..
The Shearing friction of soil mass is resisted by…………… c or Ø
The Property of the soil holding its particles together is………. c or Ø
The cell pressure in the triaxial cell is also known as 132

THIXOTROPY :

The word Thixotropy is derived from the two words:
Thixis meaning touch and
Tropo meaning to change.
Therefore, Thixotropy means any change that occurs by touch. THIXOTROPY 133

THIXOTROPY :

The loss of strength of a soil due to remoulding is partly due to change in the soil structure and partly due to disturbance caused to water molecules in the absorb layer.
Some are reversible. If a remoulded soil is allowed to stand, without loss of water,it may regin some of its lost strength.
In soil engineering, this gain in strength of the soil with passage of time after it has been remolded is called thixotropy. THIXOTROPY 134

THIXOTROPY :

Thixotropy of soils is of great practical importance in soil engineering. For example, when a pile is driven in to ground, the loss of strength occurs due to disturbance caused.
Thixotropy indicates how much shear strength will be regained after the pile has been driven and left in place for some time. THIXOTROPY 135

SENSITIVITY :

SENSITIVITY A cohesive soil in its natural state of occurrence has a certain structure. When the structure is disturbed, the soil becomes remoulded, and its engineering properties change consider.
Sensitivity of a soil indicates its weakening due to remoulding. It is defined as the ratio of undisturbed strength to the remoulded strength at the water content. 136

Classification of soil based on sensitivity :

Classification of soil based on sensitivity 137

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For most clays, sensitivity lies between 2and 4.
Clays considered sensitive have values between 4 and 8.
In case of sensitive clays, remoulding causes a large reduction in strength.
Quick clays are unsutible. These turn into slurry when remoulded.
High sensitivity in clays is due to a well-devloped flocculent structure which is disturbed when the soil is remoulded. 138

THANK YOU :

THANK YOU 139

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