Shear Strength of Soils

Transcription

1 Shear Strength of Soils

2 Soil strength Most of problems in soil engineering (foundations, slopes, etc.) soil withstands shear stresses. Shear strength of a soil is defined as the capacity to resist shear stresses, i.e., maximum value of shear stress that can be mobilized within a soil mass Soils are essentially frictional materials the strength depends on the applied stress Strength is controlled by effective stresses water pressures are required Soil strength depends on drainage different strengths will be measured for a given soil that (a) deforms at constant volume (undrained) and (b) deforms without developing excess pore pressures (drained)

3 Soil Strength Shearing strength of soil constitute basically the following components Structural resistance to displacement of the soil due to interlocking of the particle Frictional resistance of particles at their contact points Cohesion or adhesion between the surface of the soil particles

4 Soil Strength The shearing strength in cohesionless soils is due to inter granular friction alone, while For cohesive soils it is due to friction and cohesion both.

5 Stresses at a point Through every point in a stressed body there are three planes at right angles to each other which are unique as compared to all the other planes passing through the point, These planes are subjected only to normal stresses with no accompanying shearing stresses acting on the planes, These three planes are called principal planes, and the normal stresses acting on these planes are principal stresses. Once principal stresses are evaluated, the stresses on any other plane through the point can be determined

6 Stresses at a point Objective Figure 10.1

7 STRESSES AT A POINT The sum of all the forces in the x-direction is x dx tan xy dx dxseccos dxsecsin 0 Eq The sum of all the forces in the y-direction is ydx xy dx tan dxsecsin dxseccos 0 Eq Solving the two equations y y 2 2 x x y 2 x sin 2 xy cos2 cos2 xy sin 2 Eq Eq. 10.4

8 STRESSES AT A POINT By definition, a principal plane is one on which the shearing stress is equal to zero. Therefore, when τ is made equal to zero in Eq. (10.4), the orientation of the principal planes is defined by the relationship 2xy tan 2 Eq y x Equation (10.5) indicates that there are two principal planes through the point P in Fig. 10.1(a) and that they are at right angles to each other. By differentiating Eq. (10.3) with respect to α, and equating to zero, we have d d y sin2 tan 2 x 2 sin2 2 xy cos2 0 Eq xy Eq y x Equation 10.7 indicates the orientation of the planes on which the normal stresses, σ are maximum and minimum. This orientation coincides with Eq. (10.5). Therefore, it follows that the principal planes are also planes on which the normal stresses are maximum and minimum.

9 STRESSES AT A POINT and n n y y 2 2 x x y 2 x sin 2 xy cos2 cos2 The major and minor principal stress are xy sin 2 1 y 2 x y 2 x 2 2 xy Eq y 2 x y 2 x 2 2 xy Eq. 10.9

10 STRESSES AT A POINT When normal stresses are principal stresses, σ 1 and σ 3, i.e. shear forces are zero on these planes, they become principal stresses cos2 Eq and sin 2 Eq Figure 10.2

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12 Mohr s (1900) Graphical Representation of Stress From equations and 10.11, (squaring and adding) Equation of circle with coordinates, (σ 1 +σ 2 )/2, τ=0 and whose radius is (σ 1 -σ 3 )/2 General case: Center coordinates

13 Mohr s Graphical Representation of Stress Figure 10.3 Mohr s circular representation for general case A line drawn through the pole intersects the Mohr circle at a point which represents the state of stress on a plane which has the same inclination in space as the line itself Pole or Origin of planes 1. Compressive stresses are positive and tensile stresses are negative. 2. Shear stresses are considered as positive if they give a clockwise moment about a point above the stressed plane, otherwise negative.

14 Mohr s Graphical Representation of Stress Figure 10.4 Mohr s circular representation for cylindrical specimen

15 MOHR S (1900) GRAPHICAL REPRESENTATION OF STRESS 1. 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. 2. The orientation of the principal planes if the normal and shear stresses on the surface of the prismatic element (Fig. 8.2) are known. 3. The relationships are valid regardless of the mechanical properties of the materials since only the considerations of equilibrium are involved. 4. The following facts are clear from these curves: The greatest and least principal stresses are respectively the maximum and minimum normal stresses on any plane through the point in question. The maximum shear stress occurs on planes at 45 to the principal planes.

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18 Concept of Failure in soils State of failure in a soil mass upon shearing requires 1. Identifying the point on the stress-strain curve which corresponds to failure, and 2. Identifying the plane on which such a failure has occurred

19 Stress-strain relationships Most materials have a peculiar stress-strain behaviour Linearly elastic e.g. steel Non-linear elastic Visco-elastic time dependent (most soils are visco-elastic) Total deformation is sum of elastic and viscous deformations Cohesive soils at very low water content brittle (easy to identify point of failure) Dense sands and sensitive clays work-softening materials (stress decreases after the materials are strained beyond the peak stress) Loose sands and compact clays - work-hardening materials (stress increases after a strain limit) No definite yield point Strain at yield point maybe so large that structure supported ceases to perform satisfactorily

20 Best engineering way is to define failure at some acceptable % strain (engineering judgment) 5% strain for a subgrade material below a pavement 15-20% strain for an embankment dam material Identifying failure plane? Plane on which shear stress is maximum, or Plane of maximum obliquity (τ/σ) Mohr-Coulomb failure criterion most commonly used for soils Mohr s hypothesis + Coulomb law

21 Mohr s failure hypothesis The Mohr theory is based on the postulate that a material will fail when the shearing stress on the plane along which the failure is presumed to occur is a unique function of the normal stress acting on that plane. The material fails along the plane only when the angle between the resultant of the shearing stress and the normal stress is a maximum, that is, where the combination of normal and shearing stresses produces the maximum obliquity angle δ.

22 Coulomb law n The limiting shear stress (soil strength) is given by where c = cohesion (apparent) f = friction angle = c + n tan f

23 Assumed failure plane Mohr circle which is tangential to the shear strength line is called the Mohr circle of rupture Figure 10.4 Diagram for Mohr s rupture theory

24 The shear strength line M 0 N which is tangent to Mohr circle 3 is called the Mohr envelope or line of rupture. The Mohr envelope may be assumed as a straight line although it is curved under certain conditions. The Mohr circle which is tangential to the shear strength line is called the Mohr circle of rupture. Thus the Mohr envelope constitutes a shear diagram and is a graph of the Coulomb equation for shearing stress. This is called the Mohr-Coulomb Failure Theory.

25 Mohr-Coulomb failure criterion The parameters c, f are in general not soil constants. They depend on the initial state of the soil (OCR or I d ) the type of loading (drained or undrained) The Mohr-Coulomb criterion is an empirical criterion, and the failure locus is only locally linear. Extrapolation outside the range of normal stresses for which it has been determined is likely to be unreliable.

26 Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress. 3 1

27 Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress. c tan f c 3 1 The Mohr-Coulomb failure locus is tangent to the Mohr circles at failure

28 Mohr Circles (, ) f From the Mohr Circle geometry ( 1 3 ) ( 1 3 ) cos2 2 2 ( 1 3 ) sin f/ 2

29 Mohr-Coulomb criterion (Principal stresses) R f c c cot f 3 p 1 Failure occurs if a Mohr circle touches the failure criterion. Then R = sin f ( p + c cot f )

30 Mohr-Coulomb criterion (Principal stresses) R f c p c cot f 3 1 Failure occurs if a Mohr circle touches the failure criterion. Then R = sin f ( p + c cot f ) c + c cot cot f f = sin f sin f = 4 + f 2 2 tan = N f

31 Mohr-Coulomb criterion (Principal stresses) R f c p c cot f 3 1 Failure occurs if a Mohr circle touches the failure criterion. Then R = sin f ( p + c cot f ) c + c cot cot f f = sin f sin f = 4 + f 2 2 tan = N f = N + 2 c N 1 f 3 f

32 Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u n tanf u c u and f u are known as the undrained strength parameters

33 Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u n tanf u c u and f u are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil.

34 Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u n tanf u c u and f u are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil. The undrained strength is only relevant in practice to clayey soils that in the short term remain undrained. Note that as the pore pressures are unknown for undrained loading the effective stress failure criterion cannot be used.

35 Effective stress Mohr-Coulomb criterion c and f As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses c' tan f' n

36 Effective stress Mohr-Coulomb criterion c and f As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses c' tan f' = N + 2 c N n 1 f 3 f

37 Effective stress Mohr-Coulomb criterion c and f As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses c' tan f' = N + 2 c N n 1 f 3 f where N f 1 f sin 1 sin f n n u 1 1 u 3 3 u

38 Effective and total stress Mohr circles u u 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.

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41 Measurement of Shear Strength The measurement of shear strength of soil involves certain tests observations at failure with the help of which the failure envelope or strength envelope can be plotted corresponding to a given set of conditions. Shearing resistance can be determined in the lab by the following methods. o Direct shear (IS: 2720 (Part 13) 1986) o Triaxial (IS: 2720 (Part 11 & 12) 1993) o Unconfined compression test (IS: 2720 (Part 10): 1991) o Vane shear test Depending upon the drainage conditions, there are three types of shear tests.

42 o o o Undrained Test or Quick Tests: No drainage of water is permitted. Hence no dissipation of pore water pressure during the test. Drained Test: Drainage is permitted throughout the tests during the application of both normal and shearing loading so that full consolidation occurs and no excess pore pressure is set up at any stage of the test. Consolidated Undrained Test: Drainage is permitted under the initially applied normal stress only and fully primary consolidation or softening is allowed to take place. No drainage is allowed afterwards.

43 Laboratory tests Field conditions A representative soil sample vc z vc + D z hc hc hc hc vc vc + D Before construction After and during construction

44 Laboratory tests Simulating field conditions in the laboratory 0 vc hc vc + D vc + D hc 0 0 hc hc vc 0 vc Representative soil sample taken from the site Step 1 Set the specimen in the apparatus and apply the initial stress condition vc Step 2 Apply corresponding stress conditions the field

45 Schematic diagram of the direct shear apparatus (IS : 2720 (Part 13) 1986)

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

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

48 Test procedure Porous plates P Steel ball Pressure plate S Proving ring to measure shear force Step 1: Apply a vertical load to the specimen and wait for consolidation

49 Test procedure Porous plates P Steel ball Pressure plate S Proving ring to 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

50 Direct shear test Shear box Dial gauge to measure vertical displacement Proving ring to measure shear force Loading frame to apply vertical load Dial gauge to measure horizontal displacement

51 Direct shear test Analysis of test results Normal stress Area of Normal force (P) cross section of the sample Shear stress Shear resistance developedat the sliding surface (S) Area of cross section of the sample Note: Cross-sectional area of the sample changes with the horizontal displacement

52 Direct shear tests on sands How to determine strength parameters c and f Shear stress, f3 f2 Normal stress = 3 Normal stress = 2 f1 Normal stress = 1 Shear displacement Shear stress at failure, f f Normal stress, Mohr Coulomb failure envelope

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54 Direct shear tests on sands Some important facts on strength parameters c and f of sand 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

55 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) P Soil S Foundation material f c a ' tan Where, c a = cohesion, = angle of internal friction

56 Shear box test - advantages Easy and quick test for sands and gravels. Large samples may be tested in large shear boxes. Small samples may give misleading results due to imperfections (fractures and fissures) or the lack of them. Samples may be sheared along predetermined planes. This is useful when the shear strengths along fissures or other selected planes are required. Useful for determining strength parameters of freely draining soils.

57 Shear box test - disadvantages Non-uniform deformations and stresses in the specimen. The stress-strain behaviour cannot be determined. The estimated stresses may not be those acting on the shear plane. There is rotation of principal planes between start of test and failure of the soil. Hence, field loading condition cannot be simulated accurately in laboratory sample. There is no means of estimating pore pressures, so effective stresses cannot be determined from undrained tests. Directions of the principle planes are known only at failure, i.e. only after failure, Mohr envelope is known.

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59 Triaxial Shear Test Piston (to apply deviatoric stress) Failure plane O-ring Soil sample at failure Perspex cell Soil sample impervious membrane Porous stone Water Cell pressure Back pressure pedestal Pore pressure or volume change

60 Triaxial Shear Test Specimen preparation (undisturbed sample) Edges of the sample are carefully trimmed Setting up the sample in the triaxial cell

61 Triaxial Shear Test Specimen preparation (undisturbed sample) Sample is covered with a rubber membrane and sealed Cell is completely filled with water

62 Triaxial Shear Test Specimen preparation (undisturbed sample) Proving ring to measure the deviator load Dial gauge to measure vertical displacement In some tests

63 Types of Triaxial Tests Step 1 c Step 2 deviatoric stress (D = q) c c c c c Under all-around cell pressure c Shearing (loading) c + q Is the drainage valve open? Is the drainage valve open? yes no yes no Consolidated sample Unconsolidated sample Drained loading Undrained loading

64 Types of Triaxial Tests Step 1 Step 2 Under all-around cell pressure c Shearing (loading) Is the drainage valve open? Is the drainage valve open? yes no yes no Consolidated sample Unconsolidated sample Drained loading Undrained loading CD test UU test CU test

65 Advantages of the triaxial test 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

66 Consolidated- drained test (CD Test) Total, = Neutral, u + Effective, Step 1: At the end of consolidation VC VC = VC Drainage hc 0 hc = hc Step 2: During axial stress increase VC + D V = VC + D = 1 Drainage hc 0 h = hc = 3 Step 3: At failure VC + D f Vf = VC + D f = 1f Drainage hc 0 hf = hc = 3f

67 Consolidated- drained test (CD Test) 1 = VC + D 3 = hc Deviator stress (q or D d ) = 1 3

68 Compression Volume change of the sample Expansion Consolidated- drained test (CD Test) Volume change of sample during consolidation Time

69 Deviator stress, Dd Compression Volume change of the sample Expansion Consolidated- drained test (CD Test) Stress-strain relationship during shearing (D d ) f (D d ) f Dense sand or OC clay Loose sand or NC Clay Axial strain Dense sand or OC clay Axial strain Loose sand or NC clay

70 CD tests How to determine strength parameters c and f (D d ) fc 1 = 3 + (D d ) f Deviator stress, Dd (D d ) fb (D d ) fa Confining stress = 3c Confining stress = 3b Confining stress = 3a 3 Shear stress, Mohr Coulomb failure envelope 3a 3b (D d ) fa Axial strain 3c 1a f 1b 1c or (D d ) fb

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

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

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

74 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, c d and f d should be used to evaluate the long term behavior of soils

75 Consolidated- Undrained test (CU Test) Total, = Neutral, u Effective, Step 1: At the end of consolidation VC + VC = VC Drainage hc 0 hc = hc Step 2: During axial stress increase VC + D V = VC + D ± Du = 1 No drainage hc ±Du h = hc ± Du = 3 Step 3: At failure VC + D f Vf = VC + D f ± Du f = 1f No drainage hc ±Du f hf = hc ± Du f = 3f

76 Deviator stress, Dd - Du + Consolidated- Undrained test (CU Test) Stress-strain relationship during shearing (D d ) f (D d ) f Dense sand or OC clay Loose sand or NC Clay Axial strain Loose sand /NC Clay Axial strain Dense sand or OC clay

77 CU tests How to determine strength parameters c and f (D d ) fb 1 = 3 + (D d ) f Shear stress, Deviator stress, Dd (D d ) fa Mohr Coulomb failure envelope in terms of total stresses Confining stress = 3b Confining stress = 3a Axial strain f cu 3 Total stresses at failure c cu 3a 3b 1a 1b or (D d ) fa

78 CU tests How to determine strength parameters c and f 1 = 3 + (D d ) f - u f Shear stress, Mohr Coulomb failure envelope in terms of effective stresses Mohr Coulomb failure envelope in terms of total stresses u f f 3 = 3 - u f Effective stresses at failure f cu C u fb c u cu fa 3b 1b 3a 3b 3a 1a (D d ) fa 1a 1b or

79 CU tests Strength parameters c and f obtained from CU 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

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

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

82 Some practical applications of CU analysis for clays 3. Rapid construction of an embankment on a natural slope = In situ undrained shear strength 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

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84 Unconsolidated- Undrained test (UU Test) Data analysis Initial specimen condition Specimen condition during shearing No drainage C = 3 C = 3 No drainage 3 + D d 3 Initial volume of the sample = A 0 H 0 A0 A 1 Volume of the sample during shearing = A H Since the test is conducted under undrained condition, z A H = A 0 H 0 A (H 0 DH) = A 0 H 0 A (1 DH/H 0 ) = A 0

85 Unconsolidated- Undrained test (UU Test) Total, = Neutral, u Effective, + Mohr circles in terms of total stresses Failure envelope, f u = 0 c u u b u a 3a 3b 1a 1b 3 Df 1 or

86 Significance of undrained strength parameters It is often found that a series of undrained tests from a particular site give a value of f u that is not zero (c u not constant). If this happens either the samples are not saturated, or the samples have different moisture contents If the samples are not saturated analyses based on undrained behaviour will not be correct The undrained strength c u is not a fundamental soil property. If the moisture content changes so will the undrained strength.

87 Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope S < 100% S > 100% 3c 3b 1c 3a 1b 1a or

88 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa.

89 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = ( ) ( )

90 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = Failure criterion ( ) ( ) = N + 2 c N 1 f f

91 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = Failure criterion Hence 1 = 57.4 kpa, 3 = 22.4 kpa and cell pressure (total stress) = 3 + u = 65.4 kpa = N + 2 c N 1 f 3 f ( ) ( )

92 Graphical solution

93 Graphical solution

94 Graphical solution

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

96 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

97 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, c u can be used to analyze the short term behavior of soils

98 Unconfined Compression Test The unconfined compression test is a special case of a triaxial compression test in which the confining pressure σ3 = 0 The tests are carried out only on saturated samples which can stand without any lateral support. The test, is, therefore, applicable to cohesive soils only.

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102 STRESS PATH: Stress Invariants (p and q) p (or s) = ( )/2 q (or t) = ( 1-3 )/2 ( 1-3 )/2 f c 3 1 ( )/2 p and q can be used to illustrate the variation of the stress state of a soil specimen during a laboratory triaxial test

103 Mohr Coulomb failure envelope in terms of stress invariants p (or s) = ( )/2 q (or t) = ( 1-3 )/2 2 ' 2 ' ' ' 3 ' 1 ' 3 ' 1 f Cotf Sin c ( ) ( ) ' ' ' 2 2 ' 3 ' 1 ' 3 ' 1 f f Cos Sin c 3 1 ( )/2 ( 1-3 )/2 c f ' ' ' f f Cos c psin q

104 In-situ shear tests Vane shear test (suitable for soft to stiff clays) Torvane Pocket Penetrometer Pressuremeter Static Cone Penetrometer test (Push Cone Penetrometer Test, PCPT) Standard Penetration Test, SPT

105 Vane shear test This is one of the most versatile and widely used devices used for investigating undrained shear strength (C u ) and sensitivity of soft clays Applied Torque, T Bore hole (diameter = D B ) Disturbed soil Rupture surface h > 3D B ) Vane T H Vane PLAN VIEW D Rate of rotation : per minute Test can be conducted at 0.5 m vertical intervals

106 Vane shear test T = M s + M e + M e = M s + 2M e M e Assuming a uniform distribution of shear strength C u C u C u h Since the test is very fast, Unconsolidated Undrained (UU) can be expected d 2 M e (2rdr). C 0 M M e e u r d 2 2 2C u r dr 2 0 d/2 2C 3 u d 8 3 d/2 C u Cud 12 3 r 3 3 d 2 0

107 Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected C u C u M s Shaft shear resistance along the circumference 2 2 d 2 h C d dhc M u u s d C h d C T u u d h d C T u d h d T C u T = M s + M e + M e = M s + 2M e

108 Vane shear test After the initial test, vane can be rapidly rotated through several revolutions until the clay become remoulded C u h peak ultimate C u Shear displacement Since the test is very fast, Unconsolidated Undrained (UU) can be expected Sensitivity Peak Stength Ultimate Stength

109 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 C u as determined by vane shear test may be a function of the rate of angular rotation of the vane

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

111 In-situ shear tests Vane shear test Torvane (suitable for very soft to stiff clays) Pocket Penetrometer Pressuremeter Static Cone Penetrometer test (Push Cone Penetrometer Test, PCPT) Standard Penetration Test, SPT

112 Torvane Torvane is a modification to the vane

113 In-situ shear tests Vane shear test Torvane Pocket Penetrometer (suitable for very soft to stiff clays) Pressuremeter Static Cone Penetrometer test (Push Cone Penetrometer Test, PCPT) Standard Penetration Test, SPT

114 Pocket Penetrometer Pushed directly into the soil. The unconfined compression strength (q u ) is measured by a calibrated spring.

115 Various correlations for shear strength For NC clays, the undrained shear strength (c u ) increases with the effective overburden pressure, 0 c u ' ( PI) Skempton (1957) Plasticity Index as a % For OC clays, the following relationship is approximately true c u ' 0 Overconsolidated c u ' 0 Normally Consolidated (OCR) 0.8 Ladd (1977) For NC clays, the effective friction angle (f ) is related to PI as follows Sinf' log( IP) Kenny (1959)

116 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 unsaturated soils in tropical countries like Sri Lanka Water Pore water pressure, u Air Water Pore air pressure, u a Pore water pressure, u w Solid Effective stress, Solid Effective stress, Saturated soils Unsaturated soils Pore water pressure can be negative in unsaturated soils

117 Shear strength of partially saturated soils Bishop (1959) proposed shear strength equation for unsaturated soils as follows f c' ( u ) ( u u ) tan ' n a a w f Where, n u a = Net normal stress u a u w = 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 u a ) tanf' ( u a u w b ) tanf Where, tanf b = Rate of increase of shear strength with matric suction

118 Shear strength of partially saturated soils f c' ( n u a ) tanf' ( u a u w ) b tanf Same as saturated soils Apparent cohesion due to matric suction Therefore, strength of unsaturated soils is much higher than the strength of saturated soils due to matric suction f - u a

119 How it become possible build a sand castle f c' ( n u a ) tanf' ( u a u w ) b tanf Same as saturated soils Apparent cohesion due to matric suction f Apparent cohesion - u a

120 Skempton s Pore Pressure Parameters During cell Pressure During shearing load Du c = B D 3 Du d = ABD d Total pore water pressure increment at any stage, Du Du = Du c + Du d Du = B [D 3 + AD d ] Du = B [D 3 + A(D 1 D 3 )] Skempton s water equation pore pressure

121 Skempton s Pore Pressure Parameters

122 Typical values for parameter B

123 Skempton s Pore Pressure Parameters

124 Skempton s Pore Pressure Parameters

125 Typical values for parameter A

126 Skempton s Pore Pressure Parameters -Example

127 Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage.

128 Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown 3 1

129 Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown 3 1 We find that all the total stress Mohr circles are the same size, and therefore f u = 0 and = s u = c u = constant

130 Relation between effective and total stress criteria Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle c' tan f' n

131 Relation between effective and total stress criteria Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle c' tan f' n We have the following relations c u = N + 2 c N 1 f 3 f

132 Relation between effective and total stress criteria The different total stress Mohr circles with a single effective stress Mohr circle indicate that the pore pressure is different for each sample. As discussed previously increasing the cell pressure without allowing drainage has the effect of increasing the pore pressure by the same amount (Du = D r ) with no change in effective stress. The change in pore pressure during shearing is a function of the initial effective stress and the moisture content. As these are identical for the three samples an identical strength is obtained.

133 Significance of undrained strength parameters It is often found that a series of undrained tests from a particular site give a value of f u that is not zero (c u not constant). If this happens either the samples are not saturated, or the samples have different moisture contents If the samples are not saturated, analyses based on undrained behaviour will not be correct The undrained strength c u is not a fundamental soil property. If the moisture content changes so will the undrained strength.

134 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa.

135 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = ( ) ( )

136 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = Failure criterion ( ) ( ) = N + 2 c N 1 f f

137 Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kpa. Determine the cell pressure used in the test if the effective strength parameters are c = 0, f = 26 o and the pore pressure at failure is 43 kpa. Analytical solution Undrained strength = 17.5 = Failure criterion Hence 1 = 57.4 kpa, 3 = 22.4 kpa = N + 2 c N and cell pressure (total stress) = 3 + u = 65.4 kpa ( ) ( ) 1 f f

138 Graphical solution

139 Graphical solution

140 Graphical solution

141 Stress-Strain Behaviour of Soils

142 Stress-strain response of soils Triaxial tests are the standard means of investigating the stress-strain-strength response of soils. To simplify the presentation only simple shear tests will be considered.

143 Stress-strain response of soils Triaxial tests are the standard means of investigating the stress-strain-strength response of soils. To simplify the presentation only simple shear tests will be considered. The simple shear test is an improved shear box test which imposes more uniform stresses and strains.

144 Stress-strain response of soils Triaxial tests are the standard means of investigating the stress-strain-strength response of soils. To simplify the presentation only simple shear tests will be considered. The simple shear test is an improved shear box test which imposes more uniform stresses and strains. dx dz H g xz g xz = dx/h z = - dz/h = v

145 The Behaviour of Sands Depends on: Mean Effective stress (Normal effective stress in simple shear)

146 The Behaviour of Sands Depends on: Mean Effective stress (Normal effective stress in simple shear) Relative density, I d I d = e max - e e max - e min

147 The Behaviour of Sands Depends on: Mean Effective stress (Normal effective stress in simple shear) Relative density, I d I d = e max - e e max - e min g d = G s g w 1 + e

148 The Behaviour of Sands Depends on: Mean Effective stress (Normal effective stress in simple shear) Relative density, I d I d = e max - e e max - e min g d = G s g w 1 + e I d = 1 g 1 g dmin dmin g 1 g d dmax

149 The Behaviour of Sands Dense (D) Medium (M) Loose (L) e L g M v D D M g g L

150 The Behaviour of Sands For tests performed with the same normal stress All samples approach the same ultimate shear stress and void ratio, irrespective of the initial relative density Initially dense samples attain higher peak angles of friction Initially dense soils expand (dilate) when sheared Initially loose soils compress when sheared

151 The Behaviour of Sands e D 1 D 2 L L 1 g v D 1 D 2 D 1 D 2 g L 1 L 2 L 1 L 2 g

152 The Behaviour of Sands The ultimate values of shear stress and void ratio depend on the applied normal stress The ultimate stress ratio and angle of friction are independent of density and stress level Initially dense samples attain higher peak angles of friction, but the peak friction angle decreases as the stress increases Initially dense soils expand and initially loose soils compress when sheared. Increasing the normal stress causes less dilation (more compression)

153 Ultimate or Critical States All soil when sheared will eventually attain a unique stress ratio and reach a critical void ratio which is uniquely related to the normal stress.

154 Ultimate or Critical States All soil when sheared will eventually attain a unique stress ratio and reach a critical void ratio which is uniquely related to the normal stress. This ulimate state is called a Critical State A critical state is defined by d dg = d dg = d v dg = 0

155 Ultimate or Critical States All soil when sheared will eventually attain a unique stress ratio and reach a critical void ratio which is uniquely related to the normal stress. This ulimate state is called a Critical State A critical state is defined by d dg = d dg = d v dg = 0 In tests with different normal stresses different critical states will be reached.

156 Ultimate or Critical States All soil when sheared will eventually attain a unique stress ratio and reach a critical void ratio which is uniquely related to the normal stress. This ulimate state is called a Critical State A critical state is defined by d dg = d dg = d v dg = 0 In tests with different normal stresses different critical states will be reached. The locus of these critical states defines a line known as the Critical State Line (CSL)

157 The Critical State Line

158 Friction angles At critical states soil behaves as a purely frictional material f = f ult = f cs = constant f ult = F (mineralogy, grading, angularity) Typical values of f ult smectite clay 15 o clay 22 o sand 33 o angular gravel 40 o

159 Behaviour of clays Essentially the same as sands. However, data presented as a function of OCR rather than relative density. OCR is defined as OCR pc

160 Behaviour of clays Essentially the same as sands. However, data presented as a function of OCR rather than relative density. OCR is defined as e OCR pc swelling line NCL - normal consolidation line log

161 Behaviour of clays Essentially the same as sands. However, data presented as a function of OCR rather than relative density. OCR is defined as e OCR pc swelling line CSL NCL - normal consolidation line It is found that NCL and CSL have the same slope in e-log log

162 Behaviour of clays - drained response OCR = 1 OCR = 8 v OCR = 8 g NCL g OCR = 1

163 Behaviour of clays - drained response In drained loading the change in effective stress is identical to the change in total stress. In a shear box (or simple shear) test the normal stress is usually kept constant, and hence the response is fixed in the, plot. The soil heads towards a critical state when sheared, and this ultimate (or critical) state can be determined from the, plot. The change in void ratio can then be determined. Knowing the sign of the volume change enables the likely stressstrain response to be estimated.

164 Behaviour of clays - undrained response OCR = 1 OCR = 8 u +ve g OCR = 1 NCL g -ve OCR = 8

165 Behaviour of clays - undrained response In undrained loading the void ratio (moisture content) must stay constant. The soil must head towards a critical state when sheared, and knowing e the critical state can be determined from the e, plot. Once the critical state has been determined in the e, plot the ultimate shear stress is also fixed. The ultimate shear stress is related to the undrained strength. This relation can be obtained by considering a Mohr s circle.

166 Behaviour of clays - undrained response In undrained loading the void ratio (moisture content) must stay constant. The soil must head towards a critical state when sheared, and knowing e the critical state can be determined from the e, plot. Once the critical state has been determined in the e, plot the ultimate shear stress is also fixed. The ultimate shear stress is related to the undrained strength. This relation can be obtained by considering a Mohr s circle. s u ult cosf ult

167 Behaviour of clays - undrained response In undrained loading the effective stresses are fixed because void ratio (moisture content) must stay constant. The total stresses are controlled by the external loads, and the pore pressure is simply the difference between the total stress and effective stress. The CSL provides an explanation for the existence of cohesion (undrained strength) in frictional soils From the CSL it can also be seen that changes in moisture content (void ratio) will lead to different undrained strengths

168 Differences between sand and clay All soils are essentially frictional materials but different parameters are used for sands (I d ) and clays (OCR). Why?

169 Differences between sand and clay All soils are essentially frictional materials but different parameters are used for sands (I d ) and clays (OCR). Why? e Loose Dense Sand NCL log (MPa)

170 Differences between sand and clay All soils are essentially frictional materials but different parameters are used for sands (I d ) and clays (OCR). Why? e Clay Loose Sand Dense NCL NCL log (MPa)