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Shear strength of soil
- 1. Shear Strength of Soils CEL 610 – Foundation Engineering
- 2. Strength of different materials Steel Tensile strength Concrete Compressive strength Soil Shear strength Presence of pore water Complex behavior
- 3. Embankment Strip footing Shear failure of soils Soils generally fail in shear At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. Failure surface Mobilized shear resistance
- 4. Retaining wall Shear failure of soils Soils generally fail in shear
- 5. Retaining wall Shear failure of soils At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength. Failure surface Mobilized shear resistance Soils generally fail in shear
- 6. Shear failure mechanism The soil grains slide over each other along the failure surface. No crushing of individual grains. failure surface
- 7. Shear failure mechanism At failure, shear stress along the failure surface (τ) reaches the shear strength (τf). σ τ τ σ τ τ
- 8. Mohr-Coulomb Failure Criterion (in terms of total stresses) τ τf is the maximum shear stress the soil can take without failure, under normal stress of σ. σ φστ tan+= cf c φ failure envelope Cohesio n Friction angleτf σ
- 9. Mohr-Coulomb Failure Criterion (in terms of effective stresses) τf is the maximum shear stress the soil can take without failure, under normal effective stress of σ’. τ σ’ 'tan'' φστ += cf c’ φ’ failure envelope Effective cohesion Effective friction angleτf σ’ u−= σσ ' u = pore water pressure
- 10. Mohr-Coulomb Failure Criterion 'tan'' φστ ff c += Shear strength consists of two components: cohesive and frictional. σ’f τf φ’ τ σ' c’ c’ cohesive component σ’f tan φ’ frictional component
- 11. c and φ are measures of shear strength. Higher the values, higher the shear strength.
- 12. Mohr Circle of stress Soil element σ’ 1 σ’ 1 σ’ 3 σ’ 3 θ σ’ τ θ σσσσ σ θ σσ τ 2 22 2 2 ' 3 ' 1 ' 3 ' 1' ' 3 ' 1 Cos Sin − + + = − = Resolving forces in σ and τ directions, 2' 3 ' 1 2' 3 ' 1'2 22 − = + −+ σσσσ στ
- 13. Mohr Circle of stress 2' 3 ' 1 2' 3 ' 1'2 22 − = + −+ σσσσ στ Soil element σ’1 σ’1 σ’3 σ’3 θ σ’ τ Soil elementSoil element σ’1 σ’1 σ’3 σ’3 σ’1 σ’1 σ’3 σ’3 θ σ’ τ θθ σ’ τ τ σ’ 2 ' 3 ' 1 σσ + 2 ' 3 ' 1 σσ − ' 3σ ' 1σ
- 14. Mohr Circle of stress 2' 3 ' 1 2' 3 ' 1'2 22 − = + −+ σσσσ στ Soil element σ’1 σ’1 σ’3 σ’3 θ σ’ τ Soil elementSoil element σ’1 σ’1 σ’3 σ’3 σ’1 σ’1 σ’3 σ’3 θ σ’ τ θθ σ’ τ τ σ’ 2 ' 3 ' 1 σσ + 2 ' 3 ' 1 σσ − ' 3σ ' 1σ PD = Pole w.r.t. plane θ (σ’, τ)
- 15. Soil elements at different locations Failure surface Mohr Circles & Failure Envelope X X X ~ failure Y Y Y ~ stable τ σ’ 'tan'' φστ += cf
- 16. Mohr Circles & Failure Envelope Y σc σc σc Initially, Mohr circle is a point ∆σ σc+∆σ ∆σ The soil element does not fail if the Mohr circle is contained within the envelope GL
- 17. Mohr Circles & Failure Envelope Y σc σc σc GL As loading progresses, Mohr circle becomes larger… .. and finally failure occurs when Mohr circle touches the envelope ∆σ
- 18. σ’ 2 ' 3 ' 1 σσ + ' 3σ ' 1σ PD = Pole w.r.t. plane θ (σ’, τf) Orientation of Failure Plane φ’ σ’1 σ’1 σ’3 σ’3 θ σ’ τ σ’1 σ’1 σ’3 σ’3 θθ σ’ τ Failure envelope (90 – θ) Therefore, 90 – θ + φ’ = θ θ = 45 + φ’/2
- 19. Mohr circles in terms of total & effective stresses = X σv’ σh’ X u u + σv’σh’ effective stresses u σvσh X σv σh total stresses τ σ or σ’
- 20. Failure envelopes in terms of total & effective stresses = X σv’ σh’ X u u + σv’σh’ effective stresses u σvσh X σv σh total stresses τ σ or σ’ If X is on failure c φ Failure envelope in terms of total stresses φ’ c’ Failure envelope in terms of effective stresses
- 21. Mohr Coulomb failure criterion with Mohr circle of stress X σ’v = σ’1 σ’h = σ’3 X is on failure σ’1 σ’3 effective stresses τ σ ’ φ’ c’ Failure envelope in terms of effective stresses c’ Cotφ’ (σ’1+ σ’3)/ 2 (σ’1 − σ’3)/ 2 − = + + 2 ' 2 '' ' 3 ' 1 ' 3 ' 1 σσ φ σσ φ SinCotc Therefore,
- 22. Mohr Coulomb failure criterion with Mohr circle of stress − = + + 2 ' 2 '' ' 3 ' 1 ' 3 ' 1 σσ φ σσ φ SinCotc ( ) ( ) ''2'' 3 ' 1 ' 3 ' 1 φφσσσσ CoscSin ++=− ( ) ( ) ''2'1'1 ' 3 ' 1 φφσφσ CoscSinSin ++=− ( ) ( ) ( )'1 ' '2 '1 '1' 3 ' 1 φ φ φ φ σσ Sin Cos c Sin Sin − + − + = ++ += 2 ' 45'2 2 ' 452' 3 ' 1 φφ σσ TancTan
- 23. Other laboratory tests include, Direct simple shear test, torsional ring shear test, plane strain triaxial test, laboratory vane shear test, laboratory fall cone test Determination of shear strength parameters of soils (c, φ or c’, φ’) Laboratory tests on specimens taken from representative undisturbed samples Field tests Most common laboratory tests to determine the shear strength parameters are, 1.Direct shear test 2.Triaxial shear test 1. Vane shear test 2. Torvane 3. Pocket penetrometer 4. Fall cone 5. Pressuremeter 6. Static cone penetrometer 7. Standard penetration test
- 24. Laboratory tests Field conditions z σvc σvc σhcσhc Before construction A representative soil sample z σvc + ∆σ σhcσhc After and during construction σvc + ∆σ
- 25. Laboratory tests Simulating field conditions in the laboratory Step 1 Set the specimen in the apparatus and apply the initial stress condition σvc σvc σhc σhc Representative soil sample taken from the site 0 00 0 Step 2 Apply the corresponding field stress conditions σvc + ∆σ σhc σhc σvc + ∆σTraxial test σvc σvc τ τ Direct shear test
- 26. Direct shear test Schematic diagram of the direct shear apparatus
- 27. Direct shear test Preparation of a sand specimen Components of the shear box Preparation of a sand specimen Porous plates Direct shear test is most suitable for consolidated drained tests specially on granular soils (e.g.: sand) or stiff clays
- 28. Direct shear test Leveling the top surface of specimen Preparation of a sand specimen Specimen preparation completed Pressure plate
- 29. Direct shear test Test procedure Porous plates Pressure plate Steel ball Step 1: Apply a vertical load to the specimen and wait for consolidation P Proving ring to measure shear force S
- 30. Direct shear test Step 2: Lower box is subjected to a horizontal displacement at a constant rate Step 1: Apply a vertical load to the specimen and wait for consolidation PTest procedure Pressure plate Steel ball Proving ring to measure shear force S Porous plates
- 31. 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
- 32. Direct shear test Analysis of test results sampletheofsectioncrossofArea (P)forceNormal stressNormal ==σ sampletheofsectioncrossofArea (S)surfaceslidingat thedevelopedresistanceShear stressShear ==τ Note: Cross-sectional area of the sample changes with the horizontal displacement
- 33. Direct shear tests on sands Shear stress,τ Shear displacement Dense sand/ OC clay τf Loose sand/ NC clayτf Dense sand/OC Clay Loose sand/NC Clay Changeinheight ofthesample ExpansionCompression Shear displacement Stress-strain relationship
- 34. τf1 Normal stress = σ1 Direct shear tests on sands How to determine strength parameters c and φ Shear stress,τ Shear displacement τf2 Normal stress = σ2 τf3 Normal stress = σ3 She failuτf Normal stress, φ Mohr – Coulomb failure envelope
- 35. Direct shear tests on sands Some important facts on strength parameters c and φ of sand Sand is cohesionless hence c = 0 Direct shear tests are drained and pore water pressures are dissipated, hence u = 0 Therefore, φ’ = φ and c’ = c = 0
- 36. Direct shear tests on clays Failure envelopes for clay from drained direct shear tests Shearstre failure,τf Normal force, σ φ’ 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)
- 37. 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) δστ tan'+= af c Where, ca = adhesion, δ = angle of internal friction Foundation material Soil P S Foundation material Soil P S
- 38. 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
- 39. Triaxial Shear Test Specimen preparation (undisturbed sample) Sampling tubes Sample extruder
- 40. Triaxial Shear Test Specimen preparation (undisturbed sample) Edges of the sample are carefully trimmed Setting up the sample in the triaxial cell
- 41. Triaxial Shear Test Sample is covered with a rubber membrane and sealed Cell is completely filled with water Specimen preparation (undisturbed sample)
- 42. Triaxial Shear Test Specimen preparation (undisturbed sample) Proving ring to measure the deviator load Dial gauge to measure vertical displacement
- 43. 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 σc σc σc σc σcStep 1 deviatoric stress (∆σ = q) Shearing (loading) Step 2 σc σc σc+ q
- 44. Types of Triaxial Tests Is the drainage valve open? yes no Consolidated sample Unconsolidated sample Under all-around cell pressure σc Step 1 Is the drainage valve open? yes no Drained loading Undrained loading Shearing (loading) Step 2 CD test CU test UU test
- 45. Consolidated- drained test (CD Test) Step 1: At the end of consolidation σVC σhC Total, σ = Neutral, u Effective, σ’+ 0 Step 2: During axial stress increase σ’VC = σVC σ’hC = σhC σVC + ∆σ σhC 0 σ’V = σVC + ∆σ = σ’1 σ’h = σhC = σ’3 Drainage Drainage Step 3: At failure σVC + ∆σf σhC 0 σ’Vf = σVC + ∆σf = σ’1f σ’hf = σhC = σ’3f Drainage
- 46. Deviator stress (q or ∆σd) = σ1 – σ3 Consolidated- drained test (CD Test) σ1 = σVC + ∆σ σ3 = σhC
- 47. Volumechangeofthe sample ExpansionCompression Time Volume change of sample during consolidation Consolidated- drained test (CD Test)
- 48. Deviator stress,∆σd Axial strain Dense sand or OC clay (∆σd)f Dense sand or OC clay Loose sand or NC clay Volumechange ofthesample ExpansionCompression Axial strain Stress-strain relationship during shearing Consolidated- drained test (CD Test) Loose sand or NC Clay(∆σd)f
- 49. CD tests How to determine strength parameters c and φ Deviator stress,∆σd Axial strain Shear stress,τ σ or σ’ φ Mohr – Coulomb failure envelope (∆σd)f a Confining stress = σ3a(∆σd)f b Confining stress = σ3b (∆σd)f c Confining stress = σ3c σ3c σ1c σ3a σ1a (∆σd)f σ3b σ1b (∆σd)fb σ1 = σ3 + (∆σd)f σ3
- 50. CD tests Strength parameters c and φ obtained from CD tests Since u = 0 in CD tests, σ = σ’ Therefore, c = c’ and φ = φ’ cd and φd are used to denote them
- 51. CD tests Failure envelopes Shear stress,τ σ or σ’ φd Mohr – Coulomb failure envelope σ3a σ1a (∆σd)f a For sand and NC Clay, cd = 0 Therefore, one CD test would be sufficient to determine φd of sand or NC clay
- 52. CD tests Failure envelopes For OC Clay, cd ≠ 0 τ σ or σ’ φ σ3 σ1 (∆σd)f c σc OC NC
- 53. Some practical applications of CD analysis for clays τ τ = in situ drained shear strength Soft clay 1. Embankment constructed very slowly, in layers over a soft clay deposit
- 54. Some practical applications of CD analysis for clays 2. Earth dam with steady state seepage τ = drained shear strength of clay core τ Core
- 55. Some practical applications of CD analysis for clays 3. Excavation or natural slope in clay τ = In situ drained shear strength τ Note: CD test simulates the long term condition in the field. Thus, cd and φd should be used to evaluate the long term behavior of soils
- 56. Consolidated- Undrained test (CU Test) Step 1: At the end of consolidation σVC σhC Total, σ = Neutral, u Effective, σ’+ 0 Step 2: During axial stress increase σ’VC = σVC σ’hC = σhC σVC + ∆σ σhC ±∆ u Drainage Step 3: At failure σVC + ∆σf σhC No drainage No drainage ±∆uf σ’V = σVC + ∆σ ± ∆u = σ’1 σ’h = σhC ± ∆u = σ’3 σ’Vf = σVC + ∆σf ± ∆uf = σ’1f σ’hf = σhC ± ∆uf = σ’3f
- 57. Volumechangeofthe sample ExpansionCompression Time Volume change of sample during consolidation Consolidated- Undrained test (CU Test)
- 58. Deviator stress,∆σd Axial strain Dense sand or OC clay (∆σd)f Dense sand or OC clay Loose sand /NC Clay ∆u +- Axial strain Stress-strain relationship during shearing Consolidated- Undrained test (CU Test) Loose sand or NC Clay(∆σd)f
- 59. CU tests How to determine strength parameters c and φ Deviator stress,∆σd Axial strain Shear stress,τ σ or σ’ (∆σd)f b Confining stress = σ3b σ3b σ1bσ3a σ1a (∆σd)fa φcuMohr – Coulomb failure envelope in terms of total stresses ccu σ1 = σ3 + (∆σd)f σ3 Total stresses at failure (∆σd)f a Confining stress = σ3a
- 60. (∆σd)fa CU tests How to determine strength parameters c and φ Shear stress,τ σ or σ’ σ3b σ1bσ3a σ1a (∆σd)fa φcu Mohr – Coulomb failure envelope in terms of total stresses ccu σ’3b σ’1b σ’3a σ’1a Mohr – Coulomb failure envelope in terms of effective stresses φ’ C’ ufa ufb σ’1 = σ3 + (∆σd)f - uf σ’3 = σ3 - uf Effective stresses at failure uf
- 61. CU tests Strength parameters c and φ obtained from CD tests Shear strength parameters in terms of total stresses are ccu and φcu Shear strength parameters in terms of effective stresses are c’ and φ’ c’ = cd and φ’ = φd
- 62. CU tests Failure envelopes For sand and NC Clay, ccu and c’ = 0 Therefore, one CU test would be sufficient to determine φcu and φ (’ = φd) of sand or NC clay Shear stress,τ σ or σ’ φcu Mohr – Coulomb failure envelope in terms of total stresses σ3a σ1a (∆σd)f a σ3a σ1a φ’ Mohr – Coulomb failure envelope in terms of effective stresses
- 63. Some practical applications of CU analysis for clays τ τ = in situ undrained shear strength Soft clay 1. Embankment constructed rapidly over a soft clay deposit
- 64. Some practical applications of CU analysis for clays 2. Rapid drawdown behind an earth dam τ = Undrained shear strength of clay core Core τ
- 65. 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 φcu) 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 τ = In situ undrained shear strength τ
- 66. Unconsolidated- Undrained test (UU Test) Data analysis σC = σ3 σC = σ3 No drainage Initial specimen condition σ3 + ∆σd σ3 No drainage Specimen condition during shearing 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 – ∆H) = A0 × H0 A ×(1 – ∆H/H0) = A0 z A A ε− = 1 0
- 67. Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling 0 0 = + Step 2: After application of hydrostatic cell pressure ∆uc = B ∆σ3 σC = σ3 σC = σ3 ∆uc σ’3 = σ3 - ∆uc σ’3 = σ3 - ∆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 = ∆σ3)
- 68. Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load σ3 + ∆σd σ3 No drainage σ’1 = σ3 + ∆σd - ∆uc ∆ud σ’3 = σ3 - ∆uc ∆ud ∆ud = AB∆σd ∆uc ± ∆ud = + Increase of pwp due to increase of deviator stress Increase of deviator stress Skempton’s pore water pressure parameter, A
- 69. Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3, ∆uc = B ∆σ3 ∆ud = AB∆σd ∆u = ∆uc + ∆ud Total pore water pressure increment at any stage, ∆u ∆u = B [∆σ3 + A∆σd] Skempton’s pore water pressure equation ∆u = B [∆σ3 + A(∆σ1 – ∆σ3]
- 70. Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling 0 0 Total, σ = Neutral, u Effective, σ’+ -ur Step 2: After application of hydrostatic cell pressure σ’V0 = ur σ’h0 = ur σC σC -ur + ∆uc = -ur + σc (Sr = 100% ; B = 1)Step 3: During application of axial load σC + ∆σ σC No drainage No drainage -ur + σc ± ∆u σ’VC = σC + ur - σC = ur σ’h = ur Step 3: At failure σ’V = σC + ∆σ + ur - σc ∆u σ’h = σC + ur - σc ∆u σ’hf = σC + ur - σc ∆uf = σ’3f σ’Vf = σC + ∆σf + ur - σc ∆uf = σ’1f -ur + σc ± ∆uf σC σC + ∆σf No drainage
- 71. Unconsolidated- Undrained test (UU Test) Total, σ = Neutral, u Effective, σ’+ Step 3: At failure σ’hf = σC + ur - σc ∆uf = σ’3f σ’Vf = σC + ∆σf + ur - σc ∆uf = σ’1f -ur + σc ± ∆uf σC σC + ∆σf No drainage Mohr circle in terms of effective stresses do not depend on the cell pressure. Therefore, we get only one Mohr circle in terms of effective stress for different cell pressures τ σ’ σ’3 σ’1∆σ
- 72. σ3b σ1bσ3a σ1a∆σf σ’3 σ’1 Unconsolidated- Undrained test (UU Test) Total, σ = Neutral, u Effective, σ’+ Step 3: At failure σ’hf = σC + ur - σc ∆uf = σ’3f σ’Vf = σC + ∆σf + ur - σc ∆uf = σ’1f -ur + σc ± ∆uf σC σC + ∆σf No drainage τ σ or σ’ Mohr circles in terms of total stresses uaub Failure envelope, φu = 0 cu
- 73. σ3b σ1b Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope σ3a σ1a σ3c σ1c τ σ or σ’ S < 100% S > 100%
- 74. Some practical applications of UU analysis for clays τ τ = in situ undrained shear strength Soft clay 1. Embankment constructed rapidly over a soft clay deposit
- 75. Some practical applications of UU analysis for clays 2. Large earth dam constructed rapidly with no change in water content of soft clay Core τ = Undrained shear strength of clay core τ
- 76. Some practical applications of UU analysis for clays 3. Footing placed rapidly on clay deposit τ = 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
- 77. Unconfined Compression Test (UC Test) σ1 = σVC + ∆σ σ3 = 0 Confining pressure is zero in the UC test
- 78. Unconfined Compression Test (UC Test) σ1 = σVC + ∆σf σ3 = 0 Shearstress,τ Normal stress, σ qu τf = σ1/2 = qu/2 = cu
- 79. Various correlations for shear strength For NC clays, the undrained shear strength (cu) increases with the effective overburden pressure, σ’0 )(0037.011.0' 0 PI cu += σ Skempton (1957) Plasticity Index as a % For OC clays, the following relationship is approximately true 8.0 ' 0 ' 0 )(OCR cc edConsolidatNormally u idatedOverconsol u = σσ Ladd (1977) For NC clays, the effective friction angle (φ’) is related to PI as follows )log(234.0814.0' IPSin −=φ Kenny (1959)
- 80. 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 Solid Water Saturated soils Pore water pressure, u Effective stress, σ’ Solid Unsaturated soils Pore water pressure, uw Effective stress, σ’ Water Air Pore air pressure, ua Pore water pressure can be negative in unsaturated soils
- 81. Shear strength of partially saturated soils Bishop (1959) proposed shear strength equation for unsaturated soils as follows [ ] 'tan)()(' φχστ waanf uuuc −+−+= Where, σn – ua = Net normal stress ua – uw = Matric suction χ= a parameter depending on the degree of saturation (χ = 1 for fully saturated soils and 0 for dry soils) Fredlund et al (1978) modified the above relationship as follows b waanf uuuc φφστ tan)('tan)(' −+−+= Where, tanφb = Rate of increase of shear strength with matric suction
- 82. Shear strength of partially saturated soils b waanf uuuc φφστ tan)('tan)(' −+−+= 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 τ σ - φ’ ua – uw = 0(ua – uw )1 > 0(ua – uw )2 > (ua – uw )1
- 83. τ σ - ua How it become possible build a sand castle b waanf uuuc φφστ tan)('tan)(' −+−+= Same as saturated soils Apparent cohesion due to matric suction φ’ ua – uw = 0 Failure envelope for saturated sand (c’ = 0) (ua – uw ) > 0 Failure envelope for unsaturated sand Apparent cohesion