CEP 701 - Soil Engineering Laboratory Triaxial Shear Test CEP 701 - Soil Engineering Laboratory

Strength of different materials Soil Shear strength Presence of pore water Complex behavior Steel Tensile strength Concrete Compressive strength

Shear failure of soils Soils generally fail in shear Strip footing Embankment Failure surface Mobilized shear resistance At failure, shear stress along the failure surface (mobilized shear resistance) reaches the shear strength.

Shear failure of soils Soils generally fail in shear Retaining wall

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

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

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

Mohr-Coulomb Failure Criterion (in terms of total stresses)   c  failure envelope Cohesion Friction angle f f is the maximum shear stress the soil can take without failure, under normal stress of .

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

Mohr-Coulomb Failure Criterion Shear strength consists of two components: cohesive and frictional. ’f f ’  ' c’ ’f tan ’ frictional component c’ cohesive component c and  are measures of shear strength. Higher the values, higher the shear strength.

Mohr Circle of stress s’ t q s’1 s’3 Soil element q s’ t Resolving forces in s and t directions,

Mohr Circle of stress t s’

Mohr Circle of stress t s’ q (s’, t) PD = Pole w.r.t. plane

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

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

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

Orientation of Failure Plane Failure envelope q (s’, tf) (90 – q) PD = Pole w.r.t. plane f’ s’ Therefore, 90 – q + f’ = q q = 45 + f’/2

+ = t s or s’ Mohr circles in terms of total & effective stresses v X total stresses t s or s’ X v’ h’ u + = v’ h’ effective stresses u

Failure envelopes in terms of total & effective stresses h X total stresses t s or s’ X v’ h’ u + = f’ c’ Failure envelope in terms of effective stresses c f Failure envelope in terms of total stresses If X is on failure v’ h’ effective stresses u

t s’ Mohr Coulomb failure criterion with Mohr circle of stress ’1 ’3 effective stresses t s’ f’ c’ Failure envelope in terms of effective stresses X ’v = s’1 ’h = s’3 X is on failure (s’1 - s’3)/2 c’ Cotf’ (s’1+ s’3)/2 Therefore,

Mohr Coulomb failure criterion with Mohr circle of stress

Determination of shear strength parameters of soils (c, f or c’, f’) Laboratory tests on specimens taken from representative undisturbed samples Field tests Vane shear test Torvane Pocket penetrometer Fall cone Pressuremeter Static cone penetrometer Standard penetration test Most common laboratory tests to determine the shear strength parameters are, Direct shear test Triaxial shear test Other laboratory tests include, Direct simple shear test, torsional ring shear test, plane strain triaxial test, laboratory vane shear test, laboratory fall cone test

Laboratory tests z svc + Ds shc After and during construction Field conditions z svc shc Before construction A representative soil sample

Laboratory tests Simulating field conditions in the laboratory shc svc + Ds shc Traxial test Laboratory tests Simulating field conditions in the laboratory Representative soil sample taken from the site Step 1 Set the specimen in the apparatus and apply the initial stress condition svc shc svc t Direct shear test Step 2 Apply the corresponding field stress conditions

Triaxial Shear Test Failure plane Soil sample at failure Soil sample 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

Triaxial Shear Test Specimen preparation (undisturbed sample) Sample extruder Sampling tubes

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

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

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

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

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

Consolidated- drained test (CD Test) Total, s = Neutral, u Effective, s’ + Step 1: At the end of consolidation sVC shC s’VC = sVC s’hC = shC Drainage Step 2: During axial stress increase sVC + Ds shC s’V = sVC + Ds = s’1 s’h = shC = s’3 Drainage Step 3: At failure sVC + Dsf shC s’Vf = sVC + Dsf = s’1f s’hf = shC = s’3f Drainage

Deviator stress (q or Dsd) = s1 – s3 Consolidated- drained test (CD Test) s1 = sVC + Ds s3 = shC Deviator stress (q or Dsd) = s1 – s3

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

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

CD tests f s or s’ s3 Shear stress, t s3c s1c s3b s1b (Dsd)fb s3a s1a How to determine strength parameters c and f Deviator stress, Dsd Axial strain (Dsd)fc Confining stress = s3c s1 = s3 + (Dsd)f s3 (Dsd)fb Confining stress = s3b (Dsd)fa Confining stress = s3a f Mohr – Coulomb failure envelope Shear stress, t s or s’ s3c s1c s3b s1b (Dsd)fb s3a s1a (Dsd)fa

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

CD tests fd s or s’ Failure envelopes For sand and NC Clay, cd = 0 Shear stress, t s or s’ fd Mohr – Coulomb failure envelope s3a s1a (Dsd)fa Therefore, one CD test would be sufficient to determine fd of sand or NC clay

t CD tests f c s or s’ Failure envelopes For OC Clay, cd ≠ 0 OC NC s3 (Dsd)f c sc OC NC

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

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

Some practical applications of CD analysis for clays 3. Excavation or natural slope in clay t 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

Consolidated- Undrained test (CU Test) Total, s = Neutral, u Effective, s’ + Step 1: At the end of consolidation sVC shC s’VC = sVC s’hC = shC Drainage Step 2: During axial stress increase sVC + Ds shC s’V = sVC + Ds ± Du = s’1 s’h = shC ± Du = s’3 No drainage ±Du Step 3: At failure sVC + Dsf shC s’Vf = sVC + Dsf ± Duf = s’1f s’hf = shC ± Duf = s’3f No drainage ±Duf

Consolidated- Undrained test (CU Test) Volume change of sample during consolidation Volume change of the sample Expansion Compression Time

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

CU tests fcu s or s’ s3 Shear stress, t s3b s1b s3a s1a (Dsd)fa How to determine strength parameters c and f Deviator stress, Dsd Axial strain (Dsd)fb Confining stress = s3b s1 = s3 + (Dsd)f s3 Total stresses at failure (Dsd)fa Confining stress = s3a Shear stress, t s or s’ fcu Mohr – Coulomb failure envelope in terms of total stresses ccu s3b s1b s3a s1a (Dsd)fa

CU tests f’ fcu s or s’ s’3 = s3 - uf Shear stress, t s3b s1b (Dsd)fa How to determine strength parameters c and f s’1 = s3 + (Dsd)f - uf s’3 = s3 - uf uf Mohr – Coulomb failure envelope in terms of effective stresses f’ C’ Effective stresses at failure Shear stress, t s or s’ Mohr – Coulomb failure envelope in terms of total stresses fcu s3b s1b (Dsd)fa ufb ufa ccu s’3b s’1b s3a s1a s’3a s’1a (Dsd)fa

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

CU tests f’ fcu s or s’ Failure envelopes For sand and NC Clay, ccu and c’ = 0 Shear stress, t s or s’ fcu Mohr – Coulomb failure envelope in terms of total stresses s3a s1a (Dsd)fa f’ Mohr – Coulomb failure envelope in terms of effective stresses Therefore, one CU test would be sufficient to determine fcu and f’(= fd) of sand or NC clay

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

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

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

Unconsolidated- Undrained test (UU Test) Data analysis s3 + Dsd s3 No drainage Specimen condition during shearing sC = s3 No drainage Initial specimen condition 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

Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling Step 2: After application of hydrostatic cell pressure s’3 = s3 - Duc sC = s3 No drainage Duc = + Increase of pwp due to increase of cell pressure Duc = B Ds3 Skempton’s pore water pressure parameter, B Increase of cell pressure Note: If soil is fully saturated, then B = 1 (hence, Duc = Ds3)

Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load s’1 = s3 + Dsd - Duc Dud s’3 = s3 - Duc Dud s3 + Dsd s3 No drainage = Duc ± Dud + Increase of pwp due to increase of deviator stress Dud = ADsd Increase of deviator stress Skempton’s pore water pressure parameter, A

Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3, Duc = B Ds3 Dud = ADsd Total pore water pressure increment at any stage, Du Du = Duc + Dud Du = B Ds3 + ADsd Skempton’s pore water pressure equation Du = B Ds3 + A(Ds1 – Ds3)

Unconsolidated- Undrained test (UU Test) Total, s = Neutral, u Effective, s’ + Step 1: Immediately after sampling s’V0 = ur s’h0 = ur -ur Step 2: After application of hydrostatic cell pressure s’VC = sC + ur - sC = ur s’h = ur sC No drainage -ur + Duc = -ur + sc (Sr = 100% ; B = 1) Step 3: During application of axial load s’V = sC + Ds + ur - sc Du s’h = sC + ur - sc Du sC + Ds sC No drainage -ur + sc ± Du Step 3: At failure s’hf = sC + ur - sc Duf = s’3f s’Vf = sC + Dsf + ur - sc Duf = s’1f sC sC + Dsf No drainage -ur + sc ± Duf

Unconsolidated- Undrained test (UU Test) Total, s = Neutral, u Effective, s’ + Step 3: At failure s’hf = sC + ur - sc Duf = s’3f s’Vf = sC + Dsf + ur - sc Duf = s’1f -ur + sc ± Duf sC sC + Dsf 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 s’3 s’1 Dsf t s’

Unconsolidated- Undrained test (UU Test) Total, s = Neutral, u Effective, s’ + Step 3: At failure s’hf = sC + ur - sc Duf = s’3f s’Vf = sC + Dsf + ur - sc Duf = s’1f -ur + sc ± Duf sC sC + Dsf No drainage Mohr circles in terms of total stresses Failure envelope, fu = 0 t s or s’ cu s’3 s’1 s3a s1a Dsf s3b s1b ub ua

Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope t s or s’ S < 100% S > 100% s3a s1a s3b s1b s3c s1c

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

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

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

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

Unconfined Compression Test (UC Test) s1 = sVC + Dsf s3 = 0 Shear stress, t Normal stress, s qu τf = σ1/2 = qu/2 = cu

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