YULVI ZAIKA JURUSAN TEKNIK SIPIL FAK.TEKNIK UNIV.BRAWIJAYA SHEAR STRENGTH II YULVI ZAIKA JURUSAN TEKNIK SIPIL FAK.TEKNIK UNIV.BRAWIJAYA

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

Schematic diagram of the direct shear apparatus Direct shear test Schematic diagram of the direct shear apparatus

Preparation of a sand specimen Direct shear test 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 Components of the shear box Preparation of a sand specimen Porous plates

Preparation of a sand specimen Direct shear test Preparation of a sand specimen Specimen preparation completed Pressure plate Leveling the top surface of specimen

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

Direct shear test Steel ball P Test procedure Pressure plate Step 1: Apply a vertical load to the specimen and wait for consolidation P Test procedure Pressure plate Steel ball Proving ring to measure shear force S Porous plates Step 2: Lower box is subjected to a horizontal displacement at a constant rate

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

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

Direct shear tests on sands Stress-strain relationship Shear stress, t Shear displacement Dense sand/ OC clay tf Loose sand/ NC clay tf Change in height of the sample Expansion Compression Shear displacement Dense sand/OC Clay Loose sand/NC Clay

Direct shear tests on sands How to determine strength parameters c and f Shear stress, t Shear displacement tf3 Normal stress = s3 tf2 Normal stress = s2 tf1 Normal stress = s1 Shear stress at failure, tf Normal stress, s f Mohr – Coulomb failure envelope

Direct shear tests on sands Some important facts on strength parameters c and f of sand Direct shear tests are drained and pore water pressures are dissipated, hence u = 0 Sand is cohesionless hence c = 0 Therefore, f’ = f and c’ = c = 0

Direct shear tests on clays 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) Failure envelopes for clay from drained direct shear tests Shear stress at failure, tf Normal force, s f’ Normally consolidated clay (c’ = 0) Overconsolidated clay (c’ ≠ 0)

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) Where, ca = adhesion, d = angle of internal friction

Triaxial Shear Test Soil sample at failure Failure plane 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 (Dsd)fa s3b s1b 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 (Dsd)fa s3b s1b ufb ufa ccu s’3b s’1b s3a s1a s’3a s’1a (Dsd)fa

CU tests Strength parameters c and f obtained from CU 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 Increase of cell pressure Skempton’s pore water pressure parameter, B 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 = ABDsd 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 = ABDsd 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

Various correlations for shear strength For NC clays, the undrained shear strength (cu) increases with the effective overburden pressure, s’0 Skempton (1957) Plasticity Index as a % For OC clays, the following relationship is approximately true Ladd (1977) For NC clays, the effective friction angle (f’) is related to PI as follows Kenny (1959)

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 Unsaturated soils Pore water pressure, uw Effective stress, s’ Water Air Pore air pressure, ua Solid Water Saturated soils Pore water pressure, u Effective stress, s’ Pore water pressure can be negative in unsaturated soils

Shear strength of partially saturated soils Bishop (1959) proposed shear strength equation for unsaturated soils as follows Where, sn – ua = Net normal stress ua – uw = Matric suction = a parameter depending on the degree of saturation (c = 1 for fully saturated soils and 0 for dry soils) Fredlund et al (1978) modified the above relationship as follows Where, tanfb = Rate of increase of shear strength with matric suction

t s - ua Shear strength of partially saturated soils f’ 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 t s - ua (ua – uw)2 > (ua – uw)1 (ua – uw)1 > 0 f’ ua – uw = 0

t s - ua f’ How it become possible build a sand castle Same as saturated soils Apparent cohesion due to matric suction t s - ua (ua – uw) > 0 Failure envelope for unsaturated sand Apparent cohesion f’ ua – uw = 0 Failure envelope for saturated sand (c’ = 0)