Shear Strength of Soils

Name: Shear Strength of Soils
Uploaded: 2017-07-03T19:35:51+00:00
Duration: PTM38S28
Channel: Judith Ryan
Description: Shear Strength of Soils

Shear Strength of Soils
CEL 610 – Foundation Engineering

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 ’

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

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

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 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

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

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 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

Edges of the sample are carefully trimmed Setting up the sample in the triaxial cell

Sample is covered with a rubber membrane and sealed Cell is completely filled with water

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

Volume change of sample during consolidation Volume change of the sample Expansion Compression Time

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

t = drained shear strength of clay core t Core 2. Earth dam with steady state seepage

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

Volume change of sample during consolidation Volume change of the sample Expansion Compression Time

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

2. Rapid drawdown behind an earth dam t Core t = Undrained shear strength of clay core

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

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)

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

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]

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

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’

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

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

2. Large earth dam constructed rapidly with no change in water content of soft clay t = Undrained shear strength of clay core t Core

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)

THE END

Shear Strength of Soils

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