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CEP Soil Engineering Laboratory

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Presentation on theme: "CEP Soil Engineering Laboratory"— Presentation transcript:

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

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

3 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.

4 Shear failure of soils Soils generally fail in shear Retaining wall

5 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.

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

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

8 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 .

9 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 ’.

10 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.

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

12 Mohr Circle of stress t s’

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

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

15 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

16 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

17 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

18 + = 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

19 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

20 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,

21 Mohr Coulomb failure criterion with Mohr circle of stress

22 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

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

24 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

25 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

26

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

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

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

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

31 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

32 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

33 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

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

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

36 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

37 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

38 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

39 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

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

41 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

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

43 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

44 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

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

46 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

47 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

48 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

49 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

50 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

51 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

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

53 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

54 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

55 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)

56 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

57 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)

58 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

59 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’

60 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

61 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

62 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

63 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

64 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

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

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

67 Thank You


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