Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

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2 Module 4: Lecture 8 on Stress-strain relationship and Shear strength of soils

3 Contents Stress state, Mohr s circle analysis and Pole, Principal stressspace, Stress pathsin p-q space; Mohr-Coulomb failure criteria and its limitations, correlation with p-q space; Stress-strain behavior; Isotropic compression and pressure dependency, confined compression, large stress compression, Definition of failure, Interlocking concept and its interpretations, Triaxial behaviour, stress state and analysis of UC, UU, CU, CD, and other special tests, Drainage conditions; Stress paths in triaxial and octahedral plane; Elastic modulus from triaxial tests.

4 The triaxial test: Introduction Most widely used shear strength test and is suitable for all types of soil. A cylindrical specimen, generally L/D = 2 is used for the test, and stresses are applied under conditions of axial symmetry. Typical specimen diameters are 38mm and 100mm Axial stress Equal all round pressure Stress system in triaxial test

5 The triaxial test: Components Loading ram Perspex cell Porous discs Pressure supply to cell Latex sheet Soil sample To pore pressure measuring device

6 The triaxial test: Mechanism Intermediate principal stress σ 2 must be equal to major σ 1 or minor σ 3 stress, so as to facilitate representation of stress state in two dimensional Mohr s circle. A cylindrical specimen is placed inside Perspex cell filled with water. The specimen is covered with latex sheet so as to avoid direct contact with water. The specimen is loaded initially by surrounding water pressure so as to achieve isotropic loading conditions. A deviatoric stress is then applied gradually on the sample with the help of Ram axially. A duct at the bottom of the sample allows water to pass through the sample which is further monitored, or conversely, in some cases, no drainage is allowed.

7 The triaxial test: Mechanism Fine grained soil can stand the mould without any support But the coarse grained soils samples have to kept in some supporting mould until the application of negative pore pressure to the sample through drainage duct. So, u = u e (negative) σ a = σ r = 0 σ a = σ r = -u e where, σ a is the axial stress, σ r is the radial stress

8 The triaxial test: Mechanism If cell pressure increased to s cp, this isotropic pressure is taken entirely by the pore water. Thus pore pressure increases, but no change occurs in effective stresses. So, u i = σ cp + u e (negative) σ a σ a = σ r = σ cp σ r thus, σ a = σ r = -u e u u e = σ cp i.e. u = σ cp

9 Drainage conditions : Combinations in triaxial test Step 1 Step 2 Under all-around cell pressure σ c Shearing (loading) drainage valve condition drainage valve condition Open Closed Open Closed Consolidated sample Unconsolidated sample Drained loading Undrained loading CD CU UU

10 Drainage conditions : Combinations in triaxial test Unconfined Compressive test (UC) Unconsolidated Undrained test (UU) Consolidated Undrained test (CU) Applying back pressure: decreases cavitation, and reduction of voids. Consolidated Drained test (CD) Specimen is taken to failure with no confinement Specimen is taken to failure with no drainage permitted Drainage valve initially opened to allow pore pressure u i to dissipate to zero, and then closed so that specimen is taken to failure without any further drainage The drainage valve is initially opened to allow the pore pressure u i to dissipate to zero, and is kept open while the specimen is taken to failure at a sufficientlyslow rate.

11 Stresses and strains on a sample in the Triaxial compression test Axisymmetric condition, σ 2 = σ 3 or σ 2 = σ 3 ; ε 2 = ε 3 p = (σ 1 + 2σ 3 )/3 and p = (σ 1 + 2σ 3 )/3 p = p- u q = σ 1 - σ 3 ; q = σ 1 - σ 3 = (σ 1 - u) (σ 3 - u) = σ 1 - σ 3 Thus, q = q; Shear is unaffected by PWP. Deviator stress σ 1 - σ 3 = σ d = P/A Deviatoric strain ε d = 2/3( ε 1 - ε 3 ) Volumetric strain ε v = ε ε 3 Schematic of a Triaxial cell Axial total stress σ 1 = σ 3 + P/A Axial strain ε 1 = z/h o Radial strain ε r = r/r o

12 Consolidated- drained test (CD Test) σ = u + σ Step 1: At the end of consolidation σ V = σ V Drainage σ h 0 σ hc = σ hc Step 2: During axial stress increase σ V + σ σ V = σ V + σ = σ 1 Drainage σ hc 0 σ h = σ h = σ 3 Step 3: At failure σ VC + σ f σ Vf = σ V + σ f = σ 1f Drainage σ hc 0 σ hf = σ h = σ 3f

13 Consolidated- drained test (CD Test) σ 1 = σ VC + σ σ 3 = σ hc Deviator stress (q or σ d ) = σ 1 σ 3

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

15 CD Test :- Stress-strain relationship during shearing Deviator stress, σ d ( σ d ) f ( σ d ) f Axial strain Dense sand or OC clay Loose sand or NC Clay Volume change of the sample + - Loose sand /NC Clay Axial strain Dense sand or OC clay

16 CD tests : How to determine strength parameters c and φ Deviator stress, σ d ( σ d ) fc ( σ d ) fb ( σ d ) fa Confining stress = σ 3c Confining stress = σ 3b Confining stress = σ 3a σ 1 = σ 3 + ( σ d ) f σ 3 Shear stress, τ Mohr Coulomb failure envelope σ 3a σ 3b Axial strain σ 3c σ 1a σ 1b φ σ 1c σ or σ ( σ d ) fa ( σ d ) fb

17 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 Parameters are denoted as c d and φ d

18 CD tests : Failure envelopes For sand and NC Clay, c d = 0 φ d Shear stress, τ Mohr Coulomb failure envelope σ 3a σ 1a σ or σ ( σ d ) fa Therefore, one CD test would be sufficient to determine φ d of sand or NC clay

19 CD tests : Failure envelopes For OC Clay, c d 0 τ OC NC φ c σ 3 σ 1 ( σ d ) f σ c σ or σ

20 Stress paths during CD Test Stage1: Isotropic consolidation phase σ 1 = σ 1 = σ 3 = σ 3 ; σ 1 > 0; u = 0 (end of consolidation) p = p = ( σ σ 1 )/3 = σ 1 ; q = σ 1 - σ 3 = 0 q/ p = q/ p = 0 q = σ 1 -σ 3 σ 1 = σ 1 = σ 3 = σ 3 Consolidation phase σ 1 = σ 3 + P/A σ 3 = σ 1 u = 0 Shearing phase ESP = TSP p = (σ 1 + 2σ 3 )/3 p, p 1 3 p = σ 1 /3; q = σ 1 q/ p = 3 σ 3 = σ 3 σ 3 = 0 u = 0

21 Stress paths during CD Test Stage 2: Shearing phase σ 1 = σ 1 > 0 ; σ 3 = σ 3 = 0 ; u = 0; p = p = ( σ 1 )/3 = σ 1 /3 ; q = σ 1 - σ 3 = 0; = σ 1 ; q/ p = q/ p = 3

22 Consolidated- Undrained test (CU Test) Total, σ = Neutral, u + Effective, σ Step 1: At the end of consolidation σ VC σ hc 0 σ VC = σ VC σ hc = σ hc Drainage Step 2: During axial stress increase σ VC + σ σ V = σ VC + σ ± u = σ 1 No drainage X σ hc ± u σ h = σ hc ± u = σ 3 Step 3: At failure σ VC + σ f σ Vf = σ VC + σ f ± u f = σ 1f No drainage X σ hc ± u f σ hf = σ hc ± u f = σ 3f

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

24 CU Test :- Stress-strain relationship during shearing Deviator stress, σ d ( σ d ) f ( σ d ) f Axial strain Dense sand or OC clay Loose sand or NC Clay Pore water pressure varies with axial strain + u - Loose sand /NC Clay Axial strain Dense sand or OC clay

25 CU tests :- How to determine strength parameters c and φ Deviator stress, σ d Shear stress, τ ( σ d ) fb ( σ d ) fa Mohr Coulomb failure envelope in terms of total stresses Confining stress = σ 3b Confining stress = σ 3a Axial strain φ cu σ 1 = σ 3 + ( σ d ) f σ 3 Total stresses at failure c cu σ 3a σ 3b ( σ d ) fa σ 1a σ 1b σ or σ

26 CU tests: Strength parameters c and φ σ 1 = σ 3 + ( σ d ) f - u f Shear stress, τ Mohr Coulomb failure envelope in terms of effective stresses Mohr Coulomb failure envelope in terms of total stresses φ φ cu σ 3 = σ 3 - u f u f Effective stresses at failure u fb c c cu σ u fa 3b σ 1b σ 3a σ 3b σ 3a σ 1a ( σ d ) fa σ 1a σ 1b σ or σ

27 CU tests Strength parameters c d and ϕ d obtained from CD tests Shear strength parameters in terms of total stresses are c cu and φ cu Shear strength parameters in terms of effective stresses are c and φ c = c d and φ = φ d

28 Stress paths during CU Test Stage1: Isotropic consolidation phase σ 1 = σ 1 = σ 3 = σ 3 ; σ 1 > 0; u = 0 (end of consolidation) p = p = ( σ σ 1 )/3 = σ 1 ; q = σ 1 - σ 3 = 0 q/ p = q/ p = 0 q = σ 1 -σ 3 σ 1 = σ 1 = σ 3 = σ 3 ESP Con. phase σ 1 = σ 3 + P/A σ 1 = σ 1 - u σ 3 = σ 1 u = 0 u TSP q/ p = 3 p = (σ 1 + 2σ 3 )/3 p, p 1 Shearing phase 3 p = σ 1 /3; q = σ 1 σ 3 = 0 u 0 σ 3 = σ 3 - u = - u

29 Stress paths during CU Test Stage 2: Shearing phase σ 1 > 0; σ 3 = 0; σ 1 = σ 1 - u = 0; σ 3 = - u p = ( σ 1 )/3 ; q = σ 1, q/ p = 3 [For TSP] p = p - u = ( σ 1 )/3 - u; [For ESP] q = σ 1 ; q/ p = σ 1 /( σ 1 /3- u) = 3/[1-3( u/ σ 1 )]

30 Stress conditions for the UU test The purpose of UU test is to determine the undrained shear strength of a saturated soil. Quick test (Neither during consolidation and shearing stages, excess PWP is allowed to drain).

31 Mohr failure envelopes for UU tests For 100% saturated clay For partially saturated clay

32 Total stress path during UU Test Initial stage σ 1 = σ 3 ; u 0 p = σ 1, q= 0; q/ p = 0 q = σ 1 -σ 3 TSP σ 1 = σ 3 + P/A σ 3 u 0 Shearing phase σ 1 > 0 ; σ 3 = 0 p = σ 1 /3; q/ p = 3 3 p = σ 1 /3 ; q = σ 1 q/ p = 3 1 p = (σ 1 + 2σ 3 )/3 p

33 Unconfined compressive (UC) test To determine the un-drained shear strength of saturated clays quickly. No radial stress (σ 3 = 0) Deviator load is increased rapidly until the soil sample fails; Pore water can not drain from the soil; the soil sample is sheared at constant volume. σ 1 σ 3 = 0 (After

34 Stress conditions for the UC test

35 Total stress path during UC Test The effective stress path is unknown since PWP changes are not normally measured. If u is measured, it would be negative. Since σ 3 = 0, σ 3 = σ 3 - u = - u u must be ve because as σ 3 can not be ve (soils can not sustain tension). So σ 3 must be +ve. q = σ 1 -σ 3 TSP p = σ 1 /3; q/ p = p = (σ 1 + 2σ 3 )/3 p, p

36 Mohr Circles for UCS The results of from UC tests can lead to: Estimate the short-term bearing capacity of fine-grained soils for foundations. Estimate the short-term stability of slopes. Determine the stressstrain characteristics under fast (un-drained loading conditions. c u τ Total stress Mohr Circle Failure envelope 45 u Failure plane Effective stress Circle not determined UC test σ, σ

37 Typical variation of σ 1 with ε 1 (UCS Test) σ 1 (kpa) C u = 136/2 = 68 kpa ε 1 (-)

38 Consolidated undrained triaxial tests on Silty sand Property Unit Silty sand Specific gravity (Gs) - a 2.64 Particle size distribution Sand (S) % 80 Silt (M) % 10 Clay (C) % 10 Classification (Unified soil classification system) - a SM Compaction characteristics (standard Proctor) Maximum dry unit weight (MDD) kn/m Optimum moisture content (OMC) % 10.5 Co-efficient of permeability (k) m/sec 4.0 x 10-7

39 Deviator stress (kpa) σ' = 50 kpa σ' = 100 kpa σ' = 150 kpa Axial strain (%) Variation in deviator stress with axial strain

40 200 Excess pore water pressure (kpa) σ' = 50 kpa σ' = 100 kpa σ' = 150 kpa Axial strain (%) Variation in excess pore water pressure with axial strain

41 Status of silty sand sample after CU test

42 Variation in stress path at various effective stress Cambridge stress path is plotted between p or p and q. Where, q (kpa) TSP, σ' = 50 kpa TSP, σ' = 100 kpa TSP, σ' = 150 kpa ESP, σ' = 50 kpa ESP, σ' = 100 kpa ESP, σ' = 150 kpa Failure envelope p = (σ 1 + 2σ 3 )/3 p = (σ 1 + 2σ 3 )/3 q = q = (σ 1 - σ 3 ) p, p' (kpa)

43 Mohr circles for consolidated un-drained tests on silty sand Shear stress (kpa) Effective parameter (σ' = 50 kpa) Effective parameter (σ' = 150 kpa) Effective parameter (σ' = 100 kpa) Total parameter (σ' = 50 kpa) Total parameter (σ' = 100 kpa) Total parameter (σ' = 150 kpa) Failure envelopes c' = 2 kpa φ ' =35 c = 7 kpa φ = Normal stress (kpa)

44 Results of CU triaxial tests on Fine Sand Max void ratio = Min void ratio = Void ratio after consolidation stage i) e (σ = 50 kpa) = ii) e (σ = 100 kpa) = Deviator stress (kpa) Effective stress = 50 kpa Effective stress = 100 kpa Axis Strain (%) Variation in deviator stress with axial strain

45 Excess pore water pressure (kpa) Effective stress = 50 kpa Effective stress = 100 kpa Axis strain (%) Variation in excess pore water pressure with axial strain

46 q, q' (kpa) Effective stress = 50 kpa Effective stress = 100 kpa p (kpa) Variation of ESP at various effective stresses q, q' (kpa) Variation of TSP at various effective stresses Effective stress = 50 kpa Effective stress = 100 kpa p' (kpa) At failure

47 π/ /2 Status of sample after termination of CU test