Laboratory Testing Total & Effective Stress Analysis

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1 SKAA 1713 SOIL MECHANICS Laboratory Testing Total & Effective Stress Analysis Prepared by: Dr. Hetty

2 Mohr Coulomb failure criterion with Mohr circle of stress 2 ' 2 ' ' ' 3 ' 1 ' 3 ' 1 Cot Sin c ' ' 2 ' ' 3 ' 1 ' 3 ' 1 Cos c Sin ' ' 2 ' 1 ' 1 ' 3 ' 1 Cos c Sin Sin ' 1 ' ' 2 ' 1 ' 1 ' 3 ' 1 Sin Cos c Sin Sin 2 ' 45 ' 2 2 ' 45 2 ' 3 ' 1 Tan c Tan

3 Determination of shear strength parameters of soils (c, or c, Laboratory tests Field tests Most common laboratory tests to determine the shear strength parameters are, 1.Direct shear test 2.Unconfined Compressive Strength test 3.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 1. Vane shear test 2. Torvane 3. Pocket penetrometer 4. Fall cone 5. Pressuremeter 6. Static cone penetrometer 7. Standard penetration test

4 Field conditions A representative soil sample vc z vc + D z hc hc hc hc vc vc + D Before construction After and during construction

5 Simulating field conditions in the laboratory vc + D hc hc 0 vc vc + D 0 0 hc hc vc t 0 Representative soil sample taken from the site Step 1 vc Set the specimen in the apparatus and apply the initial stress condition t Step 2 vc Apply the corresponding field stress conditions

6 DIRECT SHEAR TEST

7 Direct shear test Schematic diagram of the direct shear apparatus

8 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 Porous plates Components of the shear box Preparation of a sand specimen

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

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

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

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

13 Direct shear test Analysis of test results Normal stress Area of Normal force (P) cross section of the sample t Shear stress Shear resistance Area of developed at the sliding surface cross section of the sample (S) Note: Cross-sectional area of the sample changes with the horizontal displacement

14 Shear stress, t Direct shear tests on sands How to determine strength parameters c and Normal stress = 3 Normal stress = 2 t f3 t f2 t f1 Normal stress = 1 Shear displacement Shear stress at failure, t f Mohr Coulomb failure envelope Normal stress,

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

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

17 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, t f Normally consolidated clay (c = 0) Normal force, Overconsolidated clay (c 0)

18 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) P Soil S Foundation material t f c a ' tan Where, c a = adhesion, = angle of internal friction

19 Advantages of direct shear apparatus Due to the smaller thickness of the sample, rapid drainage can be achieved Can be used to determine interface strength parameters Clay samples can be oriented along the plane of weakness or an identified failure plane Disadvantages of direct shear apparatus Failure occurs along a predetermined failure plane Area of the sliding surface changes as the test progresses Non-uniform distribution of shear stress along the failure surface

20 Example A direct shear test when conducted on a remolded sample of sand, gave the following observations at the time of failure; Normal force = 288 N; shear force = 173 N. The cross sectional area of the sample = 36cm 2. Determine the angle of frictional. Solved in 2 ways, namely graphically and analytically (31 degrees)

21 UNCONFINED COMPRESSIVE STRENGTH (UCS) TEST

22 Unconfined Compression Test (UCS Test) 1 = VC + D 3 = 0 Confining pressure is zero in the UCS test

23 Shear stress, t Unconfined Compression Test (UC Test) 1 = VC + Df τ = q u 2 = c u 3 = 0 C u Normal stress, Note: Theoritically q u = c u, However in the actual case q u < c u due to premature failure of the sample

25 TRIAXIAL TEST

26 Triaxial Shear Test Piston (to apply deviatoric stress) Failure plane O-ring Soil sample at failure Perspex cell Soil sample impervious membrane Porous stone Water Cell pressure Back pressure pedestal Pore pressure or volume change

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

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

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

30 Consists of 3 stages: 1. Saturation 2. Consolidation 3. Shearing Main stages

31 Saturation & use of back pressure 1. Reason for saturation 2. Principle of saturation 3. Maintaining saturation 4. Advantages of saturation

32 Typical values for parameter B

33 Triaxial Shear Test

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

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

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

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

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

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

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

41 CD tests Strength parameters c and obtained from CD tests Since u = 0 in CD tests, = Therefore, c = c and = c d and d are used to denote them

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

43 CD tests Failure envelopes For OC Clay, c d 0 t OC NC c 3 1 (D d ) f c or

44 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

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

46 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, c d and d should be used to evaluate the long term behavior of soils

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

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

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

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

51 Shear stress, t CU tests How to determine strength parameters c and 1 = 3 + (D d ) f - u f Mohr Coulomb failure envelope in terms of effective stresses Mohr Coulomb failure envelope in terms of total stresses u f 3 = 3 - u f Effective stresses at failure cu C u fa u fb c cu 3b 1b 3a 3b 3a 1a (D d ) fa 1a 1b or

stresses are c cu and cu Shear strength parameters

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

53 Shear stress, t CU tests Failure envelopes For sand and NC Clay, c cu and c = 0 Mohr Coulomb failure envelope in terms of effective stresses Mohr Coulomb failure envelope in terms of total stresses cu 3a 3a 1a 1a or (D d ) fa Therefore, one CU test would be sufficient to determine cu and = d ) of sand or NC clay

54 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

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

56 Some practical applications of CU analysis for clays 3. Rapid construction of an embankment on a natural slope t t = In situ undrained shear strength Note: Total stress parameters from CU test (c cu and cu ) 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

57 Unconsolidated- Undrained test (UU Test) Data analysis Initial specimen condition Specimen condition during shearing No drainage C = 3 C = 3 No drainage 3 + D d 3 Initial volume of the sample = A 0 H 0 Volume of the sample during shearing = A H Since the test is conducted under undrained condition, A H = A 0 H 0 A (H 0 DH) = A 0 H 0 A (1 DH/H 0 ) = A 0 A A 0 1 z

58 Unconsolidated- Undrained test (UU Test) Step 1: Immediately after sampling 0 0 Step 2: After application of hydrostatic cell pressure No drainage C = 3 C = 3 = Du c + 3 = 3 - Du c 3 = 3 - Du c Increase of pwp due to increase of cell pressure Du c = B D 3 Skempton s pore water pressure parameter, B Increase of cell pressure Note: If soil is fully saturated, then B = 1 (hence, Du c = D 3 )

59 Typical values for parameter A u u Axial strain OC Clay (Lightly overconsolidated) (A = ) OC Clay (Heavily overconsolidated) (A = ) Axial strain During the increase of major principal stress pore water pressure can become negative in heavily overconsolidated clays due to dilation of specimen

60 Unconsolidated- Undrained test (UU Test) Total, = Neutral, u Effective, Step 1: Immediately after sampling 0 + V0 = u r 0 -u r h0 = u r Step 2: After application of hydrostatic cell pressure No drainage C C -u r Du c = -u r c (S r = 100% ; B = 1) Step 3: During application of axial load No drainage C + D C -u r c ± Du VC = C + u r - C = u r h = u r V = C + D + u r - c Du h = C + u r - c Du Step 3: At failure No drainage C + D f C -u r c ± Du f Vf = C + D f + u r - c Du f = 1f hf = C + u r - c Du f = 3f

61 Unconsolidated- Undrained test (UU Test) Total, = Neutral, u Effective, Step 3: At failure No drainage C + D f C -u r c ± Du f + Vf = C + D f + u r - c Du f = 1f hf = C + u r - c Du f = 3f 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 t 3 1 Df

62 Unconsolidated- Undrained test (UU Test) Total, = Neutral, u Effective, Step 3: At failure No drainage C + D f C -u r c ± Du f + Vf = C + D f + u r - c Du f = 1f hf = C + u r - c Du f = 3f Mohr circles in terms of total stresses t Failure envelope, u = 0 c u u b u a 3a 3b Df 1a 1b 3 1 or

63 Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope t S < 100% S > 100% 3c 3b 1c 3a 1b 1a or

64 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

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

66 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, c u can be used to analyze the short term behavior of soils

Chapter (12) Instructor : Dr. Jehad Hamad

Chapter (12) Instructor : Dr. Jehad Hamad 2017-2016 Chapter Outlines Shear strength in soils Direct shear test Unconfined Compression Test Tri-axial Test Shear Strength The strength of a material is the