Chapter (12) Instructor : Dr. Jehad Hamad

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1 Chapter (12) Instructor : Dr. Jehad Hamad

2 Chapter Outlines Shear strength in soils Direct shear test Unconfined Compression Test Tri-axial Test

3 Shear Strength The strength of a material is the greatest stress it can sustain The safety of any geotechnical structure is dependent on the strength of the soil If the soil fails, the structure founded on it can collapse

4 Significance of Shear Strength Engineers must understand the nature of shearing resistance in order to analyze soil stability problems such as; Bearing capacity Slope stability Lateral pressure on -retaining structures Pavement

5 Shear Strength in Soils The shear strength of a soil is its resistance to shearing stresses. It is a measure of the soil resistance to deformation by continuous displacement of its individual soil particles Shear strength in soils depends primarily on interactions between particles Shear failure occurs when the stresses between the particles are such that they slide or roll past each other

6 Shear Strength in Soils (cont.) Soil derives its shear strength from two sources: Cohesion between particles (stress independent component) Cementation between sand grains Electrostatic attraction between clay particles Frictional resistance between particles (stress dependent component)

7 Shear Strength of Soils; Cohesion Cohesion (C), is a measure of the forces that cement particles of soils Dry sand with no cementation Dry sand with some cementation Soft clay Stiff clay

8 Shear Strength of Soils; Internal Friction Internal Friction angle (f), is the measure of the shear strength of soils due to friction

9 Mohr-Coulomb Failure Criteria This theory states that a material fails because of a critical combination of normal stress and shear stress, and not from their either maximum normal or shear stress alone. The relationship between normal stress and shear is given as s shear strength s c tan f c cohesion f angle of internal friction

10 General State of Stress σ 1 major principle stress σ 3 σ 3 Minor principle stress, Confining stress σ 1

12 Mohr Coulomb Failure Criterion

13 Mohr-Coulomb Failure Criterion Shear Strength,S f = f C Normal Stress, n = = g h

15 State of Stresses in Soils Normal stress σ n Shear stress σ 3 σ3 σ 1 Consider the following situation: - A normal stress is applied vertically and held constant - A shear stress is then applied until failure

17 Determination of Shear Strength Parameters The shear strength parameters of a soil are determined in the lab primarily with two types of tests; Direct Shear Test Triaxial Shear Test Normal stress σ n 1 Shear stress σ 3 Soil 3

18 Direct Shear Test Direct shear test is Quick and Inexpensive Shortcoming is that it fails the soil on a designated plane which may not be the weakest one Used to determine the shear strength of both cohesive as well as non-cohesive soils

applied through a metal platen Shear force is applied by moving one half of the box

19 Direct Shear Test The test equipment consists of a metal box in which the soil specimen is placed The box is split horizontally into two halves Vertical force (normal stress) is applied through a metal platen Shear force is applied by moving one half of the box relative to the other to cause failure in the soil specimen Shear stress σ 3 Normal stress σ n Soil

20 Direct Shear Test

21 Direct Shear Test

23 Direct Shear Test

25 Direct Shear Test Data Peak Strength Shear stress Residual Strength

26 Direct Shear Test Data Volume change DH

27 Direct Shear Test (Procedure) 1.Measure inner side or diameter of shear box and find the area 2.Make sure top and bottom halves of shear box are in contact and fixed together. 3.Weigh out 150 g of sand. 4.Place the soil in three layers in the mold using the funnel. Compact the soil with 20 blows per layer. 5.Place cover on top of sand 6.Place shear box in machine. 7.Apply normal force. The weights to use for the three runs are 2 kg, 4 kg, and 6 kg if the load is applied through a lever arm, or 10 kg, 20 kg, and 30 kg, if the load is applied directly. Note: Lever arm loading ratio 1:10 (2kg weight = 20 kg)

28 Direct Shear Test (Procedure) 8. Start the motor with selected speed (0.1 in/min) so that the rate of shearing is at a selected constant rate 9. Take the horizontal displacement gauge, vertical displacement gage and shear load gage readings. Record the readings on the data sheet. 10. Continue taking readings until the horizontal shear load peaks and then falls, or the horizontal displacement reaches 15% of the diameter.

29 Calculations Determine the dry unit weight, g d e Gs g g d w 1 Calculate the void ratio, e N ; A V A Calculate the normal stress & shear stress

30 Shear stress, s Shear Stress vs. Horizontal Displacement Peak Stress s 3 s 2 s 1 N 3 = 30 kg N 2 = 20 kg N 1 = 10 kg Horizontal displacement, DH

31 Results of a Direct shear Test Shear Stress, s (psf) C ( 1,s 1 ) ( 2,s 2 ) f ( 3,s 3 ) Normal Stress, psf

32 Vertical displacement Figures (cont) Horizontal displacement

33 The drained angle of Friction, f ', of normally Consolidated Clays Generally Decreases with the plasticity of Soil.

39 39 Triaxial Shear Test

40 Triaxial Shear Test The test is designed to as closely as possible to the actual field or in situ conditions of the soil. Triaxial tests are run by: saturating the soil applying the confining stress (called σ 3 ) Then applying the vertical stress (sometimes called the deviator stress) until failure 3 main types of triaxial tests: Consolidated Drained Consolidated Undrained Unconsolidated Undrained 40

41 41 Consolidated Drained Triaxial Test The specimen is saturated Confining stress (σ 3 ) is applied This squeezes the sample causing volume decrease Drain lines kept open and must wait for full consolidation (u = 0) to continue with test Once full consolidation is achieved, normal stress applied to failure with drain lines still open Normal stress applied very slowly allowing full drainage and full consolidation of sample during test (u = 0) Test can be run with varying values of σ 3 to create a Mohrs circle and to obtain a plot showing c and φ Test can also be run such that σ 3 is applied allowing full consolidation, then decreased (likely allowing some swelling) then the normal stress applied to failure simluating overconsolidated soil.

42 Consolidated Drained Triaxial Test In the CD test, the total and effective stress is the same since u is maintained at 0 by allowing drainage This means you are testing the soil in effective stress conditions Applicable in conditions where the soil will fail under a long term constant load where the soil is allowed to drain (long term slope stability) 42

43 43 Consolidated Undrained Triaxial Test The specimen is saturated Confining stress (σ 3 ) is applied This squeezes the sample causing volume decrease Again, must wait for full consolidation (u = 0) Once full consolidation is achieved, drain lines are closed (no drainage for the rest of the test), and normal stress applied to failure Normal stress can be applied faster since no drainage is necessary (u not equal to 0) Test can be run with varying values of σ 3 to create a Mohrs circle and to obtain a plot showing c and φ Applicable in situations where failure may occur suddenly such as a rapid drawdown in a dam or levee

44 Unconsolidated Undrained Test The specimen is saturated Confining stress (σ 3 ) is applied without drainage or consolidation (drains closed the entire time) Normal stress then increased to failure without allowing drainage or consolidation This test can be run quicker than the other 2 tests since no consolidation or drainage is needed. Test can be run with varying values of σ 3 to create a Mohrs circle and to obtain a plot showing c and φ Applicable in most practical situations foundations for example. This 44 test commonly shows a φ = 0 condition

sample O-ring impervious membrane Porous stone Water

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

46 The Real World Triaxial tests rarely run The unconfined test is very common In most cases, clays considered φ = 0 and c is used as the strength Sands are considered c = 0 and φ is the strength parameter Direct shear test gives us good enough data for sand / clay mixes (soils with both c and φ) 46

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

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

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

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

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

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

53 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

2: During axial stress increase VC + D V = VC + D = 1 Drainage hc 0 h =

54 Step 1: At the end of consolidation VC Consolidated- drained test (CD Test) Total, = Neutral, u + Effective, 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

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

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

57 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 Axial strain Dense sand or OC clay Loose sand or NC Clay Axial strain Dense sand or OC clay Loose sand or NC clay

Confining stress = 3b Confining stress = 3a 1 = 3 + (D d ) f 3 Mohr

58 Shear stress, CD tests How to determine strength parameters c and f 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 f 3a 3b 3c 1a 1b 1c or (D d ) fa (D d ) fb

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

60 Shear stress, CU tests How to determine strength parameters c and f 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 f 3 = 3 - u f Effective stresses at failure f cu C u fa u fb c cu 3b 1b 3a 3b 3a 1a (D d ) fa 1a 1b or

61 Shear stress, 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 f f cu 3a 3a 1a 1a or (D d ) fa Therefore, one CU test would be sufficient to determine f cu and f (= f d ) of sand or NC clay

62 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

63 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 3 1 D f

64 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 Failure envelope, f u = 0 c u u b u a 3a 3b 1a 1b 3 D f 1 or

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

66 Unconfined Compression Test The specimen is not placed in the cell Specimen is open to air with a σ 3 of 0 Test is similar to concrete compression test, except with soil (cohesive why?) Applicable in most practical situations foundations for example. Drawing Mohrs circle with σ 3 at 0 and the failure (normal) stress σ 3 defining the 2 nd point of the circle often called q u in this special case c becomes ½ of the failure stress 66

Laboratory Testing Total & Effective Stress Analysis

SKAA 1713 SOIL MECHANICS Laboratory Testing Total & Effective Stress Analysis Prepared by: Dr. Hetty Mohr Coulomb failure criterion with Mohr circle of stress 2 ' 2 ' ' ' 3 ' 1 ' 3 ' 1 Cot Sin c ' ' 2