1. Shear Strength of Soils

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

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

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 )   f is the maximum shear stress the soil can take without failure, under normal stress of  .  c  failure envelope Cohesion Friction angle  f 

9. Mohr-Coulomb Failure Criterion ( in terms of effective stresses )  f is the maximum shear stress the soil can take without failure, under normal effective stress of  ’. u = pore water pressure   ’ c’  ’ failure envelope Effective cohesion Effective friction angle  f  ’

10. Mohr-Coulomb Failure Criterion Shear strength consists of two components: cohesive and frictional .  ’ f  f  ’   ' c’ c’ cohesive component  ’ f tan  ’ frictional component

11. c and  are measures of shear strength. Higher the values, higher the shear strength.

12. Mohr Circle of stress Resolving forces in  and  directions, Soil element  ’ 1  ’ 1  ’ 3  ’ 3   ’ 

13. Mohr Circle of stress   ’

14. Mohr Circle of stress   ’ P D = Pole w.r.t. plane   ’ , 

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

16. Mohr Circles & Failure Envelope  c The soil element does not fail if the Mohr circle is contained within the envelope GL Y  c  c Initially, Mohr circle is a point   c +  

17. Mohr Circles & Failure Envelope Y  c GL  c  c As loading progresses, Mohr circle becomes larger… .. and finally failure occurs when Mohr circle touches the envelope 

18.  ’ Orientation of Failure Plane  ’ Failure envelope P D = Pole w.r.t. plane   ’ ,  f   –  Therefore,  –  ’ =   45 +  ’ /2

19. Mohr circles in terms of total & effective stresses = X  v ’  h ’ X u u +  v ’  h ’ effective stresses u  v  h X  v  h total stresses   or ’

20. Failure envelopes in terms of total & effective stresses = If X is on failure X  v ’  h ’ X u u +  v ’  h ’ effective stresses u  v  h X  v  h total stresses   or ’ c  Failure envelope in terms of total stresses  ’ c’ Failure envelope in terms of effective stresses

21. Mohr Coulomb failure criterion with Mohr circle of stress X  ’ v =  ’ 1  ’ h =  ’ 3 X is on failure  ’ 1  ’ 3 effective stresses  ’  ’ c’ Failure envelope in terms of effective stresses c’ Cot ’  ’    ’    ’    ’   Therefore,

22. Mohr Coulomb failure criterion with Mohr circle of stress

24. Laboratory tests How to take undisturbed samples

25. Laboratory tests Field conditions z  vc  vc  hc  hc Before construction A representative soil sample z  vc +   hc  hc After and during construction  vc + 

26. Laboratory tests Simulating field conditions in the laboratory Step 2 Apply the corresponding field stress conditions Step 1 Set the specimen in the apparatus and apply the initial stress condition  vc  vc  hc  hc Representative soil sample taken from the site 0 0 0 0  vc +   hc  hc  vc +  Traxial test  vc  vc   Direct shear test

27. Direct shear test Schematic diagram of the direct shear apparatus

28. Direct shear test Preparation of a sand specimen Direct shear test is most suitable for consolidated drained tests specially on granular soils (e.g.: sand) or stiff clays Components of the shear box Preparation of a sand specimen Porous plates

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

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

31. Direct shear test Step 2: Lower box is subjected to a horizontal displacement at a constant rate 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

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

33. Direct shear test Analysis of test results Note: Cross-sectional area of the sample changes with the horizontal displacement

34. Direct shear tests on sands Stress-strain relationship Shear stress,  Shear displacement Dense sand/ OC clay  f Loose sand/ NC clay  f Dense sand/OC Clay Loose sand/NC Clay Change in height of the sample Expansion Compression Shear displacement

35. Direct shear tests on sands How to determine strength parameters c and   f1 Normal stress =  1 Shear stress,  Shear displacement  f2 Normal stress =  2  f3 Normal stress =  3 Shear stress at failure,  f Normal stress,   Mohr – Coulomb failure envelope

36. Direct shear tests on sands 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 Some important facts on strength parameters c and  of sand

37. Direct shear tests on clays Failure envelopes for clay from drained direct shear tests 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) Shear stress at failure,  f Normal force,  ’ Normally consolidated clay (c’ = 0) Overconsolidated clay (c’ ≠ 0)

38. 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, c a = adhesion,  = angle of internal friction

40. Lets do some examples

41. Triaxial Shear Test Soil sample at failure Failure plane 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

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

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

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

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

46. Types of Triaxial Tests Is the drainage valve open? Is the drainage valve open? yes no C onsolidated sample U nconsolidated sample yes no D rained loading U ndrained loading Under all-around cell pressure  c  c  c  c  c Step 1 deviatoric stress (  = q) Shearing (loading) Step 2  c  c  c + q

47. Types of Triaxial Tests Is the drainage valve open? yes no C onsolidated sample U nconsolidated sample Under all-around cell pressure  c Step 1 Is the drainage valve open? yes no D rained loading U ndrained loading Shearing (loading) Step 2 CD test CU test UU test

48. Consolidated- drained test (CD Test) Step 1: At the end of consolidation Step 2: During axial stress increase Step 3: At failure  VC  hC Total,  = Neutral, u Effective, ’ + 0 ’ VC =  VC ’ hC =  hC  VC +   hC 0 ’ V =  VC +  = ’ 1 ’ h =  hC = ’ 3 Drainage Drainage  VC +  f  hC 0 ’ Vf =  VC +  f  = ’ 1f ’ hf =  hC = ’ 3f Drainage

49. Deviator stress (q or  d ) =  1 –  3 Consolidated- drained test (CD Test)  1 =  VC +    3 =  hC

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

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

52. CD tests How to determine strength parameters c and  Deviator stress,  d Axial strain Shear stress,   or ’  Mohr – Coulomb failure envelope  d ) fa Confining stress =  3a  d ) fb Confining stress =  3b  d ) fc Confining stress =  3c  3c  1c  3a  1a (  d ) fa  3b  1b (  d ) fb  1 =  3 + (  d ) f   3

53. CD tests Since u = 0 in CD tests,  =  ’ Therefore, c = c’ and  =  ’ c d and  d are used to denote them Strength parameters c and  obtained from CD tests

54. CD tests Failure envelopes For sand and NC Clay, c d = 0 Therefore, one CD test would be sufficient to determine  d of sand or NC clay Shear stress,   or ’  d Mohr – Coulomb failure envelope  3a  1a (  d ) fa

55. CD tests Failure envelopes For OC Clay, c d ≠ 0   or ’   3  1 (  d ) f c  c OC NC

56. Some practical applications of CD analysis for clays  = in situ drained shear strength 1. Embankment constructed very slowly, in layers over a soft clay deposit  Soft clay

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

58. Some practical applications of CD analysis for clays 3. Excavation or natural slope in clay  = 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 

59. Consolidated- Undrained test (CU Test) Step 1: At the end of consolidation Step 2: During axial stress increase Step 3: At failure  VC  hC Total,  = Neutral, u Effective, ’ + 0 ’ VC =  VC ’ hC =  hC  VC +   hC ±  u Drainage  VC +  f  hC No drainage No drainage ±  u f ’ V =  VC +  ±  u = ’ 1 ’ h =  hC ±  u = ’ 3 ’ Vf =  VC +  f  ±  u f = ’ 1f ’ hf =  hC ±  u f = ’ 3f

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

61. Stress-strain relationship during shearing Consolidated- Undrained test (CU Test) Deviator stress,  d Axial strain Dense sand or OC clay  d ) f Dense sand or OC clay Loose sand /NC Clay  u + - Axial strain Loose sand or NC Clay  d ) f

62. CU tests How to determine strength parameters c and  Deviator stress,  d Axial strain Shear stress,   or ’  d ) fb Confining stress =  3b  3b  1b  3a  1a (  d ) fa  cu Mohr – Coulomb failure envelope in terms of total stresses c cu  1 =  3 + (  d ) f   3 Total stresses at failure  d ) fa Confining stress =  3a

63. CU tests How to determine strength parameters c and   cu Mohr – Coulomb failure envelope in terms of total stresses c cu Effective stresses at failure u f (  d ) fa Shear stress,   or ’  3b  1b  3a  1a (  d ) fa ’ 3b ’ 1b ’ 3a ’ 1a Mohr – Coulomb failure envelope in terms of effective stresses ’ C ’ u fa u fb ’ 1 =  3 + (  d ) f - u f  ’   =  3 - u f 

64. CU 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 Strength parameters c and  obtained from CD tests

65. CU tests Failure envelopes For sand and NC Clay, c cu and c’ = 0 Therefore, one CU test would be sufficient to determine  cu and ’ =  d ) of sand or NC clay Shear stress,   or ’  cu Mohr – Coulomb failure envelope in terms of total stresses  3a  1a (  d ) fa  3a  1a  ’ Mohr – Coulomb failure envelope in terms of effective stresses

66. Some practical applications of CU analysis for clays  = in situ undrained shear strength 1. Embankment constructed rapidly over a soft clay deposit  Soft clay

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

68. Some practical applications of CU analysis for clays 3. Rapid construction of an embankment on a natural slope 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  = In situ undrained shear strength 

69. Lets do an example

70. Unconsolidated- Undrained test (UU Test) Data analysis 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 –  H) = A 0 × H 0 A ×(1 –  H/H 0 ) = A 0  C =  3  C =  3 No drainage Initial specimen condition  3 +  d  3 No drainage Specimen condition during shearing

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

72. Unconsolidated- Undrained test (UU Test) Step 3: During application of axial load  u d = AB  d = +  3 +  d  3 No drainage ’ 1 =  3 +  d  -  u c  u d ’ 3 =  3 -  u c  u d  u c ±  u d Increase of pwp due to increase of deviator stress Increase of deviator stress Skempton’s pore water pressure parameter, A

73. Unconsolidated- Undrained test (UU Test) Combining steps 2 and 3,  u =  u c +  u d Total pore water pressure increment at any stage,  u  u = B [  3 + A  d ]  u c = B  3  u d = AB  d Skempton’s pore water pressure equation  u = B [  3 + A(  1 –  3 ]

74. Unconsolidated- Undrained test (UU Test) Derivation of Skempton’s pore water pressure equation

75. Step 1 :Increment of isotropic stress Derivation of Skempton’s pore water pressure equation Increase in effective stress in each direction =  3 -  u c  2  3  1 No drainage  1 +  3  3 +  3  2 +  3 No drainage  u c

76. Step 2 :Increment of major principal stress Derivation of Skempton’s pore water pressure equation Increase in effective stress in  1 direction =  1 -  u d Increase in effective stress in  2 and  3 directions =  -  u d Average Increase in effective stress = (  1 -  u d -  u d –  u d )/3  2  3  1 No drainage  1 +  1  3 + 0  2 + 0 No drainage  u c

77. Typical values for parameter B

78. Typical values for parameter A NC Clay (low sensitivity) (A = 0.5 – 1.0) NC Clay (High sensitivity) (A > 1.0) Collapse of soil structure may occur in high sensitivity clays due to very high pore water pressure generation  1 –  3 u Axial strain Axial strain u  1 –  3

79. Typical values for parameter A OC Clay (Lightly overconsolidated) (A = 0.0 – 0.5) OC Clay (Heavily overconsolidated) (A = -0.5 - 0.0) During the increase of major principal stress pore water pressure can become negative in heavily overconsolidated clays due to dilation of specimen  1 –  3 Axial strain u  1 –  3 Axial strain u

80. Typical values for parameter A

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

82. Unconsolidated- Undrained test (UU Test) 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 Total,  = Neutral, u Effective, ’ + Step 3: At failure ’ hf =  C + u r -  c  u f = ’ 3f ’ Vf =  C +  f  + u r -  c  u f = ’ 1f -u r  c ±  u f  C  C +  f No drainage  ’ ’ 3 ’ 1  f

83. Unconsolidated- Undrained test (UU Test) Mohr circles in terms of total stresses  3b  1b  3a  1a  f ’ 3 ’ 1 Total,  = Neutral, u Effective, ’ + Step 3: At failure ’ hf =  C + u r -  c  u f = ’ 3f ’ Vf =  C +  f  + u r -  c  u f = ’ 1f -u r  c ±  u f  C  C +  f No drainage   or ’ u a u b Failure envelope,  u = 0 c u

84. Unconsolidated- Undrained test (UU Test) Effect of degree of saturation on failure envelope S < 100% S > 100%  3b   b  3a   a  3c   c   or ’

85. Some practical applications of UU analysis for clays  = in situ undrained shear strength 1. Embankment constructed rapidly over a soft clay deposit  Soft clay

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

87. Some practical applications of UU analysis for clays 3. Footing placed rapidly on clay deposit 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  = In situ undrained shear strength

88. Unconfined Compression Test (UC Test)  1 =  VC +    3 = 0 Confining pressure is zero in the UC test

89. Unconfined Compression Test (UC Test) Note: Theoritically q u = c u , However in the actual case q u < c u due to premature failure of the sample  1 =  VC +  f   3 = 0 Shear stress,  Normal stress,  q u

90. Lets do an example

91. Stress Invariants ( p and q ) p (or s) = (  1 +  3 )/2 q (or t) = (  1 -  3 )/2 p and q can be used to illustrate the variation of the stress state of a soil specimen during a laboratory triaxial test    3  1 (  1 +  3 )/2 (  1 -  3 )/2 c 

92. GL Stress Invariants ( p and q ) p (or s) = (  1 +  3 )/2 q (or t) = (  1 -  3 )/2  c  c   c  or q  or p Failure envelope Stress path

93. Mohr Coulomb failure envelope in terms of stress invariants p (or s) = (  1 +  3 )/2 q (or t) = (  1 -  3 )/2    3  1 (  1 +  3 )/2 (  1 -  3 )/2 c 

94. p (or s) = (  1 +  3 )/2 q (or t) = (  1 -  3 )/2 Mohr Coulomb failure envelope in terms of stress invariants Therefore, sin  =  tan   or q  or p  f = c +  tan  q = c cos  + p  sin    =  sin -1 (tan  c cos 

95. Stress path for CD triaxial test In CD tests pore water pressure is equal to zero. Therefore, total and effective stresses are equal p, p’ (or s, s’) =  3 q (or t) = 0 p, p’ (or s, s’) =  3 +  d /2 q (or t) =  d /2  3 p, p’ (or s, s’) = (  1 +  3 )/2 = ( ’ 1 + ’ 3 )/2 q (or t) = (  1 -  3 )/2  or q  or p Failure envelope Step 1  3  3 Step 2  3 +  d  3  d Stress path 45 0

96. Stress path for CU triaxial test In CU tests pore water pressure develops during shearing p, p’ (or s, s’) =  3 q (or t) = 0 p (or s) =  3 +  d /2 q (or t) =  d /2  3 = ’ 3 p (or s) = (  1 +  3 )/2 p’ (or s’) = (  1 +  3 )/2 - u q (or t) = (  1 -  3 )/2 Step 1  3  3  d Total stress path 45 0  u d Step 2  3 +  d  3  u d Effective stress path q ’ or p, p’

97. Lets do some example

100. Direct simple shear test Direct shear test Direct simple shear test  = 80 mm Soil specimen Porous stones Spiral wire in rubber membrane

102. Torsional ring shear test Peak Residual  Shear displacement  f  ’  ’ max  ’ res

103. Torsional ring shear test Preparation of ring shaped undisturbed samples is very difficult. Therefore, remoulded samples are used in most cases  N

105. Plane strain triaxial test  ’ 1 ,  1  ’ 2 ,  2  ’ 3 ,  3 Plane strain test  ’ 2 ≠  ’ 3  2 = 0  ’ 1  ’ 2  ’ 3  ’ 2 Rigid platens Specimen

108. Vane shear test This is one of the most versatile and widely used devices used for investigating undrained shear strength (C u ) and sensitivity of soft clays Rate of rotation : 6 0 – 12 0 per minute Test can be conducted at 0.5 m vertical intervals PLAN VIEW Bore hole (diameter = D B ) h > 3D B ) Vane D H Applied Torque, T Vane T Rupture surface Disturbed soil

109. Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected T = M s + M e + M e = M s + 2M e M e – Assuming a uniform distribution of shear strength C u C u d/2 d/2 C u h

110. Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected M s – Shaft shear resistance along the circumference T = M s + M e + M e = M s + 2M e C u C u

111. Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected T = M s + M e + M e = M s + 2M e M e – Assuming a triangular distribution of shear strength Can you derive this ??? C u C u h d/2 d/2 C u

112. Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected T = M s + M e + M e = M s + 2M e M e – Assuming a parabolic distribution of shear strength Can you derive this ??? C u C u h d/2 d/2 C u

113. Vane shear test Since the test is very fast, Unconsolidated Undrained (UU) can be expected After the initial test, vane can be rapidly rotated through several revolutions until the clay become remoulded C u C u h  peak  ultimate  Shear displacement

114. Some important facts on vane shear test Insertion of vane into soft clays and silts disrupts the natural soil structure around the vane causing reduction of shear strength The above reduction is partially regained after some time C u as determined by vane shear test may be a function of the rate of angular rotation of the vane

115. Correction for the strength parameters obtained from vane shear test Bjerrum (1974) has shown that as the plasticity of soils increases, C u obtained by vane shear tests may give unsafe results for foundation design. Therefore, he proposed the following correction. C u(design) =  C u(vane shear) Where,  = correction factor = 1.7 – 0.54 log (PI) PI = Plasticity Index

117. Torvane Torvane is a modification to the vane

119. Pocket Penetrometer Pushed directly into the soil. The unconfined compression strength (q u ) is measured by a calibrated spring.

120. Swedish Fall Cone (suitable for very soft to soft clays) The test must be calibrated Soil sample C u ∞ Mass of the cone ∞ 1/(penetration) 2

122. Pressuremeter Pre – bored or self – bored hole Guard cell Measuring cell Guard cell Coaxial tube Water Air

123. Pressuremeter Pre – bored or self – bored hole Guard cell Measuring cell Guard cell Coaxial tube Water Air Pressure Volumetric expansion Time Pressure Volumetric expansion Pseudo- elastic phase Elastic phase

125. Static Cone Penetrometer test Cone penetrometers with pore water pressure measurement capability are known as piezocones 40 mm 40 mm 40 mm 40 mm

126. Static Cone Penetrometer test Force required for the inner rod to push the tip (F c ) and the total force required to push both the tip and the sleeve (F c + F s ) will be measured Point resistance ( q c ) = F c / area of the tip Sleeve resistance ( q s ) = F s / area of the sleeve in contact with soil Friction Ratio ( f r ) = q s / q c ×100 (%) Various correlations have been developed to determine soil strength parameters (c,  ect) from f r

128. Standard Penetration Test, SPT SPT is the most widely used test procedure to determine the properties of in-situ soils Standard penetration resistance (SPT N) = N 2 + N 3 Number of blows for the first 150 mm penetration is disregarded due to the disturbance likely to exist at the bottom of the drill hole The test can be conducted at every 1m vertical intervals Various correlations have been developed to determine soil strength parameters (c,  ect) from N 63.5 kg 0.76 m Drill rod 0.15 m 0.15 m 0.15 m Number of blows = N 1 Number of blows = N 2 Number of blows = N 3

129. Standard Penetration Test, SPT SPT (Manual operation)

130. Various correlations for shear strength For NC clays, the undrained shear strength (c u ) increases with the effective overburden pressure,  ’ 0 For OC clays, the following relationship is approximately true For NC clays, the effective friction angle (  ’) is related to PI as follows Skempton (1957) Plasticity Index as a % Ladd (1977) Kenny (1959)

131. 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 in tropical countries like Sri Lanka Pore water pressure can be negative in unsaturated soils Solid Water Saturated soils Pore water pressure, u Effective stress,  ’ Solid Unsaturated soils Pore water pressure, u w Effective stress,  ’ Water Air Pore air pressure, u a

133. Shear strength of partially saturated soils Therefore, strength of unsaturated soils is much higher than the strength of saturated soils due to matric suction Same as saturated soils Apparent cohesion due to matric suction   -  u a ’ u a – u w = 0 (u a – u w ) 1 > 0 (u a – u w ) 2 > (u a – u w ) 1

134. How it become possible build a sand castle   -  u a Same as saturated soils Apparent cohesion due to matric suction ’ u a – u w = 0 Failure envelope for saturated sand (c’ = 0) (u a – u w ) > 0 Failure envelope for unsaturated sand Apparent cohesion

135. THE END