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Soil Strength
Soil strength ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Mohr-Coulomb failure criterion The limiting shear stress (soil strength) is given by    =  c  +   n  tan   where c = cohesion (apparent)    = friction angle   n
[object Object],[object Object],[object Object],[object Object],Mohr-Coulomb failure criterion
Effective stress failure criterion are known as the effective (or drained) strength parameters. If the soil is at failure the effective stress failure criterion will always be satisfied.
Effective stress failure criterion are known as the effective (or drained) strength parameters. Soil behaviour is controlled by effective stresses, and the effective strength parameters are the fundamental strength parameters. But they are not necessarily soil constants. If the soil is at failure the effective stress failure criterion will always be satisfied.
Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u  and   u  are known as the undrained strength parameters
Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u  and   u  are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil.
Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u  and   u  are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil. The undrained strength is only relevant in practice to clayey soils that in the short term remain undrained. Note that as the pore pressures are unknown for undrained loading the effective stress failure criterion cannot be used.
Tests to measure soil strength 1.  Shear Box Test Motor drive Load cell to measure Shear Force Normal load Rollers Soil Porous plates Top platen Measure  relative horizontal displacement, dx  vertical displacement of top platen, dy
 
[object Object],[object Object],[object Object],[object Object],Shear box test
Typical drained shear box results Normal load increasing
Typical drained shear box results    = F/A    = N/A Peak Ultimate N 1 N 2
[object Object],[object Object],[object Object],[object Object],[object Object],Interpretation of shear box tests
Shear box test - advantages ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],Shear box test - disadvantages
Tests to measure soil strength 2.  The Triaxial Test Cell pressure Pore pressure and volume change Rubber membrane Cell water O-ring seals Porous filter disc Confining cylinder Deviator load Soil
 
Stresses in triaxial specimens  r  r   =  Radial stress (cell  pressure)  a   = Axial stress F =  Deviator load  r
Stresses in triaxial specimens  r  r   =  Radial stress (cell  pressure)  a   = Axial stress F =  Deviator load  r From equilibrium we have
Stresses in triaxial specimens F/A is known as the deviator stress, and is given the symbol q The axial and radial stresses are principal stresses
Stresses in triaxial specimens F/A is known as the deviator stress, and is given the symbol q ,[object Object],[object Object],[object Object]
Stresses in triaxial specimens F/A is known as the deviator stress, and is given the symbol q ,[object Object],[object Object],[object Object],[object Object]
Stresses in triaxial specimens F/A is known as the deviator stress, and is given the symbol q ,[object Object],[object Object],[object Object],[object Object],[object Object]
Strains in triaxial specimens From the measurements of change in height, dh, and change in volume dV we can determine Axial strain Volume strain where h 0  is the initial height and V 0  is the initial volume
Strains in triaxial specimens From the measurements of change in height, dh, and change in volume dV we can determine Axial strain Volume strain where h 0  is the initial height and V 0  is the initial volume It is assumed that the specimens deform as right circular cylinders. The cross-sectional area, A, can then be determined from
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Types of triaxial test
[object Object],[object Object],[object Object],[object Object],[object Object],Advantages of  the triaxial test
Typical triaxial results q  a Increasing cell pressure
Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress.     1  3
Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress.     1  3 c The Mohr-Coulomb failure locus is tangent to the Mohr circles at failure
Mohr Circles    1  3  2  (      From the Mohr Circle geometry
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Mohr Circles
Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot   p R Failure occurs if a Mohr circle touches the failure criterion. Then   R  =  sin    ( p  +  c cot  
Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot   p R Failure occurs if a Mohr circle touches the failure criterion. Then   R  =  sin    ( p  +  c cot  
Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot   p R Failure occurs if a Mohr circle touches the failure criterion. Then   R  =  sin    ( p  +  c cot  
Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters  are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses
Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters  are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses
Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters  are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses where
   1  3 u u Effective and total stress Mohr circles For any point in the soil a total and an effective stress Mohr circle can be drawn. These are the same size with The two circles are displaced horizontally by the pore pressure, u.
[object Object],[object Object],[object Object],[object Object],Interpretation of Laboratory results
[object Object],[object Object],[object Object],Interpretation of Laboratory results
[object Object],[object Object],[object Object],[object Object],Interpretation of Laboratory results
[object Object],[object Object],[object Object],[object Object],Interpretation of Laboratory results
Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage.
Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown    1  3
Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown    1  3 We find that all the total stress Mohr circles are the same size, and therefore   u  = 0  and    = s u  = c u  = constant
Relation between effective and total stress criteria    1  3 Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle
Relation between effective and total stress criteria    1  3 Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle We have the following relations
[object Object],[object Object],[object Object],Relation between effective and total stress criteria
[object Object],[object Object],[object Object],[object Object],[object Object],Significance of undrained strength parameters
Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,   ’ = 26 o  and the pore pressure at failure is 43 kPa.
Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,   ’ = 26 o  and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 =
Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,   ’ = 26 o  and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 =  Failure criterion
Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,   ’ = 26 o  and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 =  Failure criterion Hence    ’ =  57.4 kPa,    ’ =  22.4 kPa and cell pressure (total stress) =    ’ + u =  65.4 kPa
Graphical solution   26 17.5
Graphical solution   26 17.5
Graphical solution    1  3 26 17.5
 

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Soil strength

  • 2.
  • 3. Mohr-Coulomb failure criterion The limiting shear stress (soil strength) is given by  = c +  n tan  where c = cohesion (apparent)  = friction angle   n
  • 4.
  • 5. Effective stress failure criterion are known as the effective (or drained) strength parameters. If the soil is at failure the effective stress failure criterion will always be satisfied.
  • 6. Effective stress failure criterion are known as the effective (or drained) strength parameters. Soil behaviour is controlled by effective stresses, and the effective strength parameters are the fundamental strength parameters. But they are not necessarily soil constants. If the soil is at failure the effective stress failure criterion will always be satisfied.
  • 7. Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u and  u are known as the undrained strength parameters
  • 8. Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u and  u are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil.
  • 9. Total stress failure criterion If the soil is taken to failure at constant volume (undrained) then the failure criterion can be written in terms of total stress as c u and  u are known as the undrained strength parameters These parameters are not soil constants, they depend strongly on the moisture content of the soil. The undrained strength is only relevant in practice to clayey soils that in the short term remain undrained. Note that as the pore pressures are unknown for undrained loading the effective stress failure criterion cannot be used.
  • 10. Tests to measure soil strength 1. Shear Box Test Motor drive Load cell to measure Shear Force Normal load Rollers Soil Porous plates Top platen Measure relative horizontal displacement, dx vertical displacement of top platen, dy
  • 11.  
  • 12.
  • 13. Typical drained shear box results Normal load increasing
  • 14. Typical drained shear box results  = F/A  = N/A Peak Ultimate N 1 N 2
  • 15.
  • 16.
  • 17.
  • 18. Tests to measure soil strength 2. The Triaxial Test Cell pressure Pore pressure and volume change Rubber membrane Cell water O-ring seals Porous filter disc Confining cylinder Deviator load Soil
  • 19.  
  • 20. Stresses in triaxial specimens  r  r = Radial stress (cell pressure)  a = Axial stress F = Deviator load  r
  • 21. Stresses in triaxial specimens  r  r = Radial stress (cell pressure)  a = Axial stress F = Deviator load  r From equilibrium we have
  • 22. Stresses in triaxial specimens F/A is known as the deviator stress, and is given the symbol q The axial and radial stresses are principal stresses
  • 23.
  • 24.
  • 25.
  • 26. Strains in triaxial specimens From the measurements of change in height, dh, and change in volume dV we can determine Axial strain Volume strain where h 0 is the initial height and V 0 is the initial volume
  • 27. Strains in triaxial specimens From the measurements of change in height, dh, and change in volume dV we can determine Axial strain Volume strain where h 0 is the initial height and V 0 is the initial volume It is assumed that the specimens deform as right circular cylinders. The cross-sectional area, A, can then be determined from
  • 28.
  • 29.
  • 30. Typical triaxial results q  a Increasing cell pressure
  • 31. Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress.    1  3
  • 32. Mohr Circles To relate strengths from different tests we need to use some results from the Mohr circle transformation of stress.    1  3 c The Mohr-Coulomb failure locus is tangent to the Mohr circles at failure
  • 33. Mohr Circles    1  3  2  (      From the Mohr Circle geometry
  • 34.
  • 35. Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot  p R Failure occurs if a Mohr circle touches the failure criterion. Then R = sin  ( p + c cot 
  • 36. Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot  p R Failure occurs if a Mohr circle touches the failure criterion. Then R = sin  ( p + c cot 
  • 37. Mohr-Coulomb criterion (Principal stresses)     1  3 c c cot  p R Failure occurs if a Mohr circle touches the failure criterion. Then R = sin  ( p + c cot 
  • 38. Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses
  • 39. Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses
  • 40. Effective stress Mohr-Coulomb criterion As mentioned previously the effective strength parameters are the fundamental parameters. The Mohr-Coulomb criterion must be expressed in terms of effective stresses where
  • 41.    1  3 u u Effective and total stress Mohr circles For any point in the soil a total and an effective stress Mohr circle can be drawn. These are the same size with The two circles are displaced horizontally by the pore pressure, u.
  • 42.
  • 43.
  • 44.
  • 45.
  • 46. Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage.
  • 47. Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown    1  3
  • 48. Relation between effective and total stress criteria Three identical saturated soil samples are sheared to failure in UU triaxial tests. Each sample is subjected to a different cell pressure. No water can drain at any stage. At failure the Mohr circles are found to be as shown    1  3 We find that all the total stress Mohr circles are the same size, and therefore  u = 0 and  = s u = c u = constant
  • 49. Relation between effective and total stress criteria    1  3 Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle
  • 50. Relation between effective and total stress criteria    1  3 Because each sample is at failure, the fundamental effective stress failure condition must also be satisfied. As all the circles have the same size there must be only one effective stress Mohr circle We have the following relations
  • 51.
  • 52.
  • 53. Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,  ’ = 26 o and the pore pressure at failure is 43 kPa.
  • 54. Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,  ’ = 26 o and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 =
  • 55. Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,  ’ = 26 o and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 = Failure criterion
  • 56. Example In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c’ = 0,  ’ = 26 o and the pore pressure at failure is 43 kPa. Analytical solution Undrained strength = 17.5 = Failure criterion Hence   ’ = 57.4 kPa,   ’ = 22.4 kPa and cell pressure (total stress) =   ’ + u = 65.4 kPa
  • 57. Graphical solution   26 17.5
  • 58. Graphical solution   26 17.5
  • 59. Graphical solution    1  3 26 17.5
  • 60.