CEN 512 Shear Strength of Soils

Shear Strength of Soils - soil resistance to sliding on an i nternal surface - load per unit area on a plane parallel to the direction of shear force Attributed to three components: 1) Frictional resistance to sliding between solid particles 2) Cohesion and adhesion between particles 3) Interlocking and bridging of solid particles Mohr – Coulomb Failure Criterion - From the combined principles: a) Mohr Failure in materials occur with critical combination of normal and shear stresses, and not by maximum normal and shear stress on the failure plane. failure envelope  = () () b) Coulomb Shear stress on failure plane can be expressed as a linear function of normal stress -

Figure 1. Inclination of failure plane in s oil with major principal plane (Das, 2013)

Combining,

 =    tan  Where

c = cohesion Ф = angle of friction  (internal shearing resistance)   = shear strength or shearing resistance σ = normal stress on shear plane

Applying in soil mechanics,

Figure 2. Mohr’s circle and failure envelope (Das, 2013)



 =   ( − ) tan  = ′  ′ t an an  ′ where: u = pore water pressure on shear plane σ’ = effective normal stress on shear plane c’ = cohesion intercept in terms of effective stress Ф’ = angle of shearing resistance or friction angle in terms of effective stress

Note: 1) Tangent to circle defines the Mohr-Coulomb failure envelope 2) Angle of Failure Plane - From the figure, 2Ѳ = 90o + Ф’ Ѳ = 45o + (Ф’/2)

Significance of Failure Envelope

Cohesion – resulting from bonding between particles, or a dense state of packing of particles Table 2. Typical values of cohesion Soil Type Soft normally consolidated clays Firm to stiff clays Compacted clays Stiff to hard clays Heavily consolidated clays Source: Wesley, 2010

Figure 3. Mohr’s failure envelope and Mohr -Coulomb’s failure law (Das, 2013) Case 1: If normal stress and shear stress on a plane plot as point A, shear failure does not occur. Case 2: If normal stresses and shear stress plot as point B or any point on the failure envelope, shear failure occurs. Case 3: Normal stress and shear stress plot as point C cannot exist since failure would have occurred before such condition is reached

Direct Shear Stress (Shear Box Stress) - earliest form of shear test (first used by Coulomb in 1776) - series of tests is done at varying no rmal stresses then test result is plotted: 1) stress vs displacement and 2) peak shear stress (failure value) v s normal stress a line through the points defines c’ and Ф’ - simple and easy to perform - From the figure below,

tan = tan =

Table 1. Relationship between Relative Density and Angle of Friction of Cohesionless Soils Angle of friction, Ф’ State of Packing Relative density (%) (deg) Very loose 45 Source: Das, 2013

Cohesion Intercept, c’ (KPa) Normally close to 0 Kpa, but could reach 10 KPa 10-25 KPa 12-25 KPa 25-100 KPa



if c’ = 0

  

=

 + 

if c’ > 0

σ’

Figure 4. Graph of τ vs σ’ from Direct shear Test

Triaxial test - most popular and most common method

Three Types of Tri-axial Test

3  ∆ = 1

3

3

3  ∆ = 1

Δσ σ1 σ3

Ф c

= = = = = = = = = = = =

additional pressure deviator stress maximum principal stress axial stress minimum principal stress confining pressure lateral pressure radial stress cell pressure chamber pressure angle of internal friction cohesion of soil

1) Undrained Test or Unconsolidated Undrained Test - To determine udrained shear strength of soil available on the plane when it is sheared to failure under undrained conditions - for cohesive soils (clays and silts) and not for coarse-grained soils - No drainage is permitted during consolidation stage and loading stage. Hence, Pore pressure is not normally measured - implies analysis of total stress only since undrained soil behavior is directly related to total stress - results are plotted using total stress since onl y total stresses are known - soil behaves as though it has no frictional component of shear (friction angle, Ф = 0) since soil is saturated a nd undrained - the value of c is the undrained shear strength, qu - each increase in confining stress, σ3  is accompanied by an identical increase in pore water pressure. Hence, no change in effective stress - if pore pressure in each test is measured then effective stress is calculated and plotted, circles will just coincide, hence, effective stress parameters are still unable to be determined since a single circle is insufficient to determine the failure line

cu = τf

Figure 5. Result of an unconsolidated - undrained triaxial stress on a saturated soil (Wesley, 2010)

Unconfined Compression Test - special type of unconsolidated undrained test - confining pressure σ3 is 0 and axial load is quickly applied to cause failure - undrained shear strength is independent of t he confining pressure

 = where:

1  = =  2 2

qu = unconfined compression strength

- Pore pressure is measured, hence, effective stresses are calculated and results are plotted in terms of effective stresses and σ1‘ = σ1 – u u hence, σ3‘ = σ3 – σ1‘ - σ3‘ = σ1 - σ3 - Deviator stress Δσ is the same whether expressed as total stress or effective stress

3) Drained Test - Full drainage is allowed during consolidation stage and loading stage - Commonly used on granular material like gravel and sand because of its high permeability (low permeability in clay presents problems in conducting drained tests – days required for draining) - To determine effective stress strength parameters c’ and Ф’ (results are plotted in terms of effective stress) - Pore pressure u=0, hence, effective stress = total stress Figure 6. Unconfined Compression Test (Das, 2013)

Table 3. General Relationship of consistency and unconfined compression strength of clays (Das, 2013) Consistency qu (KN/m2) Very soft Soft Medium Stiff Very stiff Hard

0-25 25-50 50-100 100-200 200-400 >400

2) Consolidated Undrained Test - Drainage is permitted during the consolidation stage until the sample is fully consolidated (all pore pressure has dissipated to 0) - No drainage is permitted during the loading stage - To determine effective stress strength parameters c’ and Ф’ (results are plotted in terms of effective stress) - Preferred for clay due to its low permeability

Figure 7. Results of a consolidated undrained triaxial test (Wesley, 2010)

Sample Problems 1) A sample of sand is subjected to direct shear testing at its normal water content. Two tests are performed. For one of the tests, the sample fails at a shear stress of 3000 psf when the normal stress is 4000 psf. In the second test, the sample shears at a stress of 400 psf when the normal stress is 6000 psf. From these, determine the f f. a) Angle of internal friction b) Cohesion of soil 2) An unconfined compression test was carried out on a saturated clay sample. The maximum load the clay sustained was 127 N a nd the vertical displacement was 0.8mm. The size of the sample was 38mm diameter x 76mm long. Determine the undrained shear strength. 3) A consolidated-undrained soil test was conducted on a normally consolidated sample with a chamber pressure of 140 Kpa. The sample failed when the deviator stress was 124 KPa. The pore water pressure in the sample at that time was 75 Kpa. Determine: a) Undrained angle of internal friction b) Drained angle of internal friction c) Drained angle of internal frictio n if soil possess a cohesion of 12 KPa 4) The sample of soil in a tri-axial test have the ff. stresses: Cell Pressure Deviator Stress Pore Pressure 25 KPa 20 KPa 12 KPa 34 KPa 31 KPa 10 KPa a) Drained angle of internal friction b) Cohesion of soil c) Angle of failure in shear

Practice Problems 1) A cohesive soil with an angle of shearing resistance of 28o has a cohesion of 32 Kpa. The shear stress at fail ure is 64 KPa. Determine: a) Normal stress  Ans. 60.1832 KPa b) Confining pressure  Ans. 21.7282 KPa c) Maximum principal stress  Ans. 166.6972 KPa

2) In a direct shear test, the soil was determined to have an angle of internal friction of 31o and cohesion of 26 KPa. If the normal stress is 150 KPa, compute the following: a) Total shear stress  Ans. 116.1291 KPa b) Force required to cause failure in shear when the sample has a dimension of 50mm x 50mm and height of 75mm.  Ans. 290.3227 KN

References Das, Braja M. (2013). Fundamentals of Geotechnical Engineering, 4th Edition. Cengage Learning Asia Pte Ltd. Wesley, Laurence D. (2010). Fundamentals of Soil Mechanics for Sedimentary and Residual Soils. John Wiley & Sons, Inc.