Lecturer: Dr. Frederick Owusu-Nimo Civil Engineering Department College of Engineering Course: Soil and Rock Mechanics (CE 260) Lecturer: Dr. Frederick Owusu-Nimo

Introduction Stress is the force per unit area on a surface Normal Stress (σ) is the component of stress acting perpendicular to a plane surface Shear Stress (𝜏) is the component which acts parallel to the plane

Introduction Principal Stress Planes: At any stress point there exist three orthogonal (i.e. mutually perpendicular) planes on which the shear stresses are zero. Example for stress point O, there can be xy, xz and yz planes where the shear stresses will be zero. These planes are called Principal Stress Planes

Introduction The normal stresses which act on the principal stress planes are called the Principal stresses The maximum of these stresses is the Major principal stress ( 𝜎 1 ) The minimum stress is called the Minor Principal Stress ( 𝜎 3 ) The intermediate stress is termed the Intermediate Principal Stress ( 𝜎 2 ) Interested in two-dimensional stress and more importantly the planes containing 𝜎 1 and 𝜎 3 The difference between 𝜎 1 and 𝜎 3 is the Deviator Stress or Stress Difference ( 𝜎 1 − 𝜎 3 )

Mohr Circle

Mohr Circle 𝜎 𝛼 = 𝜎 1 + 𝜎 3 2 + 𝜎 1 − 𝜎 3 2 cos 2𝛼 𝜏 𝛼 = 𝜎 1 − 𝜎 3 2 sin 2𝛼 Plotting this ‘circle’ in the τ-σ Space for the element gives the Mohr circle of stress. This represents the state of stress at a point at equilibrium. Any point on the circle represents the stress (normal and shear) on a plane inclined at an angle 𝛼 to the horizontal plane ( or minor principal stress)

Example Given the stress condition above, find the stresses on plane BB

Mohr Coulomb Failure Criteria When do we say a foundation or slope has failed? If the load or stress is increased until deformations become unacceptably large, we say the soil in the foundation or slope has ‘failed’ In this case we are referring to the strength of the material, which is really the max or ultimate stress the material can sustain In geotechnical engineering, we are generally concerned with the shear strength of soils and rocks, because, in most of our problems in foundations, failure results from excessive applied shear stresses.

Mohr Coulomb Failure Criteria Point A - shear failure will not occur along the plane Point B – shear failure will occur along the plane Point C – cannot exist because failure would have occurred already Illustration for one Plane

Mohr Coulomb Failure Criteria If Mohr circle touches failure line (i.e tangential) then failure occurs The condition where the Mohr circle touches the coulomb’s failure envelope is known as the Mohr-Coulomb failure condition

Characteristics of Failure Plane Angle of failure plane 𝜃=45°+ Φ 2

Shear Resistance Between Soil Particles The resistance of soil to deformation (Shear strength) is influenced strongly by the shear resistance at contacts between particles The shear strength refers to the soils ability to resist sliding along internal surfaces within a soil mass. The shear resistance is proportional to the normal force pushing the particles together

Shear Resistance Between Soil Particles Coulomb observed that for the shear strength of soils there is: Stress-independent component of shear strength Stress dependent component 𝜏=𝑐+ 𝜎 𝑛 tan ∅ Shear strength proportional to the normal force pushing the two particles together The stress independent is related to the intrinsic cohesion of the material The c and ∅ are called strength parameters and they are determined in the laboratory

Geostatic Stresses When there are no external load, stresses within a soil are caused by the weight of the soil For an almost homogenous soil with a horizontal ground surface, the pattern of stresses is known as Geostatic Stresses For such conditions, there are no shear stresses upon the vertical and horizontal planes within the soil The vertical geostatic stress is given by 𝜎 𝑣 =𝛾𝑧 For stratified soil geostatic stress is 𝜎 𝑣 = 𝛾 1 𝑧 1 + 𝛾 2 𝑧 2 + 𝛾 3 𝑧 3

Geostatic Stresses For homogenous soils, vertical geostatic stress is given by: 𝜎 𝑣 =𝛾𝑧 For stratified soil, geostatic stress is 𝜎 𝑣 = 𝛾 1 𝑧 1 + 𝛾 2 𝑧 2 + 𝛾 3 𝑧 3 𝜎 𝑣 = 𝛾𝑧

Concept of Stress in a Particulate System Soil element A with static ground water above it Total vertical stress due to soil grain and water pressure 𝜎 𝑣 = 𝜎 ′ +𝑢 𝑢= 𝛾 𝑤 𝑧 𝑤 Effective stress principle 𝜎 ′ = 𝜎 𝑣 −𝑢 Effective stress controls several aspects of soil behavior such as compression and strength

Concept of Effective Stress 𝜎 ′ =𝛾 𝑧 1 + 𝛾 𝑠𝑎𝑡 𝑧 2 − 𝛾 𝑤 𝑧 2 𝜎 ′ =𝛾 𝑧 1 + (𝛾 𝑠𝑎𝑡 − 𝛾 𝑤 ) 𝑧 2

Laboratory measurement of Shear Strength The Triaxial Test – most common test for determining the stress strain strength properties of soils Triaxial Apparatus Triaxial Cell System Base, Removable cylinder, loading ram Pressure Supplying System Measuring System Proving ring Dial gauge Pore pressure transducer Volume gauge

Laboratory measurement of Shear Strength The Triaxial Test Procedure Wrap a cylindrical soil specimen in a rubber membrane and place it in the triaxial chamber Apply a confining pressure ( 𝜎 3 ) by means of water on the sample Apply a vertical axial load and increase steadily until sample fails. The applied load is the deviator stress (Δp) The total vertical pressure at failure is 𝜎 1 = 𝜎 3 + Δ𝑝 𝑓 Repeat the procedure on another sample for a different confining pressure

Laboratory measurement of Shear Strength The Triaxial Test Procedure Undrained Triaxial Test This is when the test condition does not allow drainage of water out of sample during triaxial compression Undrained tests are constant volume test ( no volume change) Changes in cross sectional areas during testing must be taken into account When sample is consolidated in the triaxial apparatus before undrained triaxial compression, the test is Consolidated Undrained Triaxial Test Drained Triaxial Test The test condition allows for drainage of water out of sample during triaxial compression (shearing) The volume of sample continues to change during throughout the test

Laboratory measurement of Shear Strength The Triaxial Test Results

Laboratory measurement of Shear Strength The Triaxial Test Results The strength parameters can be in terms of total stress or effective stress Total stress parameters (c and Φ) Effective stress parameters (c’ and Φ’) σ3 σ1-σ3 σ1 U f σ3’ σ1' 350 319 669 256 94 413

Example Triaxial compression tests on three specimens of a soil sample were performed. Each test was carried out until the specimen experienced shear failure. Determine the soil’s cohesion and angle of internal friction. The test data are as follows: Specimen Confining Pressure, 𝜎 3 (kips/ft2) Deviator Stress at failure, Δp (kips/ft2) 1 1.44 5.76 2 2.88 6.85 3 4.32 7.50

Example

Laboratory measurement of Shear Strength Direct Shear Test Procedure Place a specimen in a relatively flat square box A normal load of specific magnitude is applied A shear stress is applied by forcing the two halves of the box in opposite directions until failure occurs in the horizontal plane Test is repeated with different normal loads

Laboratory measurement of Shear Strength Direct Shear Test results