Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

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2 Module 4: Lecture 7 on Stress-strain relationship and Shear strength of soils

3 Contents Stress state, Mohr s circle analysis and Pole, Principal stressspace, Stress pathsin p-q space; Mohr-Coulomb failure criteria and its limitations, correlation with p-q space; Stress-strain behavior; Isotropic compression and pressure dependency, confined compression, large stress compression, Definition of failure, Interlocking concept and its interpretations, Triaxial behaviour, stress state and analysis of UC, UU, CU, CD, and other special tests, Drainage conditions; Stress paths in triaxial and octahedral plane; Elastic modulus from triaxial tests.

4 Determination of shear strength parameters : Laboratory tests Field tests Direct shear test Triaxial shear test Direct simple shear test, Torsional ring shear test, Plane strain triaxial test, Laboratory vane shear test, Laboratory fall cone test Vane shear test Pocket penetrometer Pressuremeter Static cone penetrometer Standard penetration test Most common laboratory tests used to determine shear strength parameters c and φ are Direct shear test and Triaxial test

5 Direct shear test: Introduction Also called as Shear box test. Box can be of square or circular shape in plan. Used to determine soil strength and not the deformations. Different sizes of shear box can be used depending on grain size of the coarse grain soil. Sample is loaded first with normal stress with the help of dead loads. Then a lateral force is applied to split the sample in two parts. τ σ n Shearing plane

6 Direct shear test: Component parts σ n Split box τ Soil sample ` Force transducer Porous stones

7 Direct shear test: Mechanism σ h σ n At the start of the test σ n Intermediate development of stresses during the test σ hi τ i τ i σ hf τ f τ f σ ni Stress state at the end of the test Normal stress remains constant for a particular sample test

8 Direct shear test: Sample results for sand τ dense Peak shear stresses are noted down at each normal stress applied loose Displacement Volume keeps on decreasing for loose sand Volume first decreases and then increases for medium dense and dense sand Attributed to dilatancy efffect Volume change Compression Expansion dense medium Displacement loose

9 How to understand dilatancy i.e., why do we get volume changes when applying shear stresses? φ = ψ + φ i The apparent externally mobilized angle of friction on horizontal planes (φ) is larger than the angle of friction resisting sliding on the inclined planes (φ i ). strength = friction + dilatancy

10 How to understand dilatancy Bolton, 1991

11 Interlocking and dilatancy When soil is initially denser than the critical state which it must achieve, then as the particles slide past each other owing to the imposed shear strain they will, on average separate. The particle movements will be spread about mean angle of dilation Ψ In a dense sand there is a considerable degree of interlocking between particles. Before shear failure can take place, this interlocking must be overcome in addition to the frictional resistance at the points of contact.

12 Interlocking and dilatancy In general, the degree of interlocking is greatest in the case of very dense, well-graded sands consisting of angular particles. The characteristic stress strain curve for an initially dense sand shows a peak stress at a relatively low strain and thereafter, as interlocking is progressively overcome, the stress decreases with increasing strain. The reduction in the degree of interlocking produces an increase in the volume of the specimen during shearing as characterized by the relationship, between volumetric strain and shear strain in the direct shear test.

13 How to understand dilatancy The term dilatancy is used to describe the increase in volume of a dense sand during shearing and the rate of dilation can be represented by the gradient dε v /dγ, the maximum rate corresponding to the peak stress. The angle of dilation ψ is tan -1 (dε v /dγ) For a dense sand the maximum angle of shearing resistance (φ max ) determined from peak stresses is significantly greater than the true angle of friction (φ µ ) between the surfaces of individual particles, the difference representing the work required to overcome interlocking and rearrange the particles.

14 How to understand dilatancy When soil is initially looser than the final critical state, then particles will tend to get closer together as the soil is disturbed, and the average angle of dilation will be negative, indicating a contraction. In the case of initially loose sand there is no significant particle interlocking to be overcome and the shear stress increases gradually to an ultimate value without a prior peak, accompanied by a decrease in volume.

15 φ max,φ µ and φ cv Thus at the ultimate (or critical) state, shearing takes place at constant volume, the corresponding angle of shearing resistance being denoted φ cv (or φ crit ). The difference between φ µ and φ cv represents the work required to rearrange the particles. In general, the critical state is identified by extrapolation of the stress strain curve to the point of constant stress, which should also correspond to the point of zero rate of dilation on the volumetric strain-shear strain curve.

16 How to understand dilatancy If the density of the soil does not have to change in order to reach a critical state then there is zero dilatancy as the soil shears at constant volume. It is important to realize that a critical state is only reached when the particles have had full opportunity to juggle around and come into new configurations. If the confining pressure is increased while the particles are being moved around then they will tend to finish up in a more compact state.

17 φ max and φ cv In practice the parameter φ max, which is a transient value, should only be used for situations in which it can be assumed that strain will remain significantly less than that corresponding to peak stress. If, however, strain is likely to exceed that corresponding to peak stress, a situation that may lead to progressive failure, then the critical-state parameter φ cv should be used.

18 How to understand dilatancy When dense sands or over-consolidated clays are sheared they dilate Larger the particle size, greater the dilation Mohr-Coulomb idealisation implies dilation at a constant rate when soil is sheared. This is unrealistic.

19 Direct shear test: Evaluation of results Peak shear stresses are noted down at each normal stress applied There will be n numbers of normal and peak shear stresses for n numbers of samples tested. A plot of Peak shear stress vs Normal stress do gives the shear strength parameters φ and c for a particular soil. τ f φ c σ n

20 Direct shear test: Mohr s stress circle τ f (σ n3,τ f3 ) φ (σ n1,τ f1 ) (σ n2,τ f2 ) c σ n

21 Direct shear test: Stress path Initial condition :- Element on the failure plane σ n0 τ Mohr s diagrams σ h0 Pole σ h0 = K 0 σ n0 σ h0 σ n0 σ n

22 Direct shear test: Stress path During the test, before failure :- Element on the failure plane σ n τ Mohr s diagrams σ hi τ i τ i τ i σ hi σ ni σ n

23 Direct shear test: Stress path At failure :- Element on the failure plane τ Mohr s diagrams σ hi τ i σ n τ f Pole τ i σ hf σ nf σ hf σ nf σ n

24 Direct shear test: Stress path : Example A direct shear test is run on a medium dense sandy silt with σ n = 65 kpa. At failure the shear stress is 41 kpa. Draw the Mohr circles for the initial and failure conditions and determine: The principal stresses at failure The orientation of the failure plane The orientation of the plane of maximum normal stress at failure The orientation of the plane of maximum shear stress at failure

25 Direct shear test: Stress path : Solution (kpa) Initial condition: Normal stress applied before starting the test = 65 kpa (kpa)

26 Direct shear test: Stress path : Solution (kpa) Shear stress at failure is = 41 kpa. So, the point (65,41) lies on the Mohr s circle at failure (kpa)

27 Direct shear test: Stress path : Solution (kpa) As mentioned in example the soil is silty sand. So, cohesion in the material is assumed to be zero (kpa) The angle of internal friction is = 32 i.e. Slope of the line passing through origin and point (65,41)

28 Direct shear test: Stress path : Solution (kpa) To find centre of the circle: θ = 45 + φ/2 ; = 61 (kpa) Angle between horizontal and the line joining center of the circle and point (65,41) is = 180-2θ = = 58 Center of Mohr s circle

29 Direct shear test: Stress path : Solution (kpa) Drawing a circle through the centre of the circle and the point (65,41) as on the circle. (kpa)

30 Direct shear test: Stress path : Solution (kpa) Horizontal line extended through point (65,41) to the other edge of the circle gives POLE. (kpa) Lines drawn through the intersection points between circle and Normal stress axis gives Principal plane inclinations to horizontal.

31 Both the maximum stress ratio and the ultimate (or critical) void ratio decrease with increasing effective normal stress. The difference between maximum and ultimate stress decreases with increasing effective normal stress. The value of φmax for each test can then be represented by a secant parameter, the value decreasing with increasing effective normal stress until it becomes equal to φ cv. σ Due to decrease in ultimate void ratio

32 Direct shear test: Disadvantages The drainage conditions cannot be controlled. As pore water pressure can not be measured, only total stresses can be determined. Shear stress on the failure plane are not uniform as failure occurs progressively from the edges to the center of the specimen. Area under the shear and vertical loads does not remain constant throughout the test. Soil is forced to shear at predetermined plane which should not be necessarily the weakest plane. Rotation of principal planes The only advantage of direct shear test is its simplicity and, in the case of sands, the ease of specimen preparation.

33 The triaxial test: Introduction Most widely used shear strength test and is suitable for all types of soil. A cylindrical specimen, generally L/D = 2 is used for the test, and stresses are applied under conditions of axial symmetry. Typical specimen diameters are 38mm, 100mm and 300 mm Axial stress Equal all round pressure Stress system in triaxial test

34 The triaxial test: Components Loading ram Perspex cell Porous discs Pressure supply to cell Latex sheet Soil sample To pore pressure measuring device