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Prof. B V S Viswanadham, Department of , IIT Bombay Module 4: Lecture 7 on -strain relationship and of

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Contents  Stress state, Mohr’s circle analysis and Pole, Principal stress space, Stress paths in 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.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Determination of shear strength parameters :

Laboratory tests Field tests

 Vane shear test  Triaxial shear test  Pocket penetrometer  Direct simple shear test,  Pressuremeter  Torsional ring shear test,  Static cone penetrometer  Plane strain triaxial test,  Standard penetration test  Laboratory vane shear test,  Laboratory fall cone test

Most common laboratory tests used to determine shear strength parameters c′ and φ′ are Direct shear test and Triaxial test

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Introduction • Also called as Shear box test. • Box can be of square or circular shape in plan. • Used to determine 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.

σn τ

Shearing plane

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Component parts

σn

Split box

τ ` Force Soil sample transducer

Porous stones

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Mechanism

σn

σh At the start of the test

σn τi Intermediate development of σhi stresses during the test τi

σni τf σhf Stress state at the end of the test τf

Normal stress remains constant for a particular sample test

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Sample results for

τ dense • Peak shear stresses are noted down at each normal stress applied

loose

Displacement dense Expansion medium • Volume keeps on decreasing for loose sand Displacement • Volume first decreases and

then increases for medium change Volume dense and dense sand loose

• Attributed to dilatancy efffect Compression

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay How to understand dilatancy i.e., why do we get volume changes when applying shear stresses?

φ = ψ + φi The apparent externally mobilized angle of on horizontal planes (φ) is larger than the angle of friction resisting sliding on the inclined planes (φi). strength = friction + dilatancy Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay How to understand dilatancy

Bolton, 1991 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 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.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Interlocking and dilatancy  In general, the degree of interlocking is greatest in the case of very dense, -graded 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 as characterized by the relationship, between volumetric strain and shear strain in the direct shear test.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 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 dεv/dγ, the maximum rate corresponding to the peak stress.

-1  The angle of dilation ψ is tan (dεv/dγ)  For a dense sand the maximum angle of shearing resistance

(φmax) determined from peak stresses is significantly greater than the true angle of friction (φµ) between the surfaces of individual particles, the difference representing the work required to overcome interlocking and rearrange the particles.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 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. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay φmax ,φµ and φcv  Thus at the ultimate (or critical) state, shearing takes place at constant volume, the corresponding angle of shearing

resistance being denoted φcv (or φcrit).

 The difference between φµ and φcv 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.

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

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay φ max and φcv

 In practice the parameter φmax, 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 φcv should be used.

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

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

τf

φ

c

σn

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Mohr’s stress circle

τ f φ (σn3,τf3)

(σn2,τf2)

(σn1,τf1)

c

σn

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path Initial condition :-

Element on the failure plane Mohr’s diagrams τ σ n0 σ h0 Pole σ = K σ h0 0 n0

σ σ σn h0 n0

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path During the test , before failure :-

Element on the failure plane Mohr’s diagrams τ

σn τi σhi

τi τi

σh σ σ i ni n

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path At failure :-

Element on the failure plane τ Mohr’s diagrams

σn τ i τf Pole σhi

τi

σh σ f h σ σn σ f nf nf

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Example

A direct shear test is run on a medium dense sandy with σn = 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

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution

Initial condition: (kPa) Normal stress applied before starting the test = 65 kPa

(kPa)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution

Shear stress at failure is

(kPa) = 41 kPa.

So, the point (65,41) lies on the Mohr’s circle at failure

(kPa)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution As mentioned in example the soil is silty sand. (kPa)

So, in the material is assumed to be zero

The angle of internal friction is = 32° (kPa) i.e. Slope of the line passing through origin and point (65,41)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution

To find centre of the circle: (kPa) θ = 45 + φ/2 ; = 61°

Angle between horizontal and the line joining center of the circle and point (65,41) is = 180 - 2θ (kPa) = 180 - 122 = 58

Center of Mohr’s circle

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution

Drawing a circle through the centre of the circle and the

(kPa) point (65,41) as on the circle.

(kPa)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Stress path : Solution

Horizontal line extended through point (65,41) to

(kPa) the other edge of the circle gives POLE.

Lines drawn through the intersection points between circle and Normal stress axis gives Principal (kPa) plane inclinations to horizontal.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay  Both the maximum stress ratio and the ultimate (or critical) 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 φcv.

↑ σ′ Due to decrease in ultimate void ratio

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Direct shear test: Disadvantages • The drainage conditions cannot be controlled. • As 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. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 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

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay The triaxial test: Components

Loading ram

Perspex cell

Porous discs Latex sheet Soil sample

Pressure supply to To pore pressure cell measuring device

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