56 - NPTEL · 2017. 8. 4. · Laboratory tests. Field tests Direct shear test Triaxial shear test Direct simple shear test, Torsional ring shear test, Plane straintriaxial test, Laboratory

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

56


Module 4: Lecture 7 on Stress-strain relationship

and Shear strength of soils


Stress state, Mohr’s circle analysis and Pole, Principalstress space, Stress paths in p-q space;

Mohr-Coulomb failure criteria and its limitations,correlation with p-q space;

Stress-strain behavior; Isotropic compression andpressure dependency, confined compression, large stresscompression, Definition of failure, Interlocking conceptand its interpretations,

Triaxial behaviour, stress state and analysis of UC, UU, CU,CD, and other special tests, Drainage conditions; Stresspaths in triaxial and octahedral plane; Elastic modulusfrom triaxial tests.

Contents


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


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.

σn

τ

Shearing plane


`

Direct shear test: Component parts

σn

τ

Porous stones

Force transducer

Split box

Soil sample


Direct shear test: Mechanism

At the start of the test

Intermediate development of stresses during the test

Stress state at the end of the test

σn

σh

σnτi

τi

σhi

σniτf

τf

σhf

Normal stress remains constant for a particular sample test


Direct shear test: Sample results for sand

DisplacementEx

pans

ion

loose

τ

Displacement

loose

dense

dense

Com

pres

sion

Volu

me

chan

ge medium• Volume keeps on decreasing

for loose sand• Volume first decreases and

then increases for medium dense and dense sand

• Attributed to dilatancy efffect

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


How to understand dilatancyi.e., why do we get volume changes when applying shear stresses?

φ = ψ + φi

The apparent externally mobilized angle of friction on horizontal planes (φ)is larger than the angle of friction resisting sliding on the inclined planes (φi).

strength = friction + dilatancy


How to understand dilatancy

Bolton, 1991


When soil is initially denser than thecritical state which it must achieve,then as the particles slide past eachother owing to the imposed shear strainthey will, on average separate. Theparticle movements will be spreadabout mean angle of dilation Ψ

Interlocking and dilatancy

In a dense sand there is a considerabledegree of interlocking betweenparticles. Before shear failure can takeplace, this interlocking must beovercome in addition to the frictionalresistance at the points of contact.


In general, the degree of interlocking is greatest in thecase of very dense, well-graded sands consisting ofangular particles.

The characteristic stress–strain curve for an initially densesand shows a peak stress at a relatively low strain andthereafter, as interlocking is progressively overcome, thestress decreases with increasing strain.

The reduction in the degree ofinterlocking produces an increase inthe volume of the specimen duringshearing as characterized by therelationship, between volumetric strainand shear strain in the direct shear test.

Interlocking and dilatancy



The term dilatancy is used to describe the increase involume of a dense sand during shearing and the rate ofdilation can be represented by the gradient dεv/dγ, themaximum rate corresponding to the peak stress.

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

(φmax) determined from peak stresses is significantly greaterthan the true angle of friction (φµ) between the surfaces ofindividual particles, the difference representing the workrequired to overcome interlocking and rearrange theparticles.


When soil is initially looser than thefinal critical state, then particles willtend to get closer together as the soilis disturbed, and the average angleof dilation will be negative, indicatinga contraction.


In the case of initially loose sandthere is no significant particleinterlocking to be overcome and theshear stress increases gradually to anultimate value without a prior peak,accompanied by a decrease involume.


Thus at the ultimate (or critical)state, shearing takes place atconstant volume, thecorresponding angle of shearingresistance being denoted φcv (orφcrit).

The difference between φµ andφcv represents the work requiredto rearrange the particles.

In general, the critical state is identified by extrapolation of thestress–strain curve to the point of constant stress, which shouldalso correspond to the point of zero rate of dilation on thevolumetric strain-shear strain curve.

φmax ,φµ and φcv


If the density of the soil does not have tochange in order to reach a critical statethen there is zero dilatancy as the soilshears at constant volume.

It is important to realize that a criticalstate is only reached when the particleshave had full opportunity to jugglearound and come into newconfigurations. If the confining pressure isincreased while the particles are beingmoved around then they will tend tofinish up in a more compact state.



In practice the parameter φmax, which is a transientvalue, should only be used for situations in which itcan be assumed that strain will remain significantlyless than that corresponding to peak stress.

If, however, strain is likely to exceed thatcorresponding to peak stress, a situation that maylead to progressive failure, then the critical-stateparameter φcv should be used.

φmax and φcv


When dense sands or over-consolidated clays aresheared they dilate

Larger the particle size, greater the dilation

Mohr-Coulomb idealisation implies dilation at aconstant rate when soil is sheared. This is unrealistic.



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

c

φ

σn


Direct shear test: Mohr’s stress circle

τf

c

φ

(σn1,τf1)

(σn2,τf2)

(σn3,τf3)

σn


Direct shear test: Stress pathInitial condition :-

Element on the failure plane

σn0

σh0

σh0= K

0σn0

Mohr’s diagrams

σh0σn0

Pole

σn

τ


Direct shear test: Stress pathDuring the test , before failure :-

Element on the failure plane Mohr’s diagrams

τi

σnτi

τi

σhi

σhi σni σn

τ


Direct shear test: Stress pathAt failure :-

Element on the failure plane

σnτi

τi

σhi

Mohr’s diagrams

τf

σhf σnfσn

τ

Pole

σhfσnf


A direct shear test is run on a medium dense sandy siltwith σn = 65 kPa. At failure the shear stress is 41 kPa. Drawthe Mohr circles for the initial and failure conditions anddetermine:

– 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

Direct shear test: Stress path : Example


Direct shear test: Stress path : Solution

Initial condition:

Normal stress applied before starting the test = 65 kPa

(kPa)

(kPa

)



Shear stress at failure is = 41 kPa.

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

(kPa)

(kPa

)



As mentioned in example the soil is siltysand.

So, cohesion in the material is assumed to be zero

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

(kPa)

(kPa

)



To find centre of the circle:

θ = 45 + φ/2 ;= 61°

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

Center of Mohr’s circle

(kPa

)

(kPa)



Drawing a circle through the centre of the circle and the point (65,41) as on the circle.(k

Pa)

(kPa)



Horizontal line extended through point (65,41) to the other edge of the circle gives POLE.

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

(kPa)

(kPa

)


↑ σ′

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 asecant parameter, the value decreasing with increasingeffective normal stress until it becomes equal to φcv.

Due to decrease in ultimate void ratio


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.


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

Stress system in triaxial test

Equal all round

pressure

Axial stress


The triaxial test: Components

Loading ram

Perspex cell

Latex sheet

Soil sample

To pore pressure measuring device

Porous discs

Pressure supply to

cell

Documents

56 - NPTEL · 2017. 8. 4. · Laboratory tests. Field tests Direct shear test Triaxial shear test Direct simple shear test, Torsional ring shear test, Plane straintriaxial test, Laboratory