Critical State Soil Mechanics ---- By Jishnu R B

Introduction to CSSMIntroduction to CSSM

Why Geotechnical Engineering ?Why Geotechnical Engineering ?

• Soil and Rocks are still one of the most important construction materials used in natural or re compacted state.

• Water content heavily influences the way in which soil masses behave. For instance clays.

• In Slopes and retaining walls soils apply the loads as well as provide strength and stiffness.

• Thus it is important to have a knowledge of strength and stiffness of these natural geologic materials.

• Thus GT engineers need to play with material they have which requires knowledge even in geology.

• The Bridges must not fall down, slopes and foundations must not fail nor they should move, for which theories has to be developed

• The theories must be dealing with strength and stiffness which gives us ‘ultimate states’ and we must make sure that ‘working states’ got from these theories are acceptable

The Beauty of MechanicsThe Beauty of Mechanics• In any case mechanism changes once force is applied, rubber band

stretches once you pull it and so building sways in wind.• Unequal Forces causes system to accelerate by inducing stresses

and unequal strains causes incompatibility by inducing strains.• The Fundamental requirements in any system are “equilibrium and

compatability” ie material should not disappear and gaps should not appear!!

• Thus it ultimately comes to concept of “stress and strain” which are interlinked by “material behaviour”.

• Thus material becomes important and Hence “soil ,rock or fluid mechanics”.

• Some examples may be

Material BehaviourMaterial Behaviour

• If the body is rigid strains are zero even under infinite stress and movements can occur only under mechanism “rigid body motion”

• But materials do compress swell and distort• When you compress some material it goes on compressing in a

stable manner without any failure which means K is increasing.• Converse happens in shear case when shearing causes a decrease

in G and the point at which G turns zero, you call it shear strength.• Thus K and G are nothing but they are ‘stiffnesses’ and they are

related to strength (Its simplest theory is Theory of Elasticity).• When shear strength is constant you call the material ‘cohesive’ and

when it depends on confining pressure you call the material ‘frictional’

• Thus be it cohesive or frictional Both stiffness (K & G) and strength parameters (s for cohesive and nu for frictional) are material dependent

Basic Characteristics of SoilsBasic Characteristics of Soils• External loads and water pressure interact each other to produce a

stress which is effective(very important in soil mechanics).• Compressibility of soil is mainly due to rearrangement of grains and

hence void spaces changes.• Soil shearing is mainly frictional which increases with the normal

stress which it is subjected to.• Thus with increase in water content or increase in water pressure

decrease the soil strength and stiffness.• Due to rearrangement of grains which is unrecoverable soils are

generally inelastic.• It is important to understand that there is no fundamental change in

behaviour in clay and sand and it has to do more with influence of pore pressures and seepage of water in to the void spaces.

Factor of safety and Load factorsFactor of safety and Load factors

• To deal with uncertainties in determination of strength and stiffness parameters we have ‘Factor of safety’.

• Mostly in Geotechnical Engineering the ‘Factor of safety’ is used for accounting uncertainties and ‘Load factor’ is used for limiting settlements or ground movements.

• Thus in most of the cases Higher Load factors are applied than FOS• Load factors give allowable Load (Fall) and FOS gives a load higher

than that.• Both Load Factors and FOS are applied to Fc (Collapsible Load).

Principles of MechanicsPrinciples of Mechanics

• A stress is basically the intensity of loading given by a force acting on a unit area.

• Strain is basically the intensity of deformation given by a displacement over a unit gauge length.

• In soil mechanics, this unit area or gauge length should be representative of a finite number of soil particles neither they can sustain tensile stresses. Hence comp stresses are positive(δσ= -(δFn/δA)).

• In order to represent the problems of stresses and strains rather simply the common states of stresses may be classified in to two:

ContinuedContinued……

• Strains in one direction mostly perpendicular to plane of paper is zero (Plane Strain). Eg: Long Walls ,embankments, Dams.

• For Axi symmetric problems radial stresses are equal which are different from axial stresses Eg: Circular foundations and excavations.

• Hence for plane strain stress coordinates may be (σz,σh) and plain strain may be (σa,σr).

• When soil fails, they develop slip surfaces which on a geologic scale appears as faults.

• When such faults develop soil tends to fail along the slip planewhere internal strains become less significant compared to relative movement of blocks. Rigid body Mechanics!!!!.

Continued..Continued..• Closure of Polygon of forces ensures Force equilibrium, provided

length of Force quantity represents magnitude of force.• Closure of displacement diagram or hodograph represents

compatability condition.• Main striking Difference between a force polygon and hodograph is

the direction of corresponding components.• In a Hodograph, single subscript corresponds to rigid body motion of

that block where δxy corresponds to relative movement of y with respect to x .

Analysis of StressesAnalysis of Stresses• Stresses are different in different planes. One to one corres -

pondence hence is necessary.• This is done through Mohr circle method, where in soil mechanics

counter clock wise shear stresses and compressive stresses are positive.

• To construct the Mohr Circle it is preferable to know the stresses at orthogonal planes.

• Pole ‘P’ is one point from where if a line is drawn to stress point willgive the direction of the plane at which the stress acts.

Analysis of Strains..Analysis of Strains..

• Analysis of strains can be done using Mohr circle of strains except for some difference.

• In stresses we talk about absolute quantities and strains we talk about increments. (If small δε else ∆ε).

• Stresses are always positive in particulates where strains may be positive or negative.

• Line from P to the point δεz fives the plane across which strain is δεz.

Stress ratio and DilationStress ratio and Dilation

• Soils which are mostly frictional can sustain a maximum shear stress corresponding to a maximum normal stress.

• Thus more than shear stress alone or the normal stress the stress ratio(ζ/σ = tanФmob) is more significant.

• To express in terms of principal stress we have t/s=sinФmob=(σz-σh)/(σz+σh) σz/σh=(1+sinФmob)/(1-sinФmob)σz/σh=tan2(45+Фmob/2)

• Thus in the figure the planes shown by double lines are where most critical conditions occur ie failure.ie where α=β=45+Фmob/2.

•

ContinuedContinued……• Along the same planes α and β which are shown by broken lines the

corresponding normal strains become zero.• These planes are defined by angle of dilation(Ψ).• Thus v = ½(δεz+δεh) ; g = ½(δεz-δεh) and volumetric strain is given

by δεv=δεz+δεh.• Thus Angle of dilation is given by tanΨ = -(δεv/δγ); δγ is the

increment of shear strain across the plane.• To express in terms of principal strains sinΨ = -((δεz-δεh)/(δεz+δεh))• The quantity -(δεv/δγ) is called strain ratioα=β=45+Ψ/2.

Slip surfacesSlip surfaces• Now the focus is on body where homogenous straining causes the

‘slip surfaces’ to develop or along direction of zero ext lines.• They are not ‘surfaces’ but slip ‘zones’ which has a small but finite

thickness ( in the order of grains in soils).• Because of rigidity of the materials on either side we have AB as a

zero extension line and slip surface starts developing along thedirection(α) of double lines discussed.(α=45+Ψ/2).

• Thus δγ = δh/Ho and δεv = δv/Ho tanΨ=δεv/δεh = δv/δh• Thus movement across the slip surface AA1 and BB1 is at

angle Ψ to the direction of slip surface.

Essentials of Material BehaviourEssentials of Material Behaviour• Analysis of stresses and strains using Mohr circles are not dependent

on the material rather equally useful for all materials.• To analyze any structure it is necessary to have relationships between

stresses and strains.• These relations are called constitutive relations and their form depends

on nature of material and its loading.• In soils and other granular materials it is necessary to discuss about

stress which is effective in between the soil grains(σ’).• ‘Stiffness’, the quantity which is mostly referred as ‘Modulus’ is the

slope of stress-strain curve.

Continued..Continued..

• If the gradient (Stiffness) is linear it is easy to determine else if it is curved the stiffness at a point A can be ‘tangent’ or ‘secant’.

• Thus tangent stiffness = dσ’/dε and secant stiffness = ∆σ’/∆ε.• The stiffness actually determines strains and displacements in soils

when they are loaded or unloaded.• Another term used in soil mechanics is ‘compressibility’ which is

reciprocal of ‘stiffness’.• Thus ‘stiffness’ and ‘strength’ are two different aspects Stiffness

governs maximum displacement under working loads where as strength governs maximum load a structure can sustain.

• Steel is stiff and strong, margarine is soft and weak, black board chalk is stiff and weak, rubber is weak and strong.

Choice of Parameters for stress and strainChoice of Parameters for stress and strain

• For metals we do uniaxial extension tests, For concrete we use uniaxial compression test because tests are parameter dependent.

• For soils shearing and volumetric strain responses (parameters) for shearing and normal loading or unloading is important.

• The two common tests used are Triaxial tests and Shear tests.• In direct shear we can compress the soil with normal stress(σn) and

zero shear stress which results in a normal strain(εn).• Similarly in direct shear we can have shear stress(ζ) and zero normal

stress resulting in a strain(γ).• Thus from Direct shear under different conditions we have following

parameters, Shear Modulus G’ and compression Moduli M’

G’ = dζ’/dγM’ = dσn / dεn

ContinuedContinued……• From Triaxial Test we need G and compression Modulus (M), Here it is

more like 3D compression which gives a parameter (K).3G’ = (dq’ / dεs) and K = (dp’ / dεv)

Here ‘q’ must be maximum shear stress which is nothing but diameter of circle Thus q = σa’ – σr’.

εs is nothing but equivalent shear strain εs = 2/3(εa – εr).

p’ is the average compressive stress p’ = 1/3(σa’+2σr’).εv is the volumetric strain εv = εa +2εr.

It may be noted that G’ obtained in both cases are same where K and M are approximately similar.

Thus during the increment of straining Work done per unit volume of soil δW must be an invariant.


• Thus according to principal of work we have δW = q’δεs + p’δεv .• Substituting for δεs, δεv , q’ and p’ from the previous slide δW = σa’εa + 2σrδεr.

• So ultimately factors like 2/3 are coming for consistency of equations.

• So For triaxial tests use parameters q’,p’,εs,εv and for shear tests use parameters like ζn’,σn’,γ,εn.

Constitutive equationsConstitutive equations

• Due to particular nature of soil an ‘odd’ behaviour arises in soils.Thisis called ‘shear and volumetric coupling’.

• It is nothing but ‘shear stress’ cause ‘volumetric strains’ and ‘normal stress’ cause ‘shear strain’.

• Most simple constitutive equation relating shearing and volumetric stress – strain behaviour can be written as below where [S] is a stiffness matrix containing stiffness moduli.


• If you observe parameters J1’ and J2’ it is well understood we are talking about shear volumetric coupling.

• For Elastic and isotropic materials J1’=J2’ =0 ie decoupled.• For Elastic, Isotropic and perfectly plastic materials J1’=J2’ which

means [S] is symmetric.• The inverse of stiffness matrix is called Compliance matrix [C].• The relation ship between [S] and [C] are not simple but it is

relatively easy for decoupled states.• For such cases C12 = C21 =0 ;C11=1/S11=1/3G’ & C22 = 1/S22=

1/K’.• [S] and [C] don’t have constants rather in soils due to non linearity it

is strain dependent.

StrengthStrength• Strength of the material is the ultimate state of stress that it can

sustain before it fails (Not applicable for soils which fail at large strains where failure definition is tricky).

• There are so many talks about tensile, compressive and shear strengths as though they are different

• All this should be related to a fundamental characteristic strength.• The ‘link’ is nothing but ‘maximum shear stress’ or the largest Mohr

circle material can sustain.

Continued..Continued..• Thus it is clear that materials that have strength can sustain shear

stresses and ‘strength’ is nothing but ‘maximum shear stress’ that can be sustained.

• Only materials with ‘strength’ can have ‘slopes’ because ‘shear stresses’ are required to maintain the ‘slope’.eg water cannot sustain a slope because Mohr circle reduces to a point.

• For soils there are two principal criteria of failure (1).Tresca criteria (2).Mohr Coulomb criteria (Both cases have tangent failures).

• Note that Tresca criteria will be written on ‘total stresses’ and MC on ‘effective stresses’. (s= shear strength of the material).

• Just Tresca is condition of butter and MC condition of Dry sugar !!

ElasticityElasticity

• Elastic materials are conservative that all the work done by external stresses during increment of deformation is stored and recovered during unloading.

• For isotropic, elastic materials shear and volumetric effects are decoupled so that stiffness parameters J1’ and J2’ are zero.

• For materials that are elastic and anisotropic the coupling moduli(J1’ and J2’ are equal) ie matrix is symmetric.

• Non linear Elastic materials do exist when they have Elastic modulidependent on stresses and strains.(Eg.Rubber band with recoverable strain).

• In soil mechanics G’ and K’ are given more preferance to E’ and nu’since it is required to decouple shear from change in size.

Perfect PlasticityPerfect Plasticity

• Once Loading goes past the yield point simultaneous elastic and plastic strains occur and stiffness decreases.

• Work done on an increment of plastic deformation is ‘dissipated’ and hence plastic strains are never recoverable.

• The plastic strains at failure are ‘irrecoverable’,’indeterminate’,whichcan be ‘more’ or ‘less’ which gives rise to ‘plastic flow’.

• Consider a block subjected to arbitrary stresses σx’ and σy’.• At some combination of σx’ and σy’ failure occurs and plastic flow

commences.That combination of σx’ and σy’ is (σxf’,σyf’).• This (σxf’,σyf’) are represented by a failure envelope.• Vector of failure stress(σf’) can be obtained at any point of the failure

envelope.• Since the stresses remain constant and increments of plastic strains

goes on increasing with time making origin arbitrary.


• Direction of vector of an increment of plastic straining is given by δεxp/δεyp.

• The relationship between the failure envelope and direction of vector of plastic straining is called ‘plastic flow rule’.

• That is we have to superimpose failure envelope with graph showing increment of plastic strains.

• Once it is done the direction of the vector will be normal to the failure envelope for a perfectly plastic material (Normality condition of perfect plasticity).

Continued...Continued...

• Another way of explaining the normality condition is by ‘associative flow rule’.

• For this first develop different vectors ( of course in different directions) for different stresses.

• Now we have to develop a plastic potential envelope orthogonal to all these vectors.

• If the material is ‘plastic’,’plastic potential’ itself will form the failure envelope.

• Thus plastic potential is ‘associated’ with failure envelope,hence the name ‘associative flow rule’.


• In plastic straining the striking feature is that strains depend on the ‘state of stress’ and not on ‘increments of stresses’.

• In elastic straining strains are dependent on the stress increments via ‘Stiffness matrix’.

• They are independent of Loading path ( be it from BA or CA), but only depends on ‘gradient of the failure envelope at A).

Combined Combined ElastoElasto--Plastic BehaviourPlastic Behaviour• If stress-strain curve of soil is looked we can see a portion in between first

yield and ultimate state (where it is plastic)• The in-between state is having simultaneous elastic and plastic

components of strain (Elasto-Plastic behaviour).• Consider straining of member from O1 to Y1(First yield point).After Y1

plastic irrecoverable strains occur when unloaded at O2.• When the material is reloaded at O2,elastic yielding occurs at Y2(First

yield point) and subsequently done from Y2 to Y3.• So first Yield is increasing(σx3’>σx2’>σx1’).Such increase due to plastic

straining is called hardening.• Hardening law is relation between increase in yield stress(δσx’) and

increase in plastic strain(δεxp).


• Yielding and plastic straining can cause hardening or softening ie an increase in yield stress or its decrease.

• In most of the soils softening occurs where first yield point goes down in further cycles after a peak value of yield stress.

• Since we are concerned about elasto- plastic behaviourTotal strain = Elastic + Plastic (from Hardening law).

Continued..Continued..• Thus a set of yield curves may be developed for each first yield points

which are similar to failure envelope.• The state cannot reach the failure envelope where ultimate stress is

attained which corresponds to failure.• If perfect plasticity is attained on each of those yield points we have

direction vectors normal to yield surfaces.(eg Loading from AB).• For each of those yield curves or yield points a particular irrecoverable

plastic strain is adjunct which can form the third axis (εp).• Thus for Loading and Unloading cycles OABC,the loading cycle

OA and BC are elastic where AB is elasto plastic.• Now we have assembled the flow rule,Hardening law, and elastic stress

strain equation to an explicit constitutive equation for a complete range up to failure

Time and Rate EffectsTime and Rate Effects

• Till now it is assumed that no strains occur at constant load except at failure and only we have considered relationships between changes of effective stress with changes in strain.

• But Time dependent straining at constant loads do exist in soil e.g.: Consolidation.

• In concrete, steel or soil strength and stiffness are affected by rate of loading and rate of straining.

• Materials at constant stress continues to strain, but at a rate which diminishes with time called Creep, especially in soft clays.

• From the equation of creep strain rate it is evident that the rate of straining decreases with time.

Structure of Earth Structure of Earth –– In a nut shellIn a nut shell

• Weathering of rocks in situ Residual soils, when eroded forms Deposited soils. Their engineering properties dep on geological origin.

• Radius 8000km with crust just accounting 25-50km like an eggshell.• Crust have a varying altitude of 8km above and below MSL.• Maximum depth up to which soil can occur is about 300m which mostly

are underlain by rocks.• Soils older than 2 million years Geologically old soils Stiff . • Young soils (Glacial and Post glacial) maximum depth =30m.• Top soil where plants growmaximum depth =1m.• Soils and rocks close to surface is weathered and transported which on

deep burial are converted back to rocks Soil Cycle.• Sequence of strata which represent geologic history Stratigraphy.

Continued..Continued..• To describe the chronological history strata is classified based on age.• Eocene Age – 40 to 60 Million years, Devonian – 350 to 400 Million years,

Pleistocene Age (Age of modern man) – 1Million years ago.• Materials of Cenozoic age Soils, Mesozoic age Soft rocks, Paleozoic age

Hard rocks.• Engineering properties of silts, clays, sandstones, silt stones and mudstones

appear to different like many creatures which evolved through various geologic periods.

• The environment which the soil is deposited, and subsequent geological events determine the state of soil (on denseness and fabric).

• Depositional environment may be Glacial, Lake and Marine, and desert environment. Recent geologic events play an important role ,e.g. ice age.

• Glacial carries well graded soils Terminal moraines, Lacustrine carry fine graded soils Deltas , Aeolian Sand dunes (coarse materials at bottom followed by fines).

• Deposition due to factors like sea level rise leads to normally consolidated soils while once they are eroded they give rise to over consolidated soils.

• Once depositional environment is given a GT engr knows what soil to expect

Soil ClassificationSoil Classification• Mechanical properties of soil depend primarily on the nature of grains ie

what they are and state of soil ie how they are arranged. • To classify we may need its colour, size and shape, and response to

moulding .Hence we need tests and observations to classify.• Classification of soil may be based on behaviour ie according to purpose

we may need.• Agricultural classifications How to support crops, Geological

classification Age of deposit, Engg Classification Mechanical behaviour.

• Several schemes exist for classification. In UK BS 5930-1981 for site investigations, BS 1377-1991 for soil testing etc.

• A simple universal scheme for soil description is as follows : (1) Nature of Grains (2) Current state of the soil (3).The structure of the fabric (4).Formation of soils.

Continued..Continued..• The state of soil does change near foundations and excavations

during loading or unloading, compression or swell.• The soil formation manner also influences initial state, structure and

fabric (layering, fissuring and jointing).• Structure and fabric influences soil stiffness and drainage ,ie why

testing should be done on intact samples.


• From the curve given as per BS 1377, soils can be well graded, poorly graded or of uniform soils.

• Coarse grained soils behave like different sized marbles, but clays show significant difference (due to particle size and volume change characteristics).

• In clays following action happens: Surface of grain carries ‘charge’ which depends on the ‘soil

mineral’ and ‘electrolyte’ of the ‘pore water’. 2 Forces act in clay particles (1).Self weight (2).Surface Charge. Once size of the particles decrease self weight decreases by cube

of effective dia where surface force decreases by square. Thus for fines surface charge becomes predominant. The relative importance of Surface charge to Self weight brings in

a new term ‘specific surface’.


• Mechanical property of soils depend on the degree of packing.(iesoils with dense packing will be strong else weak).

• For this ‘specific volume’ (v) is an important parameter which is nothing but V/Vs ie (‘V’ volume contains ‘Vs’ volume of grains).

• Void Ratio (e) is used sometimes instead of specific volume as Vw/Vs for saturated soils. ( V = Vw + Vs) v = 1+e.

• For Loose assembly of uniform spheres v=1.92 and for dense assembly v =1.35.Common sands have v from 1.3 to 2.0.

• For Clays v ranges from 3 to 10.• For typical clays water content will vary from 20 to 70% and unit

weight from 18 to 22KN/m3.• Soil has to be ‘consistent’. Neither brittle nor muddy. That means

water content has to be adequate. Hence the term ‘consistency’.• You need tests hence to determine the limits at which it will become

brittle or muddy Attenberg Limits!!!.

The Current state.The Current state.

• Property of soil depends on another important factor ‘Current state’,iestrength, stiffness and specific volume depends on ‘stress history’.

• Same soil which was preloaded could be later unloaded.• This ‘Current state’ can be related to the relative position with respect

to limiting states. So here we need a link.• For fine grained soils it is Liquidity Index (LI).• For coarse grained soils it is relative density (Dr).• Undrained strength of clay is related to ‘current state’ or LI.

Pore pressure, Effective stress and DrainagePore pressure, Effective stress and Drainage

• It is the overall skeleton-pore water behaviour which determines the stresses induced under a saturated deposit.

• If ‘γ’ is the total weight density (Soil + water) then ‘ γh’ is the total stress Just like stress induced by water under a glass.

• The water in pores of a saturated soil is known as pore pressure ’u’which can be represented as height of water column (hw).

Thus u = γw*hw.• If the level of water in the pipe is below the ground for equilibrium it

is called ‘water table’ or ‘phreatic surface’.• That is if water is stationary pores should have constant pressure

making ‘hw’ constant which makes ‘phreatic surface’ horizontal.• For phreatic surface to be curved there should be movement of

water across the domain (seepage). E.g. earth dam.• ‘Phreatic line’ is the line at which pore pressure becomes zero.


• Negative pore pressures exist in the regions of capillary rise though they are saturated. ie water is in tension.

• This rise generally depends on pore spaces.• In unsaturated soils pore water and gas exist at different pressures.• These kind of soils exist in compacted, hot and dry climates.• Failure of slopes may occur not only by changes in total stresses but

can be due to changes in ‘pore pressures’. e.g. saturated slope fails.• Thus it is some combination of ‘total stresses’ and ‘pore pressures’

govern the strength. Concept of Effective stress emerges .• Terzaghi(1936) defined effective stress as

σ’ = σ – u ……Remains as an axiom!!!!• Since there is a shift only in ‘X axis – Normal stress axis’ total and

effective shear stresses are equal.

ContinuedContinued……• The magnitude of this effective stress is same even at the floor of a

duck pond and that of a deep ocean.• The same effective stress acts on the skin of a submarine and a

fish( which makes their skin tough and smooth).∆σ’ = ∆σ – ∆u

• Thus soil state becomes a function of ‘effective stress’.• It can change by keeping ∆σ constant with ∆u changing or vice

versa.• First case:∆u = 0 or constant; ∆σ causes a change in ∆σ’ and

settlement . Second case : ∆σ =0 But ∆u = -γwhw. Thus ∆σ’increases and hence settlement.

Volume Change And DrainageVolume Change And Drainage• When soil is loaded or unloaded, change in effective stress results in

a change in volume.• If the pore pressure remains constant changes in total and effective

stresses will be same(∆σ’ = ∆σ).Such change in volume is only due to expulsion of water(∆V = ∆V w).

• Soil just behaves as a sponge !!!.• So if Volume is restricted to change ie if sufficient time is not given

for water to ‘seep’ through ‘pore pressures will change”.• Thus there should be a relation between rate of loading, drainage,

and pore pressure.

Drained, Undrained loading and Drained, Undrained loading and ConsolidationConsolidation

Drained Loading.• Total stress ∆σ applied over a long period.• Water seeps through such that ∆V = ∆Vw.• Total stress increases and pore pressure doesn’t change (u= u0).• Since u = u0 change in σ = change in σ’.• Final stress= σ0’+∆σ’ ; Final Volume = V0 – ∆V.• The most important feature is that pore pressure is a constant

thoughout (u0).This is Steady state pore pressure.

ContinuedContinued…… Undrained Loading• ∆σ ,the total stress is increased rapidly, so that there is no change

in volume (∆V remains constant in this phase).• Sufficient time is not given to soil grains for stress transfer ie ∆σ’=0.• For ∆σ’ = 0 ; ∆u = ∆σ.• That means there is change in pore pressure which is called initial

excess pore pressure( ui’). That is u = u0 + ui’ .• If the pore pressures are in equilibrium ui’ = 0, u0 = 0.• The most important feature is that ‘no change in volume ‘.• That finishes undrained Loading part.• Now dissipation of ui’ happens by draining ie stress transfer

happens to soil grains∆σ’ increases u decreasesAs a result total stress remains constant( Applied load never changes) Volume decreases Consolidation happens !!!!

Lab Testing of SoilsLab Testing of Soils

• Lab testing of soils have mainly 3 goals (1). Description and classification of soil (2).Understand basic mechanical behaviour of soils and to develop theories (3).Determine design parameters (values for strength , stiffness and permeability).

• Samples may be reconstituted (Ageing is removed) or intact.• Plastic Limit of getting 3mm dia thread corresponds to case where

strength is approximately 150 KPa (which corresp stiff to Vstiff clay).• The main objective of soil Loading tests are to get Strength and

stiffness and the resulting stresses and strains.• Parameters obtained from reconstituted models are called ‘intrinsic’

of which PI is one of the most important. PI = LL - PL

Principal features ofPrincipal features of Soil LoadingSoil Loading

• Total stresses are pore pressures must be controlled and measured separately so that ‘effective stresses’ may be determined.

• Drainage must be controlled so that tests should be ‘constant volume’ (Undrained) or ‘constant pore pressure’ (Drained) tests.

• For measurement of ‘stiffness’ tests must be of ‘small strain’. For ‘strength’ measurement tests must be of ‘large strain(>20% even)’.

• As soils are frictional it is important to apply both normal and shear stresses.

• Normal stresses may be applied (as confining stress as in triaxial or directly as in direct shear).

• If the ‘total stresses’ are changed Stress controlled.• For a particular test ‘one set (axial)’ may be stress controlled and

another set (radial) will be strain controlled.


o Oedometer Tests (May be Rowe’s cell or conventional) Radial strains(εr=0).Porous disc ensures flow is 1D. The axial strain(εa) is measured. Pore pressures in top drain (ut) and bottom drain (ub) are kept zero. Used for 1D straining in compression or swelling.o Shear Tests (Direct shear or Simple Shear by NGI). Direct shear is by shear along horizontal plane. σn is applied by weights ,shear stress(ζn) is applied to get constant

displacement. Coordinates may be (σn,δn),(ζn,δh). ut =ub =0 for drained tests (Done for coarse grained soils) where

these tests are not preferred for clays (even if undrained tests may be simulated ut and ub are not measurable. Thus σ’,unknown).

In simple shear,due to non uniform stresses and hence strain in the sample large strains occur at the end of the shear box which is corrected in simple shear.

Continued..Continued..• Simple shear avoids non uniform strains allowing sides to rotate.• Simple shear tests may be drained or undrained.• Main disadvantage of shear tests may be inability to get stresses on

a vertical plane (σh,ζh).• Only stresses on horizontal plane (σn,ζn) are measured.• Thus Mohr circle construction is impossible with a single known

stress coordinate(σn,ζn).

Triaxial TestingTriaxial Testing

• Most ‘Common’ and ‘Versatile’ tests for Soils.• Cell pressure(σc) is applied along with an axial force(Fa).• Axial Displacements are measured under constant rate of strain.• Tests may be ‘drained’ or ‘undrained’.• ‘Radial’ strains are measured ‘indirectly’ from axial and volumetic

strains.Thus σa = σr + Fa/A Fa/A = σa – σr = σa’ – σr’

• Here A is the ‘current area’ accounting for volumetric and axial strains .

• For σc = σr = 0 ;The test is called ‘Unconfined compression test’.• To examine the ‘soil behaviour’ it is important to have a control over

axial or radial stresses which is ‘impossible’ in strain controlled tests.

Hydraulic TriaxialHydraulic Triaxial-- Stress Path testsStress Path tests• Main ‘difference’ from conventional triaxial lies in the ‘application of

the axial stress’.• Extension tensions (σa’ < σr’) may be applied for ‘cemented soils’.ie

Fa is negative.• Axial force (Fa) is measured independently using ‘Load cells’.• In this case axial and radial stresses/strains and the pore

pressure/volumetric strains can be changed independently.• Changes of total stresses are given by

δq = δσa – δσr δp = 1/3(δσa + 2δσr).• Changes in Effective stresses are given by

δq’ = δq δp’ = δp – δu .• For variation of δσa, δσr, δu Stress paths may be plotted using q

vs p and q’ vs p’• For a triaxial test σa and σr must be positive (σa < σr ). q and q’ may

be positive or negative.


• If δσa = 0, δσr = 0 it is evident slope of q-p graph becomes 3 or -1.5.• For ‘Drained’ Tests u remains a ‘constant’ . Thus ESP and TSP will

be parallel.• For ‘Undrained’ Tests u is a variable making ESP and TSP non

parallel.


• Interpretation of a Drained Triaxial Test. A0 = Π/4D0 ^2 ; V0 = A0L0 εa = -(∆L/L0),εv = -(∆V/V0) , v = v0(1-εv) Also current Area (A) = A0(1-εv)/(1-εa) σa = σr + Fa/A σr = cell pressure. Stress path parameters q’ = (σa’- σr’) = q ; p’ = 1/3(σa’+2σr’) =

p –u• Interpretation of Undrained Triaxial Tests have following variations . Pore pressure is not a constant. εv = 0.

Compression and SwellingCompression and Swelling

• For explaining these concepts only fully drained conditions are considered where ‘excess’ pore pressures are zero.

• p0’ at O compressed to A, unloaded to B and reloaded through C( First yield point) to D, where grains are densely packed.

• Stress strain line is ‘curved’. ‘Rearrangement’ of soil grains is the reason for ‘Non linear’ behaviour.

• Loop ABC becomes ‘stiffer’, where no ‘unrearrangement’ happens or ‘unfracturing’ (Shelly sands) happen .

• Since behaviour is curved we tell ‘instantaneous’ bulk modulus.K’ = dp’/dεv ====== K’ is not a soil constant.

Continued..Continued..• Specific volume may be plotted on the ‘y-axis’ with ‘isotropic’ compressive

stress in the ‘x axis’ where its ‘logarithmic’ quantities become ‘linear’.• Line OACD corresponding to first loading is known as NCL (Normal

compression line)• NCL have a slope ‘λ’ with its value ‘N’ for specific volume ‘v’ at isotropic

compression of p’ = 1.0 Kpa. Eqn of NCL hence is v = N – λ ln p’

• Line ABC ( Swelling line- SL) with ‘К’ the gradient ,’vК ‘ value of ‘v’ at isotropic compression of p’ = 1.0 Kpa. Eqn of SL hence is

v = vК – К ln p’• NCL meets SL at yield point at C where yield isotropic stress is (py’).• For a ‘particular soil’ (λ,К,N) are regarded as CONSTANTS. • Thus soil can be unloaded from any point on NCL giving ‘n’ number of

swelling lines giving new values for vК and py’(as К is const).-(dv/v)=(λ/vp’)dp’ = dεv == K’ = (vp’/λ)=Compression

K’ = (vp’/К)= Unloading and Reloading

OverconsolidationOverconsolidation• By unloading ‘state’ of the soil can reach any point ‘below and left’ of NCL. It

is impossible to have a state ‘above and right ‘of NCL.• Hence NCL is ‘part of state boundary surface’.• Any point ‘below and left’ of NCL like B is ‘overconsolidated’.• If ‘py’’ is point obtained by intersection of swelling line through ‘B’ with ‘p0’

current stressOCR (Rp) = py’/p0.

• For NC soils ‘state of soil’ is along NCL making Rp = 1.0.• For different states R1 and R2 Rp remains same which makes line through

R1 and R2 parallel to NCL.ln Rp = (ln py1’-ln p01’) =(ln py2’ – ln p02’).

• N1 and R2 have same current state of stress(say at same depth below ground) but different stiffness λ and К.

• Similarly R2 and N2 have approximately same specific volumes but different stiffness λ and К .

• Why is it so?? Because Rp determines Soil behaviour in all this cases.


• Consider states ‘R1 and R2’ in the 3rd diagram.• They are at same depths with different Rps.• Till now it is assumed state can move from R1 to R2 only through

NCL.• But in some cases it can directly move from R1 to R2.• That is called CREEP(clays) or vibration(Sands).

States of Soils on Wet and dry state of States of Soils on Wet and dry state of criticalcritical

• Clay can be Normally consolidated or over consolidated.• The ‘degree of Over consolidation’ whether it is heavily or lightly

over consolidated is defined by ‘critical over consolidation ratio’.• Normally ‘COCR’ is said to ‘light’ if Rp < 2 else Rp >3.• Soils may be ‘dense’ or ‘loose’ depending upon the position of ‘soil

state’ with respect to COCR line.• ‘State’ is defined by combination of ‘specific volume and pressure’.• A remains ‘Dense’ B remains ‘Loose’ despite specific volume of B <

specific volume of A. That defines above statement (combination works.. Stress governs!!!).

• The ‘region’ where ‘clays’ are ‘NC or lightly OC’ or sands are loose Wet side of Critical line. (For a particular stress pc’ ,specific volume (v) > vc Soil is wetter than at critical state).

• The ‘region’ where ‘clays’ are ‘Heavily OC’ or sands are dense Dry side of Critical line. (v < vc Soil is drier than critical state).


• The ‘distance’ of initial state from critical line is measure of ‘state’(which contains 2 parameters).

• Stress state parameter(Ss) = pa’/pc’ (ln Ss = ln pa’ – ln pc’).• Volume state parameter(Sv) = va – vc.• Critical Line and NCL have same slope(λ) Sv = λ ln Ss.• For soil in critical state Sv = ln Ss = 0.• For soil in dry state ( on dry side) Sv and ln Ss are negative.• For soil in wet state Sv and ln Ss are positive.• ‘Ss’ defines ratio of ‘current to critical’ where Rp defines ratio of

‘current to yield’.

11--D compression and SwellingD compression and Swelling• Isotropic conditions hardly exist in field, Most are plain strain

conditions.(Loading is 1-D with εh =0 ).• For this case εv = εz since εh =0.• Thus 1-D compression modulus (M’) = dσz’/dεz.• Inverse of parameter M’ is used in 1-D compression often called ‘mv

(coefficient of compressibility).• Parameters ‘v’ is replaced by ‘e’ ,’N’ by ‘e0’ ,’λ’ by ‘Cc’ and ‘К’ by

‘Cs’.NCL (for OACD) == e = e0 – Cc log σz’.SL (for ABC) == e = eК – Cs logσz’.

δv = δe and log x= 0.43 ln x ==Cc = 2.3λ and Cs = 2.3К

Continued..Continued..• For 1-D loading Rp is substituted by R0 = σy’/σ0’.• Due to loading and unloading (σz changes) with constant εh

variations happen in σz’ and σh’.• This variation gets reflected as K0 (Coefficient of earth pressure at

rest) == K0 = σh’/σz’.• Along NCL ie OACD, R0 =1.0 K0 = K0nco Empirically K0nc = 1- sinФc where Фc = critical friction angle.o Incorporating OCR we get K0 = K0nc√R0.• In 1-D compression σz’ and σh’ are unequal imposing shear

stresses in soil q’ = σz’(1-K0) and p’ = 1/3σz’(1+2K0).

Critical state strength of SoilCritical state strength of Soil

• ‘Strength’ of a material is the ‘maximum shear stress’ that it can sustrain.

• After this failure shear stress material fails either suddenly or by plastic straining (Normal case in soils).

• Because the stresses and deformations are ‘non uniform’ they cannot be used directly as a ‘representative’.

• Although that is the only option we have ‘the direct shear or simple shear tests’.

• When a sample is ‘sheared’ under ‘drained’ conditions due to shear volumetric coupling following effects occur.

• These effects are solely dependent on ‘where’ the material lies with respect to Critical state line.

Continued..Continued..• The following interpretations may be drawn. Soils on W side undergoes compression when subjected to shear

under drained conditions; converse happens for soils on D side. Both then reaches at states where shear stress is constant with no

more ‘volumetric strains’ ( See both graphs!!). Soils on D side reaches ‘peak’ before ‘ultimate’ state. The slope of εv and γ graph (volume change curve) is dilation

angle(Ψ) tanΨ = -(dεv/ dγ) . Samples on W side has more ‘initial void ratio’ but both reaches an

ultimate state at the same void ratio (ef). During shearing action dense particles move apart causing

volumetric effect. Shear is the cause and εv is the effect.

Peak, Ultimate and Residual StatesPeak, Ultimate and Residual States• Seeing Soils on D side ‘Peak state’ reaches at 1% shear strain where as

‘ultimate’ reaches at 10% shear strain.• This ‘peak state’ corresponds to ‘point of maximum rate of dilation’.• It is evident that ‘rate of shear strain’ controls shear strength.• At very large shear displacements on slip planes soil still have some shear

strength which is called ‘residual strength’ (Skempton 1964).• We can see from graph ( sands and plastic clays posses some real residual

at large shear displacements of 1m on special shear box).• At ultimate state ‘grain’ movements become ‘turbulent’ where as at residual

strains get ‘localized’ to zones of intense shearing.• The ‘Lowest shear stress’ reached after large displacements is called

‘residual state’(At this flat clay grains remain parallel to rupture zone, just like laminar flow).

• So such laminar flow is not possible in soils with round grains or sands ie Ultimate state = Residual state .

• For Clays Residual state = 50% Ultimate state.• The same Ultimate state is also called “CRITICAL STATE” .

•

Critical StatesCritical States• Real Essence of soil mechanics is critical state or ultimate state.• Till now we know critical state is associated with ‘turbulent flow’.• There lies a relation between ‘ζ’ , ‘σ’ , and ‘ef’ .• Critical state line (CSL) is given by ζf’ = σf’tanФc and ef = er – Cc

log σf’. Thus CSL || NCL with gradient Cc.• At critical state soil continues to distort ie suffer shear strains at

constant σ,constant ζ, constant e.• This means during shearing soil reach critical states which are

‘independent’ of ‘initial’ state (ie D side ,W side etc.)

Undrained StrengthUndrained Strength• So CSL is defined by corresponding ef s for varying normal ‘effective’ stress.For that

ef there will be a strength which is crtical shear strength(ζf’).• So according to equation ζf’ = σf’ tanФc it is important to understand σf’ to get critical

shear strength(ζf’) which is only possible from pore pressure (u) measurements.• u is determinate only in ‘drained loading’ where other approaches are required for

‘undrained loading’.• The critical state shear strength decreases with an ‘increase in void ratio’.• For saturated soils e is dependent on water content (w) ie an ‘increase in w’

decreases ‘critical state shear strength’.(It is not due to ‘lubrication’ effect but due to ‘decrease in effective stress’).

• Thus here at ‘constant void ratio’(Undrained) we are talking of shear strengthUndrained shear strength

ζf = su ==Undrained shear strength.ζf’ = ζf = su = log (su/tanФc) = (er – e )/Cc.

• The funniest part of ‘Logarithmic relation is that ‘su’ is ‘independent’ of ‘normal total stress’. Ie No need to simulate same total normal stress in the lab to get insitu ‘su’ . Su is dependent on e only.

• For designs in ‘undrained loading’ u are ‘indeterminate’ which make calculations done on total stresses as only thing we have to make sure is the void ratio Total stress analysis!!!!!!!!!.

• Converse happens for ‘drained loading ‘ u is determinate and those calculations based on Фc is effective stress analysis…

NormalizingNormalizing• At ‘critical state’ ζf, σf , ef are uniquely related giving a single CSL for a

particular soil.• It is important to ‘normalize’ the stresses since we play with peak states and

other states before critical.• After such normalization soils with same ‘current state (OCR)’ should have

same state parameters.

• There are 2 ways to achieve such normalization

OCR line containing A and A’ = eλ = ea + Cc log σa’. Both NCL and CSL appear as single points.

We have to normalize on the basis of current state or existing stress.

At NCL ζ’/σ’ = 0 and eλ = e0; At CSL ζ’/σ’ = tanФc and eλ = er.


• The second method is to normalize based on ‘equivalent stress’. This stress σc’ lies on CSL at the same void ratio (ea). Thus log σc’ = (er – ea) / Cc. Both Lines CSL and NCL appear as single points Position of CSL may be determined by ζ’ / σ’ = tanФc’ and σ/σc’

=1.0. Position of NCL may be given by log(σe’/σc’) = (e0 – er)/Cc

Critical state strength in Triaxial TestsCritical state strength in Triaxial Tests

• In triaxial tests sample is subjected to ‘total’ axial and radial stresses while pore pressures and sample volumes may be controlled and measured independently.

• Equivalent triaxial parameters for Direct shear tests (ζ’,σ’,γ,εv) are (q’, p’,εs,εv).

• Behaviour of q’ Vs εs and εv Vs εs were similar as it was obtained in direct shear.

• In triaxial tests, ultimate or critical state is obtained when sample continues to distort at a ‘constant rate’.

• Critical state at Triaxial case may be compared to direct shear test as follows. M is equivalent to Фc’ .Gradient of CSL = λ .The parameters λ,Γ, M (or Ф’) are assumed to be CONSTANTS for a particular soil.

Continued..Continued..• Corresponding to normalizing parameters in Direct shear (σc’,eλ) we

have (critical pressure,equivalent volume) annotated by (pc’,vλ).


• Undrained Shear strength (su) is related to void ratio or specific volume (v)

• su = ½(σa’ – σf’) = ½(qf’). Also we have ln(2Su/M) = (Γ- v) / λ.• Two samples of ‘same void ratio’ are subjected to ‘confined’ and

‘unconfined’ compression tests.• Both should have same effective stress circles as effective stresses

are pore pressure ‘independent’.• Total stress circles of unconfined compression will always lie at the

left of Effective stress circle since σr=0.= u is negative.• This negative u gives rise to +ve effective stress and strength.That

is reason for the stability of vertical cuts .

Relationship between M and Relationship between M and ФФc for c for Triaxial testsTriaxial tests

• From the concept of stress ratio t’/s’ = sinФ’mob = (σa’ – σr’)/(σa’+σr’).• This gives (σa’/σr’) = (1+sinФmob)/(1 – sinФmob) = tan2(45+Фmob/2).• At critical state Фmob’ = Фc’• It is known q’ = σa’ – σr’ and p’ = 1/3(σa’ + 2σr’).• Thus Mc = (6 sin Фc’)/(3 – sinФc’) and Me = (6sinФc)/(3 + sinФc’).• Фc is same for triaxial extension and compression.• Mc is not equal to Me in fact Mc >Me.

Experimental investigations on critical statesExperimental investigations on critical states

• It is important to determine the position of CSL accurately.• This is needed to have an idea regarding ‘ultimate strength’ of the

soil sample.• It is important to know D/W states and CS parameters λ and Γ .• For soils to reach critical state it must be strained with ‘No change of

state’. Ie const shear stress, normal stress and volume. Turbulent flow results at this state.

• Such simple experiments do not control pore pressure or drainage.• Thus tests on sand will be drained which gives Фc’ or for clays it will

be undrained which gives ‘su’.• Some tests for Фc’ may be rotating cylinder test or submerged cone

test.

ContinuedContinued....• For Normal Vs Shear stress graphs it can be seen that CSL passes

through origin c’= 0.• Although this is not the case for cemented soils comparitively large

strains required to obtain CS is large enough to break bonds.• We know such c’ =0 concept hold good for sands, But for clays

‘moisture content’ will give rise to ‘pore suctions’ which will rise ‘effective stresses’ and hence the ‘strength’.

• Dry flour does have zero strength (You can blow it away). Once compacted it has strength owing to specific surface of flour.

• True Cohesion is so small a force to reckon with and it is contributed from inter particle forces.

Estimation of critical strength parameters Estimation of critical strength parameters from Classification Testsfrom Classification Tests

• It can be interpreted that critical state which occurs at large strain will be dependent on intrinsic parameters not on soil structure.

• Once water content is known, PL and LL known, LI maybe measured and corresponding critical state strength may be obtained from CSL for that particular soil.

• Compressibility( Cc or λ) may be obtained as eLL=er – Cc log(1.5/tanФc’) ;epL = er – Cclog(150/tanФc’) Cc = (PI*Gs)/200or λ = (PI*Gs)/460.

• Position of critical state line (Γ or er) can be determined by Ω point (Scofield and Wroth). vΩ = 1.25,pΩ’ = 10MPa for K0 =0.5. er = 0.25 + Cc log 15000 or Γ = 1.25 + λ ln 10000.

• It is fascinating to see Compressibility (Cc or λ) is linearly related to PI .


• Critical Friction Angle Depends on Soil nature. For fine grained Фc ↓ with ↑ plasticity. Coarse grained deposit’s Фc

is dependent on particle shape and roughness. Exceptions are there e.g.Bothkennar organic Soil (PI =40,Фc =34).• Elastic volumetric strain ratio controls behaviour of soil This is a measure of ‘recovery’ Clays have high values(0.2 – 0.5) where coarse grained ones have

low values since unloading never leads to ‘recovery’ thanks to fracturing

Peak StatesPeak States

• Here samples with different void ratios and OCRs are subjected to same effective stress(σ’) .They lie along N,W,D1,D2.

• For samples at W and N the state parameters (ln Ss and Sv) are positive, for D1 and D2 state parameters are negative.

• At critical states (or at ultimate state) sample will have same shear stress (ζf’), normal stress (σf’) and void ratio (ef).

• But at ‘peak states’ these parameters are different.

• Peak states for different σ, OCR, e etc falls at OAB.• There is no relation for peak states as that for critical states.

• There may be 3 ways for examining peak states (1) To make use of MC equation with apparent cohesion intercept (2) Fit a curved line along peak stress points (3)include contribution of strength from dilation.

MC in Shear Tests MC in Shear Tests ––The concept of apparent The concept of apparent cohesioncohesion

• For a soil sample which attained a peak state at void ratio ‘e’ should be satisfying MC criteria according to (1) is given by

ζp’ = Cpe’ + σp’ tan Фp’.• Subscript p is used for ‘peak state’ and ‘e’ to say that peak state is

dependent on void ratio (e).• Consider 2 samples which reached peak states at void ratios e1 & e2.• We say about ‘peak states’ for those soils in D side.• It is found that Фp’ < Фc’. Also peak state lines (PSL) meets CSL at A1

and A2.• It can be seen PSLs are terminated at B1 and B2 with dotted lines

being extended back.• This means that ‘Cpe’ is not shear stress at ‘zero normal stress’ but

rather it is just a parameter to explain MC criteria.• Again it is nonsense to say about existence of shear stress at zero

normal stress for sands as this equation is applicable for them at peak states as well


• To make PSLs independent of void ratios Normalizing is required based of equivalent stress (σc’) approach.

• Thus PSL gets reduced to a single line (BA) hitting at CSL at A with a gradient Фp’ and a dimensionless cohesion intercept (Cp’).

Cp’ = Cpe’/σc’ ; Cp’ = tanФc’ – tanФp’ ; log (Cpe’/Cp’) = (er –e)/Cc.• Apparent cohesion (Cpe’) hence ↓ with ↑ in void ratio.

MC in Triaxial TestsMC in Triaxial Tests

• In shear tests Peak states were dependent on void ratios (e),now it will rather depend on specific volumes (v).

• Thus region AB is given by qp’ = Gpv + Hp Pp’ where Hp forms the gradient and Gpv the intercept on q’ axis.

• The broken line OT with gradient dq’/dp’ = 3 is representing a condition σr’ = 0 ie unconfined condition.

• Thus OT forms ‘tension cutoff’ an analogy to ζ’ axis in shear tests.• As for Cpe’,Gpv forms a parameter and not shear strength at zero p

value.• To be made independent of specific volumes(v) we normalize it on

pc’ to get peak states along AB.(qp’/pc’) = Gp + Hp (Pp’/Pc’).

• The relation between Hp and Фp’ is similar to relation between M and Фc’.As test is triaxial it is important to have an idea about variation of Hp in ‘compression’ and ‘tension’.


• It can be also found that ‘Gp’ from traxial tests is twice ‘cp’’ obtained from Direct shear .(since q’/pc’ = (2ζ’/σc’)= Gp = 2Cp’).

Curved Peak state LinesCurved Peak state Lines• At low effective stress ‘PSL’ remain ‘curved’ towards the origin where

once it hit CSL ,PSL and CSL coincides and have same slope.• Once PSL is normalized based on σc’ the curved line DBA is

represented by (ζp’/σc’) = A(σp’/σc’) ln(ζp’/σc’) = ln A + b ln (σp’/σc’).• Thus ‘A’ and ‘b’ are material properties just like Cp’ and Фp’.• For Triaxial tests ‘peak states’ may be given by

ln (qp’/pc’) = ln α + β ln (Pp’ / Pc’). • Like ‘A’ and ‘b’ in direct shear , ‘α’ and ‘β’ are soil parameters.• Since PSL and CSL coincides at ‘A’,material parameter ‘A’ = tanФc’ and α = M.

• Thus task reduces just to determine ‘b’ or ‘β’ to define ‘curved’ PSLs.

Relationship between Peak states & DilationRelationship between Peak states & Dilation

• Dilation is a common phenomenon in Soils on D side of critical when sheared.

• During shearing a dilating soil must overcome friction between the grains to ‘lift’the normal loads.

• The ‘peak shear stresses’ ↑ with ↑ in the rate of dilation.

• A shear box subjected to ‘σ’ in ↓ direction ‘ζ’ in ← direction is analogous to a block on a ‘rough’ inclined plane subjected to ‘N’ and ‘T’ in the corresponding direction.

• The roughness coeff µ is analogous to critical friction angle (Фc’).• Slope angle (i) is analogous to dilation angle (Ψ).

• If i=0 for slope then T/N = tan µ. For the same inclined plane with a slope ‘i’ we have T/N = tan(µ+i).

• Extending it to shear box or soil mechanics terms ζ’/σ’ = tan(Фc’+Ψ).

Continued..Continued..• We know slope of graph (δh,δv) is Ψ .From the relationship above slope

of graph (σ,ζ) is (Фc’+Ψ at Peak state and Фc’ at critical state).• So as per figure at ‘A’ and ‘C’ Ψ = 0 and at ‘P’ ‘Ψ’ becomes maximum or

‘Ψp’.• Point ‘P’ represents a ‘peak state’ under normal stress ‘σp’ with MC

fitting parameters (Cp’,Фp’) and with a dilation angle ‘Ψp’.


• If two points have same ‘e’ they lie on ‘same’ MC line.• (B1,C1) and (B2,C2) have void ratios e1 and e2 and They have MC

parameters (Cp1’,Фp’) and (Cp2’,Фp’) giving || lines.• B1 and B2 are under same ‘normal stress’ σb’.• B2 is heavily ‘over consolidated’ than B1 and hence less ‘void ratio’.• Thus ΨB1 > ΨB2 ; B1 and B2 have same critical state at Bc.• Same happens with C1 and C2.• B1 and C2 have same Ψ which means they have same ‘OCR’.


• For Direct shear ζ’/σ’ = tan(Фc’+Ψ) but for triaxial q/p = M – (dεv / dεs).• ‘q/p’ is the ‘stress ratio’ , ‘M’ the ‘critical stress ratio’,’ (dεv / dεs)’ the dilation.

Thus qp’ = MPp’ – Pp’(dεv / dεs) Obtained by cross multiplying• Both stress ratio relations ie from direct shear and Triaxial indicates an increase

in shear or deviatoric stress with an increase in dilation,Ψ or (dεv /dεs).• It is amazing to see that such relations exist at other states other than ‘peak’.• Consider 2 soils which exist under different specific volumes (v),where soil 2 is

over consolidated (Critical state is reached at ‘C’ once sheared).• It is obivious Ψ2 >Ψ1.• At Critical state vλ = Γ and eλ = er according to normalization.• Peak states ( in fact the peak stress ratios) can be associated with state

parameters ‘Sv’.At CSL Sv= 0,Otherwise on dry side Sv < 0. At Sv =0 (q’/p’)=M.• For Fine grained soils Normalization may be done one the basis of Attenberg

limits.• Ultimately it can be said that peak stress ratio is a function of OCR and state

parameter.

Behaviour of Soil Before FailureBehaviour of Soil Before Failure• It is important to understand how ‘states’ change from ‘initial’ to

‘critical’.• We know some of those ‘states – peak states’ where void ratios less

than critical states increased (dilation) in drained shearing.• There might be ‘unique’ states between NCL and CSL.• During Drained shearing it is well understood that ‘tendancy’ of the

material to compress or dilate depends on ‘where’ sample is lying ‘with respect to’ CSL.

• That is importance of stating soil in ‘D or W’ side.• We can consider 2 cases (1).Sample subjected to undrained

shearing (2) Sample subjected to drained shearing.

Continued..Continued..• Soil at ‘W’ which is normally consolidated and soil at ‘D’ are sheared

undrained (Specific volume is constant).• Both of them reaches CSL at Fu and they have same undrained

strength (since specific volume is constant).• Consider a condition at which shearing is carried out in a triaxial

apparatus at constant mean ‘total stress p’.• Thus TSP remains constant From IF. You can see difference

between TSP and ESP is pore pressure ‘u = u0+∆u’.• From TSP ESP difference it is understood that Soils on W side have

increase in ‘u’ during ‘undrained shearing’ and converse happens for Soils on D side.

• ESP for D side soil is vertical from DP (when it reaches Fu it has a higher deviatoric stress than P).This is quite ‘unlike’ ESP of W side soil.

• Now the second case may be considered (Drained shearing).

Continued..Continued..• Here also drained shearing is considered at constant mean stress p’.• Soil at W side compresses on shearing and reaches CSL at Fw.• Shearing happens in such a way that from D state to Peak (P) it

happens at a constant volume and then it diilates to Fd.• Shear stresses at failure points Fw and Fd are different unlike Fu in

the previous as effective stresses and specific volumes are different.• Paths WFw and DPFd are vertical since all happens at

constant pore pressure (drained condition or u = 0).• Soils on D side have Rp about 3;Soils on W side have Rp < 2.• It can be viewed from perspective of ‘state parameters’.ie If Sv and

Ss are +ve (ie W side) u ↑ else ↓.

State Boundary Surface for SoilState Boundary Surface for Soil• The isotropic NCL represents ‘boundary’ of all possible states of

isotropic compression.• The states can move below or inside NCL by unloading but cannot

move outside NCL.• ‘Peak envelope’ represents ‘boundary’ of all peak states.• One peak state line is for one specific volume and simlarly for for

other volumes Family is called Peak state surface.• This Family or ‘surface’ can be normalized to a ‘single line ‘ with

respect to a critical pressure pc’.• Once normalized we can see such a defined ‘peak state line’ exists

for soil on D side.It has to be seen whether such a line if at all exists at W side (If at all it will be a like a dotted line shown).

Continued..Continued..• Consider 3 initial states on ‘W side’ They are P,V,R.• P is sheared ‘drained’ with ‘constant’ p’ and V is sheared ‘undrained’ at

‘constant specific volume’.• Both these paths cross at ‘S’ which lies along anisotropically

compressed path of R (q’/p’ = ή),Also along NCL q’/p’ = 0 where as along CSL (q’/p’ = M).

• Now the question if the stress paths cross over at S, then whether we have same specific volume at S?.

• The answer is to normalize these two diagrams (1) Normalize the stress path diagram on critical pressure (pc’) (2). Normalize the second diagram on equivalent volume vλ.


• Once Normalized Paths (CSL,NCL RS) reduces to single points.• The dry side of the critical is shown by ‘dotted lines’.• The ‘line’ which has normalized axis is really part of 3D surface with

q’:p’:v.• In that 3D surface ‘constant specific volume’ sections are

represented as ‘full’ lines.• ‘Constant stress ratio’ sections are shown by ‘broken’ lines.• The part corresponding to wet side is called ‘Roscoe’ surface and

part of dry side corresponding to peak states is ‘Hrovslev’ surface.

Elastic behaviour at States inside the state Elastic behaviour at States inside the state boundary surfaceboundary surface

• Thus we saw the 3D state boundary surface (SBS) which is the boundary of all ‘possible states’ of a ‘reconstituted soil’.

• The state cannot exist outside the SBS –by theory (exceptions exist as in the case of ‘cemented soils’).

• That means soil at a ‘particular state’ remains inside SBS after loading and unloading.

• Thus SBS is like a ‘yield surface’ which means when ‘state’ attains SBS or yield it is accompanied by ‘elastic and plastic strains’.

• The strains remain elastic inside the SBS. • A sequence of loading and unloading is considered from A→D

where OCRs are same but specific volumes are different.


• Once loading curve hits NCL from B→C soils gets yielded and hardened by δpy’ with an irrecoverable plastic strain δvp.

• Along AB and CD behaviour is taken to be ‘elastic’.• For an ‘isotropic elastic material’ stress – strain behaviour is

decoupled (ie shear and volumetric effects becomes independent).• For such a material δεs = (1/3G’)*(δq’) ; δεv = (1/K’)*(δp’).• From swelling-recompression lines we have δεv = (К/vp’)*δp’ where

‘К’ is the slope of AB and CD.• For shearing δεs = (g/3vp’)*δq’. ’g’ is ‘shear stiffness’ like ‘К’ is

‘volumetric stiffness’.• For isotropic material G’/K’ = К/g is a constant(Poisson’s ratio is

constant).• Thus if soil is considered ‘isotropic and elastic’ shear and volumetric

effects will be decoupled and also volumetric strains becomes a function of δp’ and independent of δq’.

Continued..Continued..• This means inside SBS state must remain on a vertical plane above

a particular swelling and recompression line .(ELASTIC WALL).• Elastic wall concept is entirely different from constant volume

section (except for К = 0).• When Elastic wall hits SBS ‘YIELD CURVE ‘ forms.• There are infinite swelling and compression line,so many elastic

walls and so many yield curves.• For undrained loading on Elastic wall we have δεv =0 So from

equation we have δp’ =0.• It is clear that once δp’ =0 we are telling about loading paths which

are linear and vertical. They yield or hit SBS at Yw and Yd.

Undrained Loading on the SBSUndrained Loading on the SBS

• When shearing occurs at ‘Undrained’ conditions we can see at ‘constant volume’ line hits CSL at ‘Fu’.

• It can be seen that yielding occurs ‘not’ at the point where deviatoricstress is maximum, rather at the point of ‘peak stress ratio ήp at Yd’.

• It has been seen that ‘undrained section’ of SBS has been drawn approximately ‘symmetric’ to Fu.

• This is not the case always as it can have ‘asymmetric’ sections on dry side.

• Actually such an ‘undrained’ asymmetric behaviour arises from 2 things (1) Geometry of SBS (2) Geometry of elastic wall.

• Consider the condition given in the next slide. It will have peak stress ratio at Pd,Peak deviatoric stress at Pq,Ultimate state at Fu.

• This is puzzling it is all different.But that is what happens in reality!!


• As explained….

Stress ratio and DilationStress ratio and Dilation

• We have derived following equation for peak states. That is (q’/p’) = (M – dεv / dεs).

• This equation is even applicable to states before and after ‘peak’since magnitude of plastic strains high in comparison with elastic strains.

• At A and C Ψ=0 ie no volumetric strain happens dεv = 0.Thus at A and C ,q’ / p’ =M.

• So the funda is If we terminated before ulimate/critical point C we can find it by plotting q’/ p’ vs dεv / dεs.

• You need to plot it for both NC and OC clays or Dense and Loose sands from ‘either side’ to correctly identify C.

Softening of Soil beyond peak state and Softening of Soil beyond peak state and development of Slip surfacesdevelopment of Slip surfaces

• An important feature of soil on Dry state – Development of Slip surfaces or discontinuities.

• With a rigid material on either side Slip surfaces are zones of ‘intense shearing’ with a finite thickness ( may be some grains).

• Inside this thickness soil changes its volume resulting in non uniform volume changes → Non uniform strains.

• Thus by measuring strains at surface it is not possible to have an idea about ‘straining in slip surfaces’.

• Thus in slip surface specific volume of soil is high!!!.• Such a phenomenon will not be found in soils of W side. They

compress and get hardened.• A large number of faint discontinuities can be seen without a well

defined slip surface (In metals this is called Luder’s lines).• Bottom Line → Don’t trust Specific volumes when you have slip

surfaces.

CAM CLAYCAM CLAY

• We have already seen 3D SBS and for a simple theoritical model for the ‘stress strain behaviour’ of soil; This is ‘Yield surface’.

• Elastic walls comes and hit these surfaces giving ‘Yield Curves’.• It is required to incorporate ideas of yielding,hardening and normality to

get ‘constitutive equations’ for soil.• Before We have to get ‘mathematical expression’ for the shape of SBS.• By theoritical (from the mechanics of granular materials) and

experimental (by fitting lab data) we can get equations for SBS.• Cam clay model was developed in University of Cambridge in 1960s.• This is the most simple model to explain the experimental behaviour .• Cam clay is a ‘family’ of models. They are Original Cam clay (Schofield

and Wroth -1968) and Modified Cam clay (Roscoe and Burland -1968).• Soils are ‘frictional’ with ‘Logarithmic compression’.• SBS is ‘yield surface’ and as a ‘Plastic potential surface’ and hardening

is related to ‘plastic volumetric strains’.

Continued..Continued..• The basic difference between models of Cam clay is the equations

of yield curves.• They are Logarithmic spirals for original Cam clay while Ellipses for

modified Cam clay. State Boundary Surface of CAM CLAY – Expressions.• 3D SBS of Cam Clay will be definitely a function of q:p:v space and

is given by

• When v = N – λ ln p’ the SBS meets v:p’ plane. Obiviously q’=0 at v:p’ plane. Thus if substitute for v and q’ in the above expn

• It is important to get expns for constant volume sections or undrained stress paths.

• It can be obtained as “if pc’ is the stress at which constant volume section intersect with CSL” then that point

ContinuedContinued

• So the expression of constant volume surface is • Now we have to get expression of yield curve.• It is nothing but elastic wall intersection with SBS.• Elastic wall is given by• Such an yield curve hits CSL at specific volume vc and mean stress

pc’ where • Thus yield curve forms • Thus Yield curve is different from Constant Volume section unless К

=0.• We now know yield curve equation. Yield stress (py’) happens at

v:p’ plane or q’=0. • We can see how many times of pc’ is py’• Differentiation of yield curve yields


• This tells us about a logarithmic spiral curve where the gradient dq’/ dp’ is related to gradient q’ / p’ of the radius from origin.

Calculation of plastic strainsCalculation of plastic strains

• Yield curve is taken to be ‘plastic potential’ and vector of plastic strain increment will be normal to yield curve.

• If two such lines are orthogonal then their product gradient is -1.

• Differentiating yield curve we got • Thus plastic strain increments is given by • At critical state q’ / p’ = M and hence• On wet side q’ / p’ < M and D side we have q’ / p’ >M resulting δεv

to be +ve and – ve respectively.• The result Dilation and compression !!!• This equation of plastic volumetric strains resemble• Only difference is that former is ‘Plastic strains’ and latter is ‘Total

strains’.


• The Total strain expression is derived based on ‘principle of work done’ by friction and dilation at peak states on dry side.

• Hence Cam Clay becomes an ‘equivalent work equation’ extended to wet side as well.

Yielding and HardeningYielding and Hardening

• As the state moves from ‘One yield surface’ to other there will be yielding.

• ‘Hardening’ or ‘Softening’ depends on whether state is on W side or D side respectively.

• If Loading is done on W side from A→ B it it accompanied by compression or decrease in ‘v’ with an ‘increase’ in ‘yield stress’.

• If Loading is done on D side from C→ D it is accompanied by dilation or increase in ‘v’ with a ‘decrease’ in ‘yield stress’.

• The basic SBS equation of CAM CLAY may be rewritten as

• Differentiating, dividing by ‘v’ we have

• These are the ‘total volumetric strains’.

Continued..Continued..• From the total volumetric strains ‘elastic volumetric strains’ must be

deducted to get ‘plastic volumetric strains’.• Elastic volumetric strains are given by• Plastic Volumetric strains hence are obtained as

• We also know

• Thus Plastic ‘shear’ strains are given by

Complete Constitutive Equations for Cam Complete Constitutive Equations for Cam ClayClay

• The elastic and plastic strains for both shear and volumetric effects may be added together to get ‘Complete constitutive equations for CAM CLAY’. They are given by

• These equation is called complete because of following reasons (1) These apply for ‘states’ that are ‘ on SBS’. (2) These apply for ‘states’ ‘inside SBS’.If we put λ = К we get basic

expression for elastic strain.• If we write the complete constitutive equation in the following form

Continued..Continued..• We get C11,C22, and C12 or C21 as the following

• This demonstrate that in Cam Clay basic compliances contain intrinsic soil parameters M,λ,К and g.

• It also depends on ‘current state’ given by v, p’ , ή’ = q’ / p’.• Thus CAM CLAY behaves in ‘non – linear’ manner as p’, q’ ,v all

changes with loading path.• When the failure happens or when it reaches ‘critical state’ or when ή→

M, C11→ infinity,C22 → 0.• Thus at ultimate state or failure or critical state ‘Shear strains become

large and Volumetric strains become small which is correct !!

Application of Cam clay in designApplication of Cam clay in design• Now the Cam clay equations are written for ‘shearing and volumetric

effects’.• But for real calculations these equations need to be wirtten in terms if

normal and shear stresses and corresponding strains in 3 dimensions.• Best advantage of CAM CLAY is the ‘shape of yield curves’ which are

logarithmic spirals.• In modified CAM CLAY shape of yield curves are ‘Ellipses’.They are

proposed by Muirwood(1991).• FEM analysis can be done on CAM CLAY as per Britto and Gunn(1987).• These are advanced techniques and need lot of experience.

Stiffness of SoilsStiffness of Soils• ‘Stiffness’ relates the increment of ‘stress’ with the increments of ‘strain’ which can be

used to predict ground movements.

• Simple Analysis are carried out as ‘Elastic’ where as it is known that ‘soil strains’ are often inelastic or Elasto plastic (Cam Clay).

• Stress-strain behaviour doesn’t show a great agreement with Cam Clay model when strains are small.

• States lie inside the SBS at these stages where in simple theory ‘strains’ remain elastic.

• In Cam clay classical theories of Elasticity and Plasticity were combined with soil mechanics theories of ‘friction’ and ‘logarithmic compression’.

• In Cam clay we have a set of non –linear constitutive equations developed in terms of intrinsic soil parameters λ,M,Γ,К and g.

• The Basic Equations for ‘Total Shear and Volumetric strains’ have elastic and plastic components when States are ‘ON SBS’.

• But for states inside SBS basic equations contain only elastic strains.• That is Cam Clay model holds good when states are ‘on SBS’ but fails ‘inside SBS’

where behaviour is taken to be ‘elastic and recoverable’

Continued..Continued..• Two soils are considered which are subjected to loading paths A→B

and C→D.• A is lightly over consolidated, which yields at ‘Y’ and reaches SBS

where it moves along Y→B with ‘both elastic and plastic strains’.• C is heavily over consolidated and state will not reach SBS which

causes the resulting strains to be ‘elastic’.• For Path A→B majority of path is along Y→B (on SBS) where Cam

clay gives better results for strains or ground movements.• For C→D the path is inside SBS where Cam Clay gives bad results.

StiffnessStiffness--Strain Relationships for SoilsStrain Relationships for Soils

• The general constitutive equations may be written as

where G’-Shear Modulus, K’- Bulk Modulus, J’-Moduli coupling shear and volumetric effects.

• Consider 2 cases (1) Undrained shear Loading (δεv=0) (2) Isotropic compression(δεs=0).

• For the First case if we substitute δεv=0 in constitutive relation we have dq’ / dεs = 3G’ and dp’/ dεs = J’. In this case G = Gu (Undrained)

• For Second case if we substitute δεs =0 in constitutive relation we have dp’/ dεv = K’ and dq’/dεv = J’.

Continued..Continued..• We could obtain J1’ and J2’ if δq’ was plotted against δεv and δp’

against δεs.• As seen ‘gradient’ of the first curve is shear modulus(3G’) and that

of the second curve is bulk modulus (K’).• Although CDE and RST ‘looks non-linear’ it is difficult to see how

soil is behaving especially at ‘small increments’.• It is ‘curious’ that ‘shear modulus – shear strain curve’ and ‘bulk

modulus –volumetric strain curve’ are similar (except at for large strains >1% where G’ dies down and K’ increases)

• Stiffness-strain curve for shear and bulk moduli are typical for soil.

Continued..Continued..• At ‘Large strains’ or at ‘Ultimate failure’ G’ becomes zero which is

understood.• Now we have to see whether our Cam clay is giving such ‘stiffness –

strain’ behaviour.• Consider Drained constant p’ loading path O→Y→A where state

reaches SBS at ‘Y’ and travels on SBS along Y→A.• Since p’,v and g remains constant according to Cam clay in the

loading path O→Y;G remains constant according to Cam clay.• After yield ‘Y’ stiffness ↓ in Cam clay as well as in reality.• So after Yield ‘Y’ behaviour is properly captured by Cam clay but no

so before Yield.


• Thus ‘Shear stiffness – Shear strain curve’ has 3 zones (1).Very small strain (2) Small Strain (3) Large Strain.

• At Very small strain Shear stiffness remains constant where as at large strain Cam clay holds good. For the ‘intermediate part’ the behaviour is highly ‘non-linear’.

• For normal strains ie for ‘Foundation settlements (εv = 0.1%) for maximum settlement at the surface of 10mm’.

• For retaining structures ‘outward movement at wall top of 10mm is equivalent to mean shear strain(γ = 0.1%).

Measurement of Soil stiffness in Lab Tests.Measurement of Soil stiffness in Lab Tests.

• The Best method for measuring Stiffness and Stiffness parameters will be ‘Stress path triaxial tests’.

• To get ‘Very small strain part’ of stiffness-strain curve we need to get strains of 0.001% ie for a sample of 100mm you need to measure strains of 0.001mm or 1µm.

• Errors in measurement of axial displacement ‘∆L’ may come from (1)Axial displacements at the sample ends (2) displacement where loading ram joins top platen(3)Movements in Load cell (4) Movements in cell.

• Hydraulic Triaxial tests may be an alternative to measure strains <0.01%.

• Using Conventional triaxial tests with gauge inside the cell may be used to measure strains<0.001%.


• Although Conventional Triaxial tests with gauge mounted cell may be used for strains <0.001%,Direct observation of strains are unreliable at this strain range.

• So indirect calculation of Shear modulus for dynamic waves is a better substitute to triaxial at very small strains <0.01%.

G0’ is small strain shear moduli,γ weight density of soil,Vs shear wave velocity through the sample.

• For such dynamic tests loading rates are high and experiment undrained.

• Hence G’ = Gu and since δεv=0 small strain bulk modulus (K0’) which is undrained is theoritically infinite.

• This means it is impossible to corelate undrained K0’ with Vs as we did for G.


• So…

Stiffness of Soil at Small and Very small Strains For getting very small strain Gs or G0’ samples are vibrated at strain

less than 0.001%.(Soil is linear and elastic). Small strain G is given by

A, m ,n depend on nature of soil, pr is reference pressure included to make equation dimensionless. ‘n’ varies from 0.5→1.0,’m’ from 0.2→0.3.


• Taking natural logs on the equation

Stiffness At Small Strain• The equation applicable for strain range 0.001 to 1% is given by

• where A = Gnc’/p’ where Gnc’ is the compliance term C11 of Cam clay model. or Shear stiffness of Normally consolidated soil.’m’ depends on the type of soil and strain.

Continued..Continued..• It has been found that soil behaviour at SBS can be represented by

mathematical models like ‘Cam clay’.• Inside the SBS you need highly complicated models to tackle non

linear elastic behaviour of soils.• One alternative is to use ‘curve fitting’ to obtain empirical relations

between ‘G and εs’ and ‘K and εv’. (Duncan and Chang -1970).• Another approaches may be to assume soil to be inelastic for small

strains, use moving yield surfaces for Cam clay etc.•

ConsolidationConsolidation

• Undrained Loading causes excess pore pressure (u’) which may be positive or negative wrt to static pore pressure(u0).

• This ultimately results in hydraulic gradients causing seepage flow.• Volume changes happen for soil mass, effective stresses change

finally causing excess pore pressure to dissipate.• Finally system reaches steady state when u’=0 and gradient = 0.• It is the rate at which consolidation happen is the area of interest.• So we are in Need of a Theory for Consolidation.• In 1-D consolidation seepage flow and strain in radial direction is nil.• The seepage happens upward as in a oedometer test where u’=0

thus u=u0. Theory of 1-D consolidation• It is assumed that for a time interval ‘δt’ there is change of thickness

of ‘δh’.


• Flow at top is given by ‘q’ and through bottom ‘q+∆q’.By Definition of ‘mv’ we have δh = -mvδzδσ’.It is assumed ‘mv’ is a constant.

• As ‘mv’ is constant this is applicable to low stress increments.• From continuity equation we have Aδh = -δqδt.• Combining these equations we have ∂q / ∂z = A mv ∂ σ’/∂t.• The rate of ‘seepage flow as per Darcy’ is V = q/A = ki. Where ‘V’

seepage velocity ,’i’ is (-1/γw)(δu’/δz).• Thus ∂q /∂z = (-Ak/γw)*∂/∂z (∂u /∂z) ∂q /∂z = (-Ak/γw)(∂2u’/∂z2).• Thus (k/mvγw)(∂2u/∂z2) = -(∂σ’/∂t). We know σ’=σ-(u0+u’).• Thus ∂σ’/∂t = ∂σ/∂t -∂u’/∂t. It is assumed total stress(σ) is const.• Thus where

• Possible to solve the equation to represent u’ = f( z ,t ).It is important to notice that we play with excess pore pressures (u’) here.

Isochrones.Isochrones.

• Excess pore pressures (u’) plotted with Depth for different times gives a family of curves – Isochrones.

• To see them physically insert a set of sand pipes in to a consolidating soil below rapidly constructed embankment.

• At t=0 it can be seen that u’ = ∆σ.ie entire load is taken pore water.• Then u’ starts dissipating near drains giving curves which at t reaches

infinity becomes u’=0.• It is imperative that isochrones (rep u’) satisfy 1-D consolidation eqn abd

drainage boundary conditions.

Continued..Continued..• At relatively small times ‘tn’ consolidation happens at ‘upper layers’

and below depth ‘n’ pore pressures are not fallen.• At times ‘tm’ consolidation is occuring for the entire layer and at a

critical time ‘tc’ pore pressures at base just start to dissipate.• Else you can say for depth ‘z3’ u’ stay till critical time (tc) at depth

‘z2’ u’ started falling below ‘tc’ ,where for depth ‘z1’ u’ started falling way below ‘tc’.

• Gradient of isochrones (du ’/ dz) is related to hydraulic gradient (i) ie du’/dz = -γw*i.

• This gradient will be causing seepage.Again it is imperative amount of seepage will depend on permeability V = (-k/γw) (du’/dz).

• By negative sign it indicates flow is happening against direction where gradient is decreasing ie upward .

• Once when du’/dz =0 or gradient becomes horizontal we are telling a non flow condition That is reason isochrones become vertical at impermeable boundaries.


• Rate at which settlement ‘ρ’ occurs is dependent on gradient du’/dz.∂ρ/∂t =(k/γw)(∂u’/dz).

• If two isochrones are considered at time ‘t1’ and ‘t2’.If a thin slice ‘δz’ is considered change of thickness ‘δh’ = -mvδzδσ’.

• By the spring analogy model it is well known that increase in σ’ is shared as decrease to excess pore pressure (u’). ‘δσ’ = ‘-δu’.δh=mv δzδu’.

• This ‘δzδu’ is the area which gives total settlement δρ = mv * Area OAB.

Solution of 1Solution of 1--D equations by Parabolic D equations by Parabolic IsochronesIsochrones

• To get ‘rate of settlement’ we can consider 2 cases for parabola (1) t< tc (2) t>tc.

• For the 1st case ‘no consolidation’ happens below depth ‘n’ and the surface settlement ∆ρt = mv * ∆AEN = (1/3)mvn∆σ

• Thus rate of settlement ∂ρ/∂t = 1/3mv∆σ(dn /dt)---(1)• We also know ∂ρ/∂t = (k/γw) (∂u’/ ∂z) From geometry ∂u’/ ∂z @ A is 2∆σ/n

∂ρ/∂t = (k/γw)(2∆σ/n)---(2)• Equating (1) and (2) n dn/dt = 6(k/mvγw) n dn/dt = 6cv.• Applying BCs @ t=0 n=0 n=√12cvt.• Thus for surface settlement (∆ρ(t<tc)) =(1/3)mv∆σ√12cvt.• Final settlement given by consolidation equation is (∆ρ) mv H ∆σ.• Thus (∆ρ(t<tc)) / (∆ρ) = (2/√3)(√cvt/H2) where cvt/H2 = Tv• Thus Degree of Consolidation U = (2/√3)(√Tv)• This solution is valid till t = tc where n = H = √12cvt • At this point Tv =1/12 U =33%.• Ie This derivation is valid till 33% consolidation


• For the second case tm > tc isochrone intersects orthogonally at ‘M’where CM =m∆σ.

• Thus as previously ∆ρ = mv∆σH(1-2/3m).Thus ∂ρ/∂t = (-2/3) mv ∆σ H dm/dt ∂ρ/∂t = (k/γw)(2m∆σ/H). m dm/dt = (-1/t) 3Tv.

• To get m integrate within limits with BCs (m=1 @ Tv=1/12 m=0 @ t is infinity) Thus m = exp(1/4 – 3Tv).

• Thus surface settlement(∆ρ) = mv H ∆σ(1-2/3 exp(1/4 – 3Tv). And U= 1-(2/3)exp(1/4 – 3Tv).

• Thus for Tv<1/12 U=U = (2/√3)(√Tv) For Tv >1/12 U =1-(2/3)exp(1/4 –3Tv). For complete consolidation Tv =1.

•


• Till now One way drainage is considered where as for two way drainage ‘H/2’ is substituted for ‘H’.

• The main limitation of parabolic isochrones is that it is applicable only when u0= constant.

• If we solve differential equation analytically we solve it as Fourier series which is normally used where

• For U< 60% U =(2/√3.14)√Tv according to this equation which is close to parabolic isochrones.

• For different u0 cases you have different solutions for U as shown

Continued..Continued..• Cv can be obtained by Square root time or Logarithmic Time method. Continuous Loading and Consolidation• When Loading happens u’ develops but at the same time drainage

happens.• Thus problem is neither drained nor fully undrained. Rather it is coupled

problem of drainage and loading. Coupled 1-D consolidation are rather easy to tackle with.

• Pore pressures at top and bottom are given as ‘u0’ and ‘ub’ with total stress being ‘σ’ and settlement ‘ρ’.

• Thus u’ at base = ub –u0. From concept of effective stress(σ’) Shaded area = σ’H. Trends of σ,u0,ub,ρ during test is shown below.


• Basic equation for Coupled consolidation and Loading is

• If loading is at slow rate ub’<<σ-u0.• From this ‘mv’ =(-1/H)(dH/dσ’).For a parabolic isochrones excess

pore pressure at any depth

• Differentiating twice we now know Coupled equation From that

• Also

• Thus following gradients dH /dt, dσ’/dH ,,dσ’/dt becomes important for measurement of important parameters like ub’ making these gradients to be measured at suitable intervals

Ageing and Structure in Natural SoilsAgeing and Structure in Natural Soils• We have already seen soil behaviour for ‘reconstituted soils’.But natural

soils are far from reconstituted.• Due to deposition (by agents) we are likely to have different materials.• Such soils which are deposited undergo physical and chemical changes.• They are collectively termed ‘Ageing’ which includes ‘cementing’ and

‘weathering’.• This results in ‘structure’ formation and reconstituted soils ‘destructured’.• We have to now see How the natural soil behaves ?.• For this we need a sample of ‘minimum’ disturbance• It is worth to consider the behaviour of intact soils within frame work of

‘reconstituted soils.• Soils deposited as layers undergo compression and swelling due to

erosion and deposition.• When such erosion and deposition happen it is worth to notice ‘e’ or ‘v’

changes and as a result water content (w) changes ( as e =w G).


• At ‘A’ and ‘B’ we have soils as NC and at ‘C’ OC. • It is seen that ‘A’ and ‘C’ have similar stress (depths) but different

water contents (w).• Vertical and Horizontal stress ratio(K0) can be also noticed at these

salient points.• For NC σh’<σz’ K0 <1 while for OC σh’>σz’ K0 >1.• If we know K0 of NC soil (ie K0nc = 1-sinФc) K0 for OC soils may be

estimated as K0oc =K0nc√R0• Thus state of soil depends on current stresses (on depth) and OCR

(ie current depth and depth of erosion).• Consider 2 soils A and B at different depths. They are lightly eroded.

It can be seen that there is a huge diff between their water contents.• Consider 2 soils C and D at different depths. They are heavily

eroded. Difference in water content is less.• Thus it can be concluded that If we see soils whose water content

does not wary much depth they might be subjected or heavy erosion.

Continued.Continued.

• So..

AgeingAgeing

• By the simple theories it is known ‘plastic volume changes’ can happen for loading along SBS.

• Thus as plastic straining happens during loading along SBS we are telling about change in ‘e’ and hence ‘w’ This is called ‘irrecoverable plastic water content change (δwp).

• This is due to change in yield stress(σya’ to σyb’) and change in OCR( Ra to Rb).

• By time as for London Clays(60 million years old) such changes can happen with out any loading or unloading.

• This is collectively called Ageing and most important of them are Creep,, weathering, compaction and changes of salinity of pore water.

• Ultimately we are telling a change in OCR at const effective stress

Continued..Continued.. Vibration and Compaction.• In vibration or compaction we get irrecoverable plastic volume changes

at constant effective stress(σ’) with change in OCR.• So stress(σ0’) is constant and plastic volume change happens(δwp)

giving path A→B vertical.• OCR ↑ from Ra to Rb with yield point increased from σya’→σyb’.• It is worthy to note all this happened without any loading/unloading.• State path normalized to σc shows a decrease from A→B as σc ↑ and

volume ↓.

ContinuedContinued

Creep• The main difference between creep and compaction is the ‘ time factor’.• Compaction occurs ‘instantaneously’ where as creep occurs at a rate

which diminishes with time.• The basic constitutive equation is δw = Cα ln (t/t0).• The water content or void ratio decreases with logarithm of time and the

effect is more apparent for soft clays.

Continued..Continued.. Cementing

• During Creep and Compaction current states were changing where in cementing along with ‘current states’ SBS itself changes.

• The basic principle of cementing is deposition of CaCO3.

• As a result e↓,v↓ NCL changes SBS changes• .But nothing happens to CSL as material has to fracture to reach ultimate or

Critical state( Leaving material to be of its original identity) .• Hence we have a unique CSL, intrinsic NCL (for reconstituted soil) and a

structured NCL.

• During A→B cementitious material deposits at const stress. Along B→Y→C compression (w↓) happens with first Yield at INCL.

• Y→C zone indicates brittle fracture and sudden volume decrease.

• First yield will be at σys’ and second at σyi’.• Since there is only one CSL normalizing parameter may be chosen easily(σc’).

• A and C have same states of consolidation ‘B’ remains Over consolidated.

• The distance of structured SBS to intrinsic SBS depends on ‘cementing’.


• So..

Weathering• At ‘constant effective stress’ physical and chemical properties of soil

changes.• Thus ‘intrinsic’ boundary itself changes and so the CSL.• If water content changes you can expect a change in ‘current state’.• Net effects are like that of Ageing only. Changes in Pore water Salinity• Intrinsic properties of soil changes giving two NCLs.

Ground InvestigationsGround Investigations

• BS 5930:1981 covers lab testing procedures for GI studies.• By seeing the nature and state it can be seen whether soil is OC or

NC .• Above statement can be substantiated as Stiff clays are OC due to

low ‘w’.Hence state plays an important role.• After any GI you should be knowing the following about each of

principal strata.(1). Classification in terms of nature and state.(2). Boundaries of each strata.(3). Stress History.(4).Structure and Fissuring.(5).Parameters for strength, stiffness and permeability.

• There are 3 Stages for any GI (1) Desk studies (2) Preliminary Investigations (3) Detailed Investigations.

Continued..Continued..• Excavations are usually done by (1) Test pitting (2) Drilling (3)

Sampling.• Test pits are done by excavators, drilling by rigs or augers ,and

samplers for sampling (Tube sampler U100 is popular in UK).• Tests to determine strength, stiffness ,permeability are mostly done

insitu.• Probing tests like SPT, Dutch cone (with or without pore pressure

measurements) are used for stiffness ascertaining.• Loading tests used for Plotting load deformation behaviour for

ultimately getting stiffness.• Some of the Loading tests are Vane shear ,Pressure meter etc. States of Soil in the Ground.• As the states are often governed by deposition and erosion, In case if

geological history is known it is possible to predict the engg properties of soil.


• Consider a case where a particular soil is compressed from LL →PL• If we consider expression for CSL ζf =σf tan ϕc taking a value for ϕc

=26 we have σc’ as 300KPa.• Distance of NCL from CSL depends on type of soil(σe’ with σc’ relation

for some clays σe’ = 2σc’).• We also know with the decrease in ‘w’ we have an increase in σz’.• For the same soil we have σz’=6Kpa at LL and σz’=600Kpa for PL.• Amazingly this is stress that soil would experience at 0.6 and 60m.• We can hence see OC clays exist at close to PL where NC clays water

content decreases with depth.• Now how critical state undrained shear strength(su) varies for NC and

OCs ?.• For OCs with ‘w’ close to PL we have su =150KPa where for NC ‘su’

increases with depth.


• Thus only for NCs ‘su’ increases with depth given by su/σz’ =(σc’/σz’) tanϕc.

• That is the reason of this commonly seen relation (su/σz’) = (0.11 + 0.0037PI).This is used for undrained shear strength of ‘soft clays’.

ContinuedContinued

• Permeability is an important parameter for getting consolidationparameters.

• Due to low permeability especially in clays ‘steady state’ will be attained after some time.

• For getting ‘insitu permeability’ we need pumping tests.• The rate of flow ‘q’ definitely depends on potential(P) and its gradient

(dP/dr) q= Aki = 2πPk dP /dr dr/r = (2πk/q)P dP.• On integration ln(r2/r1) = (πk/q)(P2^2 – P1^2)• Under a pressure of u= γwhw there will be flow from a spherical cavity

which is given by q = 4πrkhw(1 + r/√πcst) where ‘cs’ is the coefficient of spherical consolidation.

• At ‘infinite’ time we have q = 4πrkhw. In all this assumption is that cavity is spherical and hence the ‘4πr’.

Stability of soil structures using bound Stability of soil structures using bound methodsmethods

• We have already developed simple constitutive equations relatingincrements of stress and strain.

• Time has now come to apply these relationships to geotechnical structures like foundations ,slopes ,retaining walls etc.

• The ‘solutions’ must satisfy conditions of equilibrium and compatibility.• Obtaining complete solutions satisfying the above 2 conditions and

material properties are difficult (even for simple footings).• First of all we are considering condition of ultimate collapse.• In ultimate collapse the major material property of soil is shear

strength.• In determining shear strength we have to specify the condition ie

drained or undrained ?.• For undrained cases shear strength(ζf =su) where as for drained

loading (ζ’ = σ’tanϕc’).


• In drained loading ‘pore pressures’ can be measured from hydrostatic ground water conditions or flow nets(Hence effective stress parameters).

• With these 2 simple expressions itself it is ‘impossible’ to reach at complete solutions .

• Thus we need approximations For that we have 2 methods (1)Bound method (2)Limit equilibrium method.

• Theorems of Plastic Collapse• So how this approximations can be made? Definitely by ignoring some

conditions of equilibrium and compatibility and to make use of theorems of plastic collapse.

• It turns out by ignoring equilibrium condition we get upper bound for the collapse load.

• By Ignoring the compatibility condition we get Lower bound for the collapse load

ContinuedContinued

• If structure is loaded to upper bound it results in collapse where as it does not when loaded to lower bound.

• In between these bounds lie TRUE COLLAPSE LOAD.• The advantage of this method may be that the upper and lower

bounds are fairly close so that level of uncertainty is low.• Now we have to see the plastic theorems.We are not going for

proofs rather we quote some results.• For these theorems to be valid material must be plastic.• That is soil must be straining at a constant rate(which happens at

ultimate state) with an associate fllow rule(vector of plastic increment must be normal to failure envelope).

• At ultimate states soil behaves according to equations already given for drained and undrained cases(ζf =su, ζ’ = σ’tanϕc).


• In both drained and undrained cases elastic strains are zero.Hencetotal strain = Plastic strain.

• For undrained loading Failure envelope is horizontal. • Also Volumetric strain is zero hence vector of plastic strain,δεp is

perpendicular to failure envelope.• For drained loading, flow rule is associated the angle of dilation at

the critical state(ψc).

• At critical or ultimate state, soils in drained loading strains at a constant rate.(ψc = 0).

• This means soils in drained loading are not plastic since associative flow rule is not applicable at ‘Ultimate state’.


• But this is not a huge issue as it can be proven that upper bound of a material ‘ψc=ϕc’ is true.

• But in real practice based on experimental observations ψc = ϕc may be used for upper and lower bounds.

• The statements of bound theorems may be :• Upper Bound :If you take any compatible mechanism of slip

surfaces and consider an ‘increment of movement’ ,if you show work done by stresses in soil = work done by external loads ,thestructure must collapse and external load are upper bound to true collapse load.

• Lower Bound: If a set of stresses can be determined which are in equilibrium with external loads and it does not exceed strength of soil the structure cannot collapse and external loads are lower bound to true collapse load


• So ultimately in upper bound we have to satisfy conditions of compatibility and material properties(which govern the work done by stresses in soil).

• Here in Upper bound Nothing is said about Equilibrium.• In Lower bound Equillibrium(with external loads) must be satisfied

along with the material properties(which govern strength).• Here in Lower Bound Nothing is said about Compatibility.• Load obtained in Upper bound is called ‘unsafe’ and in Lower bound is

called ‘Safe’ (based on Collapsing conditions).• Compatible Mechanisms of Slip Surfaces:• To calculate an upper bound ‘mechanism of slip surfaces’ must satisfy

compatibility condition. These compatibility condition determine allowable shape of slip surfaces and its arrangement.


• On a slip surface on one side the material is ‘stationary’ and on the other side there is an increment of displacement ‘δw’.

• This increment (δw) occurs at an angle of dilation(ψ).• The length along the slip surface is a ‘constant’.• If this slip surface makes an angle ‘α’ & ‘β’ to major ppl planes we

already have α = β= 45+ψ/2.• From geometry dr/rdθ = tanψ and hence rB/rA = exp(∆θ tanψ).• Here ∆θ is the angle between rB and rA. When ψ>0 that is for

drained loading this is expn for a logarithmic spiral.• But for undrained loading when ψ = 0 we have rB/rA = exp(0) =1.• That is for ‘undrained loading’ We have a circular arc .• Similarly when rA is ‘infinite’ we have a straight line.

Continued..Continued..• Thus for ‘undrained loading’ ie for ψ=0 we have slip surfaces

‘circular arcs’ or ‘straight lines’.• For ‘Drained loading’ ie for ψ = ϕc’ we have slip surfaces as

‘logarithmic spirals’ or ‘straight lines’.• Another possibility is that radii intersect slip surfaces at 90-ψ making

radii also possible candidates for slip surfaces.• How we can ensure system is compatible?.By constructing

hodograph .

Work done by Internal stresses and Work done by Internal stresses and External LoadsExternal Loads

• In Upper bound we have to calculate ‘work done by internal stresses and external loads’ during increment of movement.

• It should be also noted that mechanism should be ‘compatible’.• Work done by a force = Magnitude of Force * Increment of

Displacement resolved in the direction of Force.• External loads can be (1) Concentrated loads below footings

(2)Distributed loads below embankments (3) Self weight of Soil.• External loads due to Concentrated forces are easy to determine and

are same for drained and undrained loading.• For the next two cases drained and undrained loading must be

considered separately.


• Consider the soil element with unit weight ‘γ’ which is acted upon by a total stress ‘p’ a concentrated load ‘F’ at the top surface where pore pressure is ‘u’.

• For an increment ‘δw’ in the direction of ‘F’ and ‘p’. We have the following work done for undrained loading. δE = Fδw + pAδw+γVδw

• For Drained loading water remains stationary and work is done by effective stresses δE = Fδw + (p-u)Aδw+(γ-γw)Vδw .For dry soil u=γw= 0.

• The Work done by Internal stresses is work done while plastic straining happens at slip surfaces.

• For the work done by internal stresses Drained and Undrained Loading must be taken separately.

• When dilation happens in drained loading it can be seen that ‘normal displacement(δn)’ happens against applied ‘normal effective stress(σn’)’ .

Continued..

• Thus work done by normal stress is negative. Thus for drained loading work done by effective stress is given by δW = ζ’Lδl –σn’Lδn.

• At the same time Volume of Slip plane ‘V’ = Ly.• If work done(δw) is presented in terms of strain energy δW =

V(ζ’δγ+σn’δεn) = V ζ’δγ(1 – tanΨc/tanФc’)• If we want to calculate work dissipated by internal stresses during

drained loading,we have to put Ψc=Фc δW= 0 !!!.• The result is an absolute Nonsense!!!• Bottom line flow rule for a frictional material cannot be associated .• Now consider undrained loading case. We have to calculate the

work done by total stresses ζ and σ. δW = ζLδw = su L δw.• It is understood that for undrained loading there is no effect for ‘σ’ on δW as there is no displacement normal to slip surface (δn).

Continued..

• Thus for upper bound calculations, for undrained loading this δW has to be evaluated for all slip planes.

• Now we will see simple upper bounds for different examples. Simple Upper Bounds for a Foundation • First of all let us derive upper bound solution for Bearing capacity for a

foundation subjected to undrained loading.• For this consider a foundation ‘B’ width with unit length perpendicular to

surface of paper (Area = B).• Once we increase the foundation load F, contact (bearing) pressure ‘q’↑,settlement ρ ↑.

• This increase will continue till we reach over burden load = Collapse Load (Fc) or the bearing capacity (qc).

• It is also assumed that foundations is too smooth and hence no shear stresses being developed.

• Let us Obtain the Solutions

Continued..

• We can obtain such upper bounds for foundations by (1) A simple mechanism (2) By 2 stress discontinuities (3) Using Slip fan and a stress fan.

• Consider the Simple Mechanism Case :

Continued..

• For the simple mechanism we consider ‘3 triangular wedges’ and ‘corresponding displacement diagram’.

• So Let us calculate increment of ‘work done’ by the ‘self weight forces’.

• It sums up to zero since Block B moves horizontally where A and C have equal and opposite displacements.

• Hence external Work is only due to External forces which is δE = Fu *δwf.

• Now Let us calculate internal stresses

Continued..

• Thus internal work (δW) = 6suBδwf• We know external Work (δE) = Fuδwf • To get upper bound Collapse Load we have to equate these

quantities δW = δE Fu = 6BSu. Discontinuous Equilibrium Stress States• To calculate a Lower Bound it is necessary to check ‘equilibrium’

state of stress.• At the same time we have to make sure that it does not exceed

failure criteria given by ζf=su or ζf = σ’tanФc’.• Along the ‘discontinuities’ sudden stress changes may happen but

conditions of equilibrium is met along that discontinuity.• Consider one such Discontinuity which splits a domain to ‘A’ and ‘B’.• The magnitude and direction of principal stresses are shown .• For Domain A it is ‘σ1a’ and for Domain B it is ‘σ1b’.

Continued..

• So..

• So there is a rotation in the ‘direction’ of principal stresses across the discontinuity. That is δθ =θb – θa.

• Mohr Circle for the total stresses are also shown above.On that Mohr circle Point ‘C’ may be noted.

• Point ‘C’ represents normal and shear stresses on the discontinuity.• If I draw a line through ‘C’ parallel to the discontinuity it will hit the

circles at Poles ‘Pa’ and ‘Pb’.• Once Poles are obtained we can see the direction of principal

planes are shown by broken lines.

Continued..

• We can see angles ‘2θa’ and ‘2θb’ in the Mohr circle.• These θa and θb are nothing but angles subtended by normal stresses σ1a and σ1b on the discontinuity.

• Now Let us see how such changes of stresses are happening across the discontinuity for (1) Undrained (2) Drained conditions.

• Let us consider ‘undrained’ loading First.

Continued..

• We know for undrained loading ζf = su.• From Mohr Circle AC = su Thus change of stress across the

discontinuity, δs = 2 su sinδθ.• Now it is curious to see that ‘change of total stress’ across the

discontinuity becomes just a function of rotation(δθ).• Now Let us see Drained Loading Case .

Continued..

• It is obvious that during this condition our Mohr circles at Domain A and B will hit failure envelope given by ζf = σf’ tanФc’.

• Angle ρ’ is the angle made by Point C (Discontinuity coordinates) with origin. (It defines ratio ζn’/σn’ on the discontinuity).

• Some geometrical extrapolations are needed for some derivations.Hence

Continued..

• The following derivation is not of much importance but the result is interesting.

• P = 90 – δθ . Also sin P = A’D’/ta’, sinρ’ = A’D’/sa’ sinρ’ =sinPsinФc’ sin (P+ρ’) = O’E’/sa’, sin(P – ρ’) = O’F’/sb’

• Thus finally we get sb’/sa’ = cos( δθ – ρ’)/cos( δθ +ρ’)• We know sinρ’ =sinP sin Фc’ . Thus ρ’ = f(P,Фc’) • Thus for drained case the change in effective stress across a

discontinuity is simply related to the direction of rotation (δθ) of principal stress.

• We have already got upper bound for Fu for a foundation as 6suB.• Now let us see the simple lower bounds for a foundation . Simple Lower Bounds for a Foundation:• For the derivation of simple lower bounds we need the concept of

discontinuities. Let us see how it is done .

Continued..

• Let us consider a foundation loading of ‘q1’ distributed along ‘B’.• Consider 2 vertical discontinuities ‘α’ and ‘β’ below the footing.• On either side of ‘α’ we have elements A and B and for ‘β’ we have

C and D.• We know shear stresses in horizontal and vertical planes are zero.

Continued..

• Vertical stresses on the elements ‘A’ and ‘C’ is given by σz =γz.• Vertical stresses on the elements ‘B’ and ‘D’ is σz = q1+γz.• We have seen total stress ‘Mohr Circles’ for 2 sets of elements on

either side of discontinuity ‘α’ and ‘β’.• On those circles ‘a’ and ‘b’ represent stresses along discontinuity ‘α’

and ‘β’.(It is imperative shear stress = 0 at the discontiuity).• We know total stress circles have a diameter ‘su’.• Thus change of total stress across the discontinuity (δs) =2su.• We also have derived for undrained loading δs = 2su sinδθ.Here for

discontinuity there is rotation in direction of 90.Thus δs = 2su.• Thus from geometry of Mohr circles • q1+γz = 4su + γz Fu = q1*B Fu = 4suB

Continued..

Upper and Lower Bound Solutions Using Fans• We already have seen a combination of straight and curved slip

surfaces to have a compatible mechanism.• This is rather called ‘Fan of slip surfaces’.• Mechanisms and Hodographs of Slip fans are shown below for 2

cases (1) Undrained Loading (2) Drained Loading.

Continued..

• From the undrained case : From figure rb= ra and δwb = δwa.• For the drained case : From figure rb = ra exp(θf tanΨ) and δwb = δwa exp(θf tanΨ). Θf if nothing but ‘fan angle’.

• We already have seen for undrained loading mostly we get circlesand drained we get logarithmic spirals.

• The increment of work done by internal stresses through fan = Work done on Circular Slip surface + Work done on radial slip surfaces.

• For this we use following diagram •

Continued..

• Thus δW = ∑su R(δwδθ) +∑su (Rδθ) δw = 2su R δw∑δθ.• Now this is integrated all along the slip fan that is from 0→θf.• That is δW = 2 su R δw ∫ dθ = 2 su R δw θf δW = 2 su R δw ∆θ.• Now consider the change of stress ‘δs’ across the fans for (1)

Undrained Loading (2) Drained Loading.• See the figure below:

• It is well sightable that rotation of major principal stress across the fan (∆θ) = Fan Angle (θf).

Continued..

• Now consider the Undrained loading. Outer most limits of fan aredefined by θa = θb =45 . As δθ→0 we have ds /dθ = 2su.

• Integrating from A→ B we get ∆s = 2 su ∆θ. • That is for undrained loading there is a change in total stress of

‘2suθf’ when we come from A→ B across the fan.

• Now consider drained loading. Outer most limits of fan is defined by θa = θb = 45 + Фc’/2

Continued..

• It is possible to get ds’ / s’ = 2 sinδθ sinρ’ /cos(δθ+ρ’).• As δθ→0 we have ρ’ = Фc’ ds’/dθ = 2s’tanФc’.• On integration from region ‘A’ to ‘B’ we have sb’/sa’ =exp(2tanФc’∆θ).• Thus for drained loading sb’/sa’ = exp(2tan Фc’θf).

• Thus it can be seen that for drained and undrained loading the stress change across the fan ( ∆s or sb’/sa’) is a function of fan angle ( θf).

Continued..

• Again it can be seen that for drained loading the ‘stress change’ is a function of critical angle(Фc’).

• But for undrained loading the ‘stress change’ is a function of undrained shear strength (su).

Bound Solutions for Bearing Capacity Using Fans• Earlier we obtained upper bound and lower bound solutions for the

bearing capacity using simple mechanism. Now we use fans.• Upper Bound with a Slip Fan:• Consider the following Compatible Mechanism and its corresponding

hodograph.

Continued..

• Here also work done by self weight forces sum to zero.• Thus external work will be only due to super imposed load (Fu).That

is δE = Fu δwf.• Now what about the work done for internal stresses ? We have

already derived it as δW = 2suRδwθf.----------(1)• Here in this case R = B/√2 ,θf =(1/2)π ,δwa = √2δwf. Substituting all

these quantities in (1) we get δW = π su Bδwf.• So we have seen internal work done around the fan. What about

those edges of the 2 triangles? For that use the table below.

Continued..

• Thus total internal work done by stresses is δW = (2+π)su B δwf.• We know δE = Fu δwf. = Fu =(2+π)B su. Lower Bound Solutions Using Slip Fans:

• Here we define fan regions in II and IV. Stress state is symmetric about the central axis.

• Equation σz= q1+γz is applicable for region III and σz = γz is applicable for region I and V.

• If I take Mohr circles for total stresses for elements at ‘A’ and ‘C’ ‘a’ and ‘c’represents stress states at discontinuity adjunct to layer 2.

Continued..

• From geometry fan angle θf =90 = π/2. Also we know for undrained loading ∆s =2 su ∆θ = 2 su θf. = ∆s = π su.

• From geometry of Mohr circles it is clear that q1+γz = (2+π)su +γz Hence Lower bound of Collapse load is

F1 = (2+π ) B su• If you check for stresses in region VI analysis is lengthy and

cumbersome.• But amazingly Lower and Upper bounds coincide to give an exact

solution for bearing capacity determination using fans .

Limit Equilibrium Method

• This is the most commonly used method used for analysis of stability of geotechnical structures.

• It involves 3 steps.• (1) You may draw an ‘arbitrary’ collapse mechanism of SLIP SURFA -

CES (it can contain straight lines, curves etc).• (2) Calculate the statical equilibrium of the components of the

mechanism. This is by resolving forces or moments and calculate strength mobilized in the soil.

• (3).You can examine statical equilibrium of other such mechanisms and Load which causes the most critical mechanism is called Limit equilibrium Load.

• Here also as we did for Bound methods we have to distinguish between drained and undrained conditions.

• These are just Block mechanisms!!!

Continued..

• The Limit equilibrium method is a combination of upper and lowerbound methods.

• In upper bound we ensured collapse occurs for a compatible mechanism by formation of slip surfaces.

• Limit equilibrium method borrows ‘slip surfaces’ and ‘collapse’ out of it but ‘need not meet all conditions of compatibility’.

• Equilibrium must be satisfied on forces ON blocks within the mechanism but local stress states INSIDE blocks are not investigated.

• There are NO FORMAL PROOF for this approach .But It holds good .

Simple Limit equilibrium Solutions• For illustrating how the method works we take two problems (1) An

infinite slope in drained condition (2) A shallow foundation under undrained condition.

Continued..

• Infinite Slope under Drained condition:• The problem is to find critical slope angle (ic) at which slope fails ?

• The most critical failure slip surface may be a parallel one at depth ‘z’ .(Common it is obvious ).

• Now consider the static equilibrium. For that consider forces in the block mechanism. Now for a drained infinite slope F1= F2 and pore pressure (u)=0.

• So T’ = ζ’L ,N’ = σ’L, W = γL z cos I

Continued..

• To check the static equilibrium considering ‘polygon of forces’ T’/N’ = ζ’/σ’ = tan ic.

• For drained condition ζ’= σ’tanФc’ ic = Фc’.• The same solution can be also obtained as an upper bound or lower

bound making it an exact solution. A Shallow Foundation Under Undrained Condition.

Continued..• The problem is to obtain Collapse load (Fc) and hence the ultimate

bearing capacity (qc) ? ‘qc’ = Fc/A.• Slip surface taken is a ‘circular’ surface with centre ‘O’ at the edge of

the foundation.• Now static equilibrium may be ensured by taking moments about

‘O’.• We know undrained shear strength of soil is ‘su’• Taking moments about O Fc * (1/2)B = su *B* Arc ST.• Arc ST = π B Fu = 2πB su.• For the same problem we got an upper bound of 6Bsu and lower

bound of 4Bsu using simple mechanism.• Using Fans we got an upper and lower bound of (2+π)B su ie

5.14Bsu.• That means the critical slip surface is highly conservative.We need

to modify it .

Continued..

• We assume a slip surface drawn at a height ‘h’ above the surface with centre ‘O’. Look at the figure below.

• If we take h/B =0.58 we get Fc = 5.5Bsu. That can be taken as critical slip or you may do more trials .

Coulomb Wedge Analysis :• Loads required to maintain stability of a retaining wall provides an

excellent illustration to principles of limit equilibrium.• The calculation was first developed by Coulomb ,in 1770 one of the

oldest engineering calculations used in practice.

Continued..

• First is the assumption of Slip surface. It is a straight line at some angle ‘α’ measured in clockwise direction to horizontal.

• Now we have to consider 2 cases (1) Undrained (2) Drained.• First of all for both the cases we consider smooth wall which means

there is no friction between the wall and soil.• So First case is an undrained smooth wall.

• Above we have shown the ‘polygon of forces’ We know their directions very well. Let us see whether we know their magnitude.

Continued..

• We know magnitude of ‘W’ and ‘T =su L ‘.Thus Pa and N remains as unknowns and T is the force resisting sliding.

• Also it should be noted that there is no relation between T and N in undrained loading.

• Thus with only 2 unknowns and all its direction known problem becomes statically determinate.

• To get limit equilibrium solution vary the ‘α’ and get the critical value of ‘Pa’. It is found that critical α =45.

• If you resolve the force polygon parallel to slip surface we get Pa = 1/2γH^2 – 2suH.

• Pa or active force is nothing but the minimum force required from the excavation side to keep it stable.

• That means put Pa =0 for this undrained condition to get critical height of vertical cut . Hence we get Hc = 4su/γ

Continued..

• Consider the second case now when soil is drained when pore pressures u=0.

• Here we have 3 unknowns (1) T’ (2) N’ (3) Pa’. But the best part is that T’ and N’ are related here.

• In the figure the resultant of T’ and N’ are shown which is at Фc’ to N’.• Now vary ‘α’ to get critical value of Pa which occurs at α = 45+Фc’/2.At

that value Pa = (1/2)γH2 tan2(45-Фc/2).• This is the same equation by Rankine in 1850 and he derived it

differently.• Now we understand Coloumb and Rankine were not dissimilar rather

same and different ways of same Limit equilibrium analysis for wedges.

Continued..

• Again the same drained analysis can be extended with other load combinations (1)External Loads (2)pore pressures (3)shear stresses between soil and walls.

• The forces and its polygon is shown below.

• Here we assume water table is same on both side of ‘cut’ so that there is no seepage.

• The pore pressure force (U) can be obtained by summing them overslip surface ( u= γH).T’ hence will be known = (N-U) tanФc’.

• Shear force between wall and soil is Tw’ = Pa’ tanδc’.Obviously0<δc<Фc’.

• So finally Pa’ and N’ are indeterminates and statically solvable.

Continued..• In Simple cases for drained and undrained wedge analysis where

load combinations are not there our principal planes are horizontal.• This is because wall friction is not considered.• We already found zero extension lines are at 45+Ψ/2 to major

principal plane.• If you know the stress ratio(ζ’/σ’) is tanρ’ then it was at 45+ρ’/2 to

major principal plane.• For undrained condition we know dilation angle(Ψ) =0 • Similarly for drained condition ρ’ = Фc’• Thus we can see zero extension lines are coinciding with critical

surfaces drawn from Limit equilibrium analysis.• Two concepts coincide and confirms !!Soil mechanics is beautiful • Definitely If load combinations are coming then we don’t have same

45 or 45+Фc/2 for respective cases.

Continued..

Simple Slip circle Analysis for undrained Loading• Already we have seen circular slip surface for undrained footing. Now

we are going to apply it for slopes.• So here we consider an undrained slope. Not surprisingly with our

experience in footings we consider a slip circle.

• Taking Moments about O, the foundation and slope are just stable when Wxw + Fxf –Pwxu = su Arc ABR.

• The attractive part is ‘N’ (normal force) don’t have effect as they have zero moments about O. You can deal with W and su by dividing geometry into slices.

Continued..

Slip circle method for drained Loading- Method of Slices• In the previous case ‘su’ was independent of normal forces and we

got rid of them easily by taking circular slip circle whose moments are zero about O.

• Consider a slope where draining happens to a toe drain and flow net is shown.

• Take moments about O like we did previously ;we get W x = R∫ζ’dl where ζ’ = (σ-u) tanФc’. See we have unwanted σ here.

• We got rid of that in undrained condition ,but not here. Thus we have a new method…The method of Slices !!.

Continued..

• So what is method of slices? Here we subdivide the mechanism into a number of slices, definitely after assuming slip surface.

• Now we check for statical equilibrium of individual slices and by summation the whole mechanism.

• Consider the slope divided in 4 slices and a typical slice FGHJ is shown below.

• So what about the forces? You have weight (W), normal and shear forces N and T at base FJ.

• There are also interslice forces F1 and F2.As they are not equal and opposite they have a resultant F at ‘a’ above base of slice at an angle ‘θ’.

Continued..

• This interslices are not slip surfaces and so nothing can be told about magnitude and direction of ‘F’.

• For the slice FGHJ we know magnitudes, direction and point of application for ‘W’ and ‘U’(U=ul, u will come from flow net and ‘l’ length of slip surface).

• We know direction and point of application of T and N but not sure about their magnitudes as they are related in drained condition T=(N-U) tanФc’. (we just know U and Фc’ here).

• Thus we have 5 unknowns (1) T (2) N (3)F (4) a (5)θ . We have 3 equations from statics and one equation T=(N-U) tanФc’.Thus it makes 4 equations.

• This means all slice problems are statically indeterminate which forces us to make atleast one simplifying assumption.

• The more ways of taking such assumptions the more the methods !!We go for the 2 commonest of them all.

Continued..

The Swedish Method of Slices (Fellenius,1927).(E=0,X=0)• See we have 5 unknowns T ,N ,F ,a ,θ. ‘F’ is causing all the problems

by bringing ‘a’ and ‘θ’ along. What is the simple way? Ignore F. Put F=0

• Thus T=W sinα and N= W cosα It is important to understand this ‘α’ is not a constant for entire slope but it is constant for a single slice.

• Hence for the whole slope ∑T = ∑(N – U) tanФc’.• Thus for this case ∑W sinα = ∑(W cosα – ul)tanФc’ Here it should be

seen u is the ‘average pore pressure’ at base. So do this for different slip circles and get the critical one the one which above eqn satisfies .

Continued..

The Bishop Routine Method (Bishop -1955) (X=0)• Here also we play with interslice force F.• We assume interslice forces are horizontal that is θ=0.That makes it

a statically determinate problem.

• Finally from the basic equation for drained loading ∑T = ∑(N-U)tanФc’ you get

Other limit equilibrium methods

• So far we have considered mechanisms to be either straight line or circular arc.

• There is no hard and fast rule for the shape of slip surface. We will discuss two other commonly used geometries.

• It can be seen that this geometry here has vertical failure surfaces The shear and normal stresses at the vertical failure surface confirms to our normal calculation of shear stresses.

Stability of Slopes.

• The geometry of slope is characterized by slope angle (i) and Slope height (H).

• The loads on the slope are due to self weight of soil and external loads. (For a vertical cut ‘i’ =90).

• To maintain slope, the slope material must be able to sustain shear stresses which is the reason why we don’t see a slope in water.

• During Slope excavation you may have a decrease in mean normal stresses.

• At the same time construction of embankment can cause the same to increase.

• In both excavation or embankment construction induced shear stresses ↑ as ‘H’ or ‘I’ ↑.

• The above statements say that slope construction loading may increase induced shear stresses irrespective of what happens fornormal stresses.

Continued..

• If the slope is too steep, it will result in failure.You continue reducing this height to particular height (Hc) and slope angle (ic) at which slip just stops.

• These are critical height and critical slope angle where FOS of slope is unity.

• Instability may be due to many reasons (1)By pore pressures (2)Influence of strong/weak layers (3)Due to fissures etc.

Continued…

Stress Changes in Slopes • It is important to understand Natural slopes gets enough time to get

drained where as Man made slopes particularly in clays will be in undrained condition.

• Let us consider the changes in total and effective stresses on a typical element on slip surface during an undrained slope excavation

• Consider an element on the slip surface which experiences σ and ζas normal and shear stresses.

• We connect a stand pipe to this element to measure the pore pressure.

Continued..

• Due to construction of slope we have an increase in shear stress.• At the same time we know method of construction of slope is by

excavation which implies ↓ in normal stress.• That is why we see total stress path A→B like that.• Now what about effective stress path A’-B’? We are discussing

undrained loading Const Volume Const water content.• That is why A’→B’ is having same water content ‘w’.• The stress path A’→B’ in (σ,ζ) space is a function of various soil

parameters ,its state, OCR etc.• So ‘B’ represents a stage immediately after construction. • So at ‘A’ start of construction we had steady state pore pressure

(u0).• Pore pressure immediately after construction that is at B is given by

‘ui’.

Continued..

• From figure it is clear that ui <u0. So here initial excess pore pressure ui’ is negative.

• That is why stand pipe has water level below the phreatic surface.• So now we are going to discuss in long term. What will happen to

these stresses and pore pressures in long term?.• Long term condition is represented by ‘C’.• As there is no loading or no change happens to material weight or

geometry total stress at B = total stress at C.• Now what about effective stress path B’→C’?• We know at B we had negative initial excess pore pressure ui’.It will

dissipate and rise.• The result is a reduction in mean normal effective stress.• At final state C’ we have steady state pore pressure (uc) and we

know as a result of reduction in mean normal stress we have a consolidation swelling

Continued..• Since we have our water level still there submerging the slope uc=

u0.• If the excavation was drained of water then uc will be different from

u0.• So when will our slope fail? If say at B’ effective stress path hits CSL

we can say slope fails immediately after construction.• If say at C’ effective stress path hits CSL we can say slope fails in

long term .• Position of effective stress points B’ and C’ wrt CSL gives a measure

of FOS of slope.• We can also see the effective stress path heads towards CSL with

time. It means FOS ↓ with time .• Thus permanent slopes must be designed drained.• Temporary slopes are designed undrained (In mean time we may

pray God that ESP never hits CSL .If so can result in failure)

Continued

• Now once slope fails there is change in angle and height (Infact the height reduces).

• As result total stress experienced by element in slip surface also changes.

• In the figure just after the excavation slope fails at B’ where undrained shear strength is ‘su’.

• Total stress mean while could not continue to ‘X’ corresponding to initial slope angle ‘ix’ due to failure.

• Corresponding to ‘su’ we get a slope geometry ‘ic’ and ‘Hc’ which we earlier called as critical quantities.

•

Continued..• Consider this figure where slope failure occurs at C’ ie in long term.

• The failure occurs at C’ where pore pressure is ‘uf’. • In TSP after failure at C slope angle ↓ slope height decreases which

result in a decrease shown in ‘ζ’ under constant normal stress.• That is why in TSP C→D is vertical.• Now what about ESP C’-D’ that is after failure?• With the decrease in TSP stress quantities decreases in ESP but

along the CSL.• Slope finally reaches a stable state when pore pressure ‘uc’ is final

steady state pore pressure at ‘ic’ and ‘Hc’.

Continued..

Influence of water on Stability of Slopes • Water has a significant effect on slope stability.• We have already seen slopes failing after completion of excavation

due to dissipation of negative excess pore pressures which result in consolidation swelling and softening of soil.

• Also slopes in river banks, lakes and trenches fail if external water level is quickly lowered.

• It is common sight of slope failures post heavy rainfall.

• If we see the figure free water is inducing total stress ‘σw’ on a vertical slope which in turn helps to retain the slope.

• It is common practice to use such technique when excavations happen for construction of retaining walls and piles. There we use bentonnite slurry whose unit weight >water

Continued..• It is common sense to interpret why slope failure happens after

rainfall.• During rainfall pore pressure ↑ resulting in low effective stresses.• An ↑ in effective stress causes a decrease in shear strength. Slope

fails .• To calculate pore pressures in a slope under steady state conditions

we have to draw a flownet

• At equi potential line pore pressures or level in stand pipe rises to same height. That is hw = mz (cosi)^2

• For dry soil m= 0 and if phreatic surface is at ground level m =1.

Continued..

Choice of Strength Parameters and Factor of Safety• There are a different types of possible criteria for defining soil

strength.• Hence it is important to distinguish between undrained strength (su)

and the drained or effective stress strength.• At the same time it is also important to distinguish between peak,

ultimate or critical strength and residual strength.• The choice between undrained and drained strength is relatively

easy.• For temporary slopes with fine grained soils and low permeability

choose ‘su’ and carry out analysis in total stresses.• For permanent slopes critical conditions occur at the end of

consolidation swelling and analysis has to be effective.• Now the choice should be between peak, critical and residual

strength after applying FOS to them.

Continued..• Careful geotechnical investigation will reveal whether there is any trace

of preexisting land slides.• In such a case strength should be based on residual friction angle(Фr’)

as soil may have already reached that state.• If such history is not there the choice is between peak or ultimate

strength.

• It is well explicit from the figure that peak strength (which is associated with dilation occurs at small displacements(1% or 1mm).

• The ultimate strength occurs at strains or displacements(10% or 10mm).• These ground movement criteria is mainly used for design of

foundations and retaining walls.• So what is that strength we should use for slope designs?

Continued..

• If you consider a steep slope it will come to rest when geometry of slope is in equilibrium with critical state strength with FOS =1.

• Thus for slopes we choose the critical strength parameters su or Фc’.• Normally slope designs based on critical strength gives safe designs.• So we got the available strength ‘su’ or Фc’. Now an FOS should be applied to

this available strength to get strength mobilized(ζa or ζa’)ζa = su / Fs = sua ; ζa’ = σ’ (tanФc’/Fs) = σ’tanФa’

• Why such an FOS is required ?• (1) To reduce the soil strength so that slope is in equilibrium with lower

strengths.• (2)To take account of uncertainties( Weight of soil, soil strengths,pore pressure

conditions).• (3) The greatest uncertainty is determination of steady state pore pressure in

drained analysis .• (4) Also problem lies in the assumption of constant volume in undrained

analysis.• (5) So in such worst combination FOS > 1 will be adequate.

Continued..

Stability of Infinite Slopes • So we have chosen our parameters used for slope designs to be

critical parameters.• The slip surface may shallow slip surface parallel to the slope.• The stability for an infinite slope problem can be solved by UB or LE

approach• We consider 3 cases (1) Undrained Loading (2) Drained Loading

with no seepage (3) Drained Loading with steady state seepage. Undrained Loading ( Using UB and LB approach)• Consider an infinite slope with an angle ‘iu’ as upper bound .• The plastic collapse consist of slip surface through soil at rock level.

Continued..

• First of all consider the external work done (1) F1 and F2 are equal and opposite hence having no effect (2) Effect is due to self weight.

• W = γHl cos iu Displacement in direction of W,δv = δw sin iu.• Thus external work (δE ) = γHl cos iu δw sin iu.• Now what about internal work done (δW )?• It is internal shear force ( sul) * displacement in that direction (δw) =

su lδw.• δE = δW sin iu cos iu = (su/γH) • Now consider LB approach for the same problem ie equilibrium

conditions.•

Continued..

• For the lower bound angle ‘i1’ W = γHl cos i1. • Thus resolving normal and along the slope σs = γH (cos i1)^2 and ζs

= γH sini1cos i1.• If we consider Point ‘b’ on the Mohr circle we get σ = γH (cos i1)^2

and ζ = γH sini1cos i1.• That means ‘b’ corresponds to plane of (σs , ζs ).Thus draw a line

parallel to slip surface it intersects Mohr circle at Pole P.• Thus if I take ‘2i1’ from ‘b’ I get stresses corresponding to horizontal

surface which at Mohr circle is ‘a’.• By simple geometry tan i1 = ζs /σs = su/(γH (cosi1)^2) • Thus UB and LB coincide to get an exact solution.• That is for an undrained infinite slope critical angle

Continued..

Drained Loading – No Seepage • A straight slip surface is assumed whose angle with horizontal is an

upper bound ‘iu’.• As we know the direction of displacement(δw) with the slip surface

will be at dilation angle (Ψ).• Now what is the external work? It is only due to self weight and not

due to forces F1 and F2(As they are equal and opposite).• Work done by external forces or self weight(δE) = Wδv = (γV) δv.• For drained loading work done by internal stresses for an increment

of plastic collapse (δW) = 0.

Continued..

• Also we know V = zl cos iu . We know δE = δW That means δv γV = 0.

• Since Volume, V is non zero we have δv =0.If we put δv = 0 in hodograph shown in previous slide we get ‘iu’ = Фc’.

• Let us try LB approach now. Here we consider equilibrium of forces and may be the Mohr circles.

• Resolving Normal and parallel with base AB the normal and shear forces N and T are

• For dry soil where pore pressures are zero total and effective stresses are equal at AB

• Thus we have .We know limiting condition is • Hence lower bound slope angle which implies lower and

upper bound have similar values ic = Фc’

Continued..• Drained Loading –Steady state Seepage • Here we are discussing a condition where flow occurs downhill

parallel to slope.• This induces seepage stresses on the slope making it less stable

and hence ic < Фc’.• Results here will be obtained by LE approach as bound approaches

are tedious.

Continued..

• Forces on slip surfaces are T’,N’,U and N. T’ = ζ’l, N = σl , N’ = σ’l and U = ul.

• From force polygon T’ = N tan ic = (N-U) tan Фc’. From this critical slope angle ‘ic’

• For steady state seepage pore pressure at a depth ‘z’ is u =γw mz ( cos i)^2

• Thus for dry soil m= 0 it reduces to the ic = Фc’ case • If phreatic line is at surface m =1.Also we know γ = 1/2γw Thus

gives

• Bottom Line we know for undrained slope critical angle is a function of depth ‘H’ below slip surface. For drained case for both dry and submerged slope ic = Фc’ .For steady state seepage ic < Фc’

Continued..

Stability of Vertical Cuts • We know vertical cuts can be only made in undrained conditions

where negative pore pressures generate positive effective stresses.• We will analyze the cut stability by UB and LB methods. For UB you

need a compatible slip mechanism and is shown below.

• Now going for work energy principle. External work is only due to self weight ‘W’ Thus δE = W δv.

• ‘δv’ = (1/√2) δw ; W = (1/2) Hu^2 δE = (1/√2)δwγ * (1/2) Hu^2.• Work done by internal forces is Shear force * slip length = (su L)*

(δw) δW = (su√2 Hu)*δw

Continued..

• Equating δW = δE , we get Hu = 4su/γ.• For LB analysis We take Mohr circles of two stress elements ‘A’ and

‘B’.

• Consider element A alone where σh =0 and σz = γz.• Maximum shear strength of material in undrained condition is ‘su’

which is given by radius of Mohr circle.• Thus γH1 = 2su Lower bound Height of cut(H1) = (2su/γ)• Here we can see a UB and LB are separated appreciably. So in

design we normally use Hc = 3.8su /γ

Continued..

• If it is filled with water we have Hc = (3.8 su)/(γ – γw).• Hence once you fill excavation with water you can raise the height of

cut by twice the amount.• So basically we use LE approach for slope stability.• For undrained loading the problem is statically determinate and

solution is simple.• For drained loading problem is statically indeterminate and problem

involves method of slices (Bishop, Janbu. Morgenstern and Price etc).

• For slopes with simple geometries standard solutions are available using non dimensional charts and tables.

• People call them stability numbers .Actually what are they ? .• The basic fact is that these numbers should be non – dimensional.

How to get them ?

Continued..

• Consider 2 cases for deriving our stability numbers (1) Undrained loading (2) Drained loading

• For undrained infinite slope we have• Thus we can write where Ns is a stability number

dependent on geometry of slope.• This concept has been extended by Taylor (1948) for any arbitrary

slope (from infinite slope .The concept is as above).

Continued..

• Now we go for second case. For drained loading we are going to deduce stability numbers.( This includes steady state seepage).

• If I substitute allowable friction angle Фa’ for Фc’ we have

• Thus we have

• This can be rewritten as where m and n depend ongeometry of slope and Ф’.

• ‘ru’ = u/σz. In practical cases ‘ru’ = 1/3.

Continued..• Now from our understanding How simple excavations behave ? • They behave like this

Earth pressures and stability of Retaining walls

• A retaining structure is used for supporting slopes or vertical cuts when space is not available to form adequate slope.

• A retaining structure may be propped sheet pile wall, cantilever wall, gravity wall etc.

• When excavations are made props are used to support such retaining structures.

• During excavation or filling earth pressures develop causing wall to move from active to passive side.

• Development of earth pressure with its corresponding displacement is an interesting food for thought.

Continued..

• So what does this diagram mean? The force P just helps to maintain the wall in equilibrium.

• If I increase my ‘P’ wall will move to passive side increasing ‘δp’. As a result horizontal stresses ↑.

• After some time horizontal stresses(σh) reach a limiting value ‘σp’ called passive earth pressure.

• If I decrease the same ‘P’ wall will move to active side increasing the magnitude of ‘δa’ and limiting value of ‘σh’ ie ‘σa’ is called active earth pressure.

• The horizontal stress developed when there is no wall movement is called earth pressure at rest (Corresponds to K0).

• Design of retaining wall involves calculation of active and passive earth pressures, calculation of depth of embedment, Loads on props etc.

• Some of recently encountered retaining structures are shown below:

Continued..

• Like …

• Permanent retaining walls are seen at highway cuttings, bridge abutments etc.

• Temporary structures are used in excavations.• Different types of failures can occur for retaining structures.

Continued..

• It can be like wall remains intact with soil fails.

• Gravity walls can slide, overturn or fail by bearing.

• Base of excavation can fail by erosion and piping or slip circle failure can happen.

Continued..

• Or even structural elements in wall may fail.

Stress changes in soil near retaining walls.• To understand whether undrained or long term drained case in

critical it is necessary to draw TSP and ESP of soil near retaining walls.

• It is also important in this case to separate between loading due to excavation and loading due to filling.

• What ever happens to normal stress loading is assumed to have anincrease of shear stress.

• Now we say first case (1) Retaining wall loaded by excavation

Continued…

• Consider 2 slip surfaces for this retaining wall creating 2 elements one on passive and other on active side.

• Now what will happen to normal total stress and shear total stress.• Due to excavation normal total stress ↓ and due to ‘loading by

excavation’ shear total stress ↑.• Now what about the ESP? A’B’.It will correspond to undrained

loading (Characteristics will depend on soil, OCR etc)• Pore pressure immediately after construction (ui) < final steady

state pore pressure (uc).• Thus initial excess pore pressure is negative. Stand pipe level is

lowered as you can see

Continued..

• Now as the time goes on pore pressure ↑ ESP moves away from TSP at B hence .(TSP at B and C remains same as no change in total stress happens).

• B represents short term stage and C long term.• B’C’ hence corresponds to consolidation swelling where finally at C’

we have steady state pore pressure after swelling (uc).• The time when ESP hits CSL we say excavation failled and distance

of ESP from CSL is measure of margin of safety.• This means FOS for retaining wall supporting excavation ↓ with time.• Now consider the second case (2) Wall embedded in soil retaining

coarse grained fill.

Continued..

• As it is again ‘loading by filling’ shear total stress ↑ and because of the same reason normal total stress also ↑.

• In this case ui is greater than final steady state pore pressure and hence initial excess pore pressure is positive.

• Thus you can see rised level in stand pipe.• This pore pressure which is seen immediately after construction ie at

B has to dissipate with time and hence ESP B’C’ moves in positive X direction.

• As ESP moves away from CSL FOS for a wall retaining fill always ↑with time.

• So bottom line If you consider long term behaviour of walls supporting excavation and fill one becomes safe due to consolidation swelling while other becomes safe and strong due to consolidation .

Continued..

Influence of water in Retaining walls.• Water can influence retaining wall in different ways as shown below.

• First one is a coffer dam wall embedded in soil and retaining water causing total stress on wall Pw = 1/2(γw)(Hw)^2.

• In the second case there is a prop and water total stress acting on wall where prop will prevent wall from rotation. (TS = Pw + Pa).

• In the third case we have wall retaining coarse soil in drainedcondition. Pore pressure not only ↑ total stress on wall but also it reduces strength of slip surface reducing effective stresses.

Continued..

• So how we can manage it ? Which is shown in the 4th case.Providea drain at toe.There is no pore pressure action on wall. TS↓ and effective stress along slip surface ↑ which strengthens the slip surface .

• Now see the final case :Due to water piping and erosion can happen when exit hydraulic gradient becomes close to unity.

• Now as an engineer it is important to calculate earth pressures.Weconsider 2 cases (1) Drained (2) Undrained

Continued..

Calculation of Earth pressures – Drained Loading:• What is importance of CSSM in retaining structures?• When retaining wall moves horizontal stress changes and they

reach limiting active and passive pressures.• At that time soil is assumed to reach Critical state and hence use

critical state parameters.• From LE solution for wall retaining dry soil ;from the mechanism of

polygon of forces we have• Thus Ka = For the same case from LE it is possible to

show passive pressure Pp = (1/2)Kp γ H^2 where Kp =• This is for a simple problem ie a smooth vertical wall with a level

ground surface. • But there can be more complicated cases like back of wall and

retaining surface both inclined with wall being rough.• For such a rough wall shear stress ζs’ is induced along wall where ζs’=σn’ tan δc’(δc’=critical angle of wall friction)

Continued..

• It is very much clear that 0< δc’< Фc’. Usually δc’ = (2/3)Фc’.• Ka and Kp for such complicated cases may be obtained from

published charts for various combinations of Фc’,δc’,α and β.

• Now Let us do the second case ie calculation of earth pressures in undrained loading.

• For undrained loading LE solution from Coulomb wedge analysis already gave us solution as

• Assumin stress increase linearly with depth we can easily state the following , . We can write in generalized terms as

•

Continued..• Kau and Kpu are earth pressure coefficients for undrained loading .T

hus for a smooth vertical wall we have Kau =2 and Kpu =2.• For more complicated cases of rough walls with shear stress ‘sw’ we

have tables are charts available.• Thus from ,when σa<0 tension happens and from this

critical depth of tension crack is 2su/γ.The critical depth of a water filled crack will be (2su/(γ – γw)).

• So by poring some water into the crack we can reduce depth of tension crack.

• Now we want to make Hc=0 .From the stress should be 2su.That means if we load our tension crack with a surcharge (q > 2su) tension crack closes.

Continued..

Overall stability : • Forces and moments acting from all pressures must be in

equilibrium.

• In most of retaining walls unknown variable (depth of embedment) is increased till we achieve suitable margin of safety.

• In retaining structures we consider at point of collapse which means that horizontal stresses everywhere are full active and passive earth pressures. There are different approaches for different retaining structures. Let us see them.

• (1) Anchored or Propped Walls.• Here we have active and passive pressures on either side which are

not in equilibrium. So you need a prop ‘P’.

Continued..

• ‘Pa’ will be obtained in terms of unknown ‘d’ similarly Pp can be also obtained in terms of ‘d’.

• ‘za’ and ‘zp’ are also obtained in terms of ‘d’• To solve ‘d’ take moments about ‘P’ point as Pa za = Pp zp.• We get d from here and get numeric values of Pa and Pp.• Difference of them will give required prop force (P) to maintain the

wall in equilibrium (P = Pa – Pp).

• In the figure toe of wall rotates .This is called free earth mechanism.• If you increase ‘d’ there will be no such translation and is called fixed

earth support condition.

Continued..

• Now there will be cases when we cannot give prop ‘P’. Such cases give rise to our second case ,the cantilever retaining walls .

• (2) Cantilever retaining walls: Such stiff cantilever wall rotates about an unknown point at depth ‘d’ to satisfy equilibrium.

• So how this ‘d’ can be obtained?. For that we place a fictitious force Q at ‘d’.

• Now to get rid of Q we take moments about Q for Pa and Pp.• We get Pa ha = Pp hp (All Pa, ha, Pp ,hp are in terms of ‘d’ and

hence get d from here).• In order to mobilize the pressure below ‘d’ it is a common practice to

increase the depth by 20%. These walls just behave like beam carrying transverse loads of which SFD and BMDs are easy.

Continued..

• Another method is to resist the forces by self weight of wall which is used in concept of Gravity walls.

• They fail by sliding, overturning ,bearing etc.

• To estimate failure by sliding we need sliding resistance T.• For undrained case T= sw B ; sw = undrained shear strength

between soil and wall.• For drained case T = (W-U) tan δc’.• Up to now we considered failures using critical parameters which

will result in huge displacements. This demands need of FOS.

Bearing capacity and settlement of shallow foundations

• The main criteria for design of foundation is that settlement should be of limited amounts.

• We already know strength and stiffness of homogenous soil ↑ with depth as there is an ↑ of mean effective stress with depth.

• That is advantage we take by using deep foundation. The term ‘deep’ is quite tentative in civil engineering hence following rule ofthumb may be used.

Continued..

• Another advantage of using deep foundation is that you can take advantage of side resistance.

• So what about the bearing pressure ‘q’? ‘q’ = (F + W)/ A. The bearing stress is nothing but the ‘contact stress at founding level’.

• W = γc AD ;where γc is slightly greater than γ of soil. In some cases like basement it can be even zero.

• Total stress at D out side foundation is ‘p0’ = γD. Thus net bearing pressure (qn) = q – p0.

• For foundations it all about ‘qn’.If qn > 0 foundation will settle otherwise which it will heave.

• That is to arrest settlements we can make qn =0 which is concept of floating foundations

Foundation Behaviour:• Consider a shallow foundation with a bearing pressure ‘q’ which

results in a settlement ‘ρ’.

Continued..

• So..

• If foundation is rigid settlement ‘ρ’ will be uniform and bearing pressures wlll vary.

• If foundation is flexible bearing pressures will be uniform but settlements will vary.

• It is very easy to understand the fact that pattern of settlements will be dependent very much on loading ie drained or undrained.

• As we ↑ the bearing pressure settlements ↑ and it fails at ‘qc’.• The bearing pressure at failure is the gross not the net bearing

pressure.• You don’t want bearing pressure to reach ‘qc’ hence limit it to some

‘qa’ after applying some FOS on bearing capacity.

Continued..

• So let us come back to our drained and undrained loading and their corresponding settlements.

• During drained loading if you ↑ q settlement will increase in parallel reaching a final settlement ‘ρd’ at ‘qa’.

• Now consider the loading happening so suddenly ie in undrained condition. At the point where load =qa some settlement ‘ρi’ happens which is immediate undrained settlement.

Continued..

Stress Changes in Foundations:• In foundations loading happens which increase total shear and

normal stresses. This give TSP , A→B.• At the same time settlement(ρi) happens in undrained condition and

hence water content remains constant along A→B.

• Now we know ESP A’→B’ depends on OCR, soil etc.• We know just after construction pore pressure ,ui > u 0 thus initial

excess pore pressure ui’ is positive.• At A before all started pore pressure is steady (u0) at B we have ui’.• We know after B consolidation happens u↓ ESP moves in positive

X direction.

Continued..

• As ESP moves away from CSL FOS always increases for a foundation which is consolidating.

• Bottom line when consolidation swelling happens Less FOS and vice versa.

Bearing Capacity for Shallow Foundations:• Bearing capacity for foundations can be obtained from LE or UB

approaches. There can be 2 cases (1) Undrained (2) Drained.• For Undrained condition BC is obtained as qc = su Nc +p0 .From

this we get Fc + W = su Nc B + γDB.• This Nc is called ‘bearing capacity factor’. It can be recalled that we

got Nc =2 + π using bound methods for a long foundations.• Skempton did work on undrained BC for different LE solutions and

got

Continued..

• Now let us see the second condition (2) Drained BC.• It is given by This will deduce •• Nγ and Nq are bearing capacity factors. The above expression is also

coming from LE solution of wedge spiral and pressure zones.• In the expression shown above water table is assumed at surface and

hence γw is deducted from γ.• This Nc derived by Skempton and Nγ and Nq derived from Terzaghi can

be used for BC calculations.• See If I replace soil with water in the above equations either put su=0 or Ф’ =0 ,we get Fc + W = γw B D …What is this ?? Archimedes principle…

Continued..

How to choose soil strength and Load factor for foundations ?• It is notable that we did not use critical state parameters for arriving

BC of foundations. Why is it so?• It is because like retaining walls we are not concerned about

ultimate stability.• For foundations main criteria for design is magnitude of settlements.• See we can see the problem in this way.• I take the same soil. One at wet side of critical (loose or lightly over

consolidated) and other on dry side.• We know both of them have same critical strength depending on Фc’.

• There is a catch here .But their shear stiffness are not same .

Continued..

• The sample on dry side even has got peak .• But we know both soils have same Фc’ which have same value of

qc.• But their ultimate BC will be different.• If you think from settlement perspective for the same allowable

settlement (ρa) we have different allowable pressures ‘qd’ and ‘qw’.• So we need to apply different FOS on ‘qc’ for these different cases.• It is all confusing…• So FOS in bearing capacity is not something which you apply to

account for uncertainties of soil.• It is just a factor to reduce the bearing pressure to limit settlements..• It is the Load Factor not FOS.• Now to which bearing pressure it has to be applied ? To that bearing

pressure which causes settlement or heave. That is net bearing pressure (qn).

Continued..

• So ideally that means we need not apply any load factor for floating foundations…Normally we take 2 to 3.

Foundations on Sand • So we saw how dry side of critical and wet side affects allowable

pressure ‘qa’. (we had qw and qd for same ρa). • That means relative density has got to do something with ‘qa’.• What is method to find relative density? Of course any probing test

will give you and indication (SPT ,CPT etc).• Terzaghi gave one such relation qa = 10N kpa.• The ρa for this ‘qa’ will be 25mm(1inch).Below this small values of

settlement you can well see q-ρ curve is linear.• What is advantage of having linear curves? Half the load you can

halve the settlement.

Continued..• Foundations on Elastic Soil• Soil in reality has been non linear neither it is elastic. This means

there will be limitations to what ever deduced in this method.• It has been found that stresses developed in this way could be close

to actual field stresses.• This case hardly applies for settlements.• Consider a load ‘δQ’ applied at the surface. This results in vertical

stress ‘δσz’ and settlement ‘δp’ at a point (R, z).

• This gives a singularity at z =R =0 giving infinite stresses and infinite settlements.

Continued..

• Poulos and Davis (1974) gives a lot of elastic solutions and onesuch solution is shown below.

• This eliminates singularity in the previous case and is strictly for circular or rectangular foundations on elastic soil.

• A vertical stress ‘δσz’ causes a settlement ‘δp’ at depth ‘z’.

• It is interesting from both (Poulos and Bouissenesq) that vertical stress at some depth is independent of material properties.

• For settlements we see they are dependent on materials properties( E and ν). We expect that .

Continued..

• So now you need to estimate settlements using elastic approach.• The questions zeroes down to the matter of choosing material

parameters (E and ν). How to choose them ?• So as usual we will have different E and ν for different cases (1) Drained

(2) Undrained.• For drained loading we choose parameters E’ and ν’ corresponding to

effective stresses.• For undrained loading we choose parameters Eu and νu corresponding

to total stresses.• It is obvious that ν =0.5 for undrained/constant volume loading.• Again we are estimating settlements for an elastic soil. That means our

shear and volumetric effects will be decoupled.• That is G’ = Gu .For an elastic material G = E/2(1+ν).• This means but vu = 0.5.

Continued..

• That means finally • That means if we know drained parameters E’ and ν’ we can find

Eu.• So if we use our settlement equation for circular/rectangular

foundation for elastic soil we will get

• That means for a typical drained ν’ =0.25 we get ρu = 0.67 ρd. • Now we know how to estimate consolidation settlement .We will see

how to estimate radial consolidation settlement.• We know complete consolidation will happen when Tr or Tv → 1.0

or when Ut =100% (1)

Piled Foundations

• We will not discuss pile foundations in detail rather we will review some concepts.

• We know applied load Q = Qs + Qb.• In all pile analysis we will never take the weight of pile in calculation

as it is assumed that weight of pile = weight of soil displaced.• After all these forces are very small when compared to applied

loads.• TSP and ESP of pile group are similar to that of a shallow

foundation.• We know end bearing resistance of pile Qb = qb Ab.• The mechanism of slip surface at the tip of pile will be as below.

Continued..

• Now how to get this ‘qb’? For undrained loading qb = su Nc (Skempton say for piles adopt Nc =9).

• For drained loading qb = σz’Nq. This ‘σz’ is vertical stress at the pile tip.

• The mechanism at the tip failure is different for pile which makes bearing capacity factors for a shallow and pile different.

• Hence we have a different ‘Nq’ here adopted from Berezantev’s curves. (See the previous slide)

• So to use this curve we need Ф’.Which Ф’ to use ?.• For a bored cast in situ pile stress relief happens and for driven pile

high straining happens. So logically we should use Фc’.• But in practice it leaves us with over conservative designs. So use Фp’ for bearing capacity factor Nq from Berezantev’s curves.

• We are done with end bearing. What about shaft friction?

Continued..

• If ζs is the shear stress mobilized between pile and soil with D being diameter of pile we get

• Like ‘qb’ in the previous case here ‘ζs’ has to determined. It depends on type of loading (1)Drained (2) Undrained.

• For undrained or piles in clay ζs = α su with 0 < α < 1 .Typical value for α may be 0.5.

• For drained loading ζs = σh’ tanδ = Kσz’tanδ.’• K in this expression is nothing but ratio of horizontal to vertical

stress.• For a pile Ka < K < Kp and for rough pile Фr’ < δ’ < Фp’.• These parameters K and δ gets much influenced by pile installation.• Once you drive a pile ,horizontal stresses ↑ and hence shaft friction.• That is why we say driven piles have more capacity .

Continued..

• Now let us just glance what happens in group action of piles ?.• Due to interaction between neighboring piles capacity of each pile in

group will get reduced.• If I have n piles closely spaced If I give ‘F’ total load ideally for

individual pile load carried will be F/n.• But due to the earlier stated pile interaction capacity of individual

pile will get reduced by a factor ‘ή.• That means only ήQ will be available where we expect Q due to

interaction effects. Thus F = n ή Q.

Geotechnical Centrifuge Modeling

• We wonder if tunnels can be modeled retaining walls may be modeled to just confirm our theoretical studies.

• We have alternative in geotechnical engineering with out making full scale models. We just enumerate concepts.

• Simply if you want to simulate stress at some depth just take some sample and accelerate in centrifuge in vertical direction.

• By intuition this will solve our problem like hell .• We establish similarity between model and prototype by establishing

scaling laws and by tweaking dimensional analysis.• Remember Reynolds number? It is a dimensionless number which

will explain a phenomena ‘flow’.• So if we get such a dimensionless group for all problems,it is just the

matter of making sure that both for your model and protype they match .

Continued..

• We take simple 3 or 4 cases to illustrate the same.• You may want to simulate stress at some high depth ‘zp’.How to do

in a model? The answer Use a centrifuge

• For the prototype σp = γzp = (gρ) zp where g is mass density of soil.

• Now what about the model ? I accelerate my centrifuge vertically at ‘ng’. I need same σp to be there. So change ‘zm = zp/n’. Thus what about stress in model σm = (ng) ρ zm = (ngρzp)/n

• That is σm = σp which is our goal.

Continued..

• Let us see a undrained slope stability problem.• For the prototype slope with height ‘Hp’ slope angle ‘i’ stability

depends on undrained shear strength su.• We know some dimensionless number the stability number related

with slope. Thus Ns = (γH)/su = (gρH)/su.• Now to simulate the slope failure phenomenon our stability number

for both model and prototype should be same.

• If we scale the height of model Hm = Hp/n and accelerate at ‘ng’what will be Ns for model? = (ng) ρ Hm /su = (ng) ρ Hp/(n su) = Ns for the prototype

Continued..

• Now we solve a consolidation problem.• Here our non dimensional factor will be Tv .So Tv for the model and

prototype should be same.• Thus

• We have to maintain cv same in both model and prototype. We scale down the thickness of model as Hm = Hp /n.

• Thus for such a model • For a scaling factor n =100 tm of 1 hour will represent 1 year

prototype consolidation time as • Geotechnical centrifuges can be of different types capacity etc and

are represented by pay load and acceleration (tonnes).• The next slide shows some details of centrifuge. Capacity of

centrifuge is product of acceleration and payload.

Continued..

• So..

Thanks

Jishnu R B

Geotechnical EngineerHalcrow Consulting India Pvt LtdIndia Phone +918527059628

Critical State Soil Mechanics ---- By Jishnu R B

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Transcript of Critical State Soil Mechanics ---- By Jishnu R B