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    Soil MechanicsProf. B.V.S. Viswanathan

    Department of Civil EngineeringIndian Institute of Technolog! Bom"a

    #ecture $ %&

    Shear Strength of Soils#ecture 'o.(

    Students we have gradually come to the very last lecture on the shear strength of soils. We haveseen 6 different aspects connected to the shear strength of soils over the last 6 lectures and todayis the concluding lecture on whatever items are still remaining with respect to shear strength ofsoils which a typical student should know. As usual lets take a quick look at whatever we hadpassed through in the last lecture.

    (Refer Slide ime! "!#$ min%

    &n the last lecture we saw quiet in detail what are the drained test' what are undrained test othdirect shear and tria)ial shear and where they are used and we even saw some numericale)amples. We also saw that it is possile to look at the topic of shear strength in a totallydifferent way in terms of the soil. hat is how does cohesionless soil ehave under shear andhow cohesive soil ehaves under shear. We saw in particular the application of the differenttypes of drained and undrained test with a typical e)ample of the earth dam. *uring the course of

    life of an earth dam there are several stages it passes through starting from construction toreservoir filling and drawdown to long term staility. And under each conditions the shearparameters which must e used in order to test the safety of the earth dam will vary and the typeof that will give the appropriate parameters will also vary.+or e)ample the ,* test which we saw was good for a slow construction and also for long termstaility where in consolidation would have already taken place that mean drainage would haveoccurred' shear stress also does not produce any pore pressure.

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    (Refer Slide ime! -!-# min%

    n the other hand suppose a dam has already een filled' the reservoir is full and the reservoir isemptied somewhat suddenly rapidly relatively for certain purpose. hen the test that is applicaleand the corresponding parameter would e the consolidated undrained test ecause the earth damhas already undergone the consolidation during its life time filling up the reservoir. /ut nowsudden depreciation of reservoir produces shear stresses under undrained condition. And ifsuppose we are uilding a dam very rapidly then pore pressures get developed even duringconstruction they do not have time to dissipate and therefore the staility of the dam even duringconstruction has to e kept in mind and that short term staility is est determined underundrained conditions using parameters that derived from a 00 test.

    (Refer Slide ime! 1$!"6 min%

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    So if you take construction' if it is slow and pore pressures are allowed to dissipate there isconsolidation in drainage' s is equal to c dash plus sigma dash tan phi dash where all theparameters involved are drained parameters and effective parameters. &f the construction is onthe other hand rapid' pore pressures are not allowed to dissipate' the shear strength equationecomes s is equal to cuplus sigma tan phi u where c u and phiu corresponds to undrained

    conditions. &f drawdown is eing considered the parameters which will e applicale will e c cuand phicuand the shear strength equation would e s is equal to ccu plus sigma tan phicu. 0nderlong term conditions once again drained parameters will apply and s is equal to c dash plussigma dash tan phi dash. here are a numer of factors which affect shear strength' we mustknow them and their influence and so we spent quiet sometime in the last lecture to understandthe impact of many factors which affect the shear strength.2ere is simply a list which & am reproducing to remind what are the factors which influenceshear strength. he first three factors shape and gradation of particles' relative density thecompaction the denseness with which the soil is packed and the confining pressure that is appliedto the soil or the confining pressure under which the soil is at any given depth. hese are all

    parameters which are particularly important from point of view of shear strength of cohesion lesssoils. n the other hand for cohesive soils what are more important oviously ecause of theirlow permeaility are drainage conditions' degree of saturation' rate of loading.

    (Refer Slide ime! 16!13 min%

    &n addition to this there is another important parameter which during the course of todays lecturewe will pay some attention to that is the stress history. he history which shows what kind ofstress the soil has undergone in the past' this is particularly important for clayey soils. 4et uslook at the shear strength from point of view of the type of soil. &f you have a cohesionless soiland su5ected to direct shear test' this is a kind of relationship you will get etween shear stressand the shear displacement. &f it is a dense soil it will undergo a little it of compression and thene)pansion.

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    (Refer Slide ime! 16!#" min%

    &f it is a loose sand it will under compression continuously. here will e a peak shear stress andan ultimate shear stress as the displacement increases. he same thing can e oserved in atria)ial test also' for dense sand there is a peak' for loose sand there is no marked peak. 4oosesand undergoes compaction or compression' dense sand undergoes a little it of compressioninitially ut then undergoes e)pansion susequently' it ecomes loosened.

    (Refer Slide ime! 1!17 min%

    So with respect to ehavior of a soil whether it is cohesionless or cohesive under shear we cansee a few special points which are summari8ed here.

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    (Refer Slide ime! 1!#1 min%

    ne there is a peak and an ultimate which we saw in the previous diagram. *ense sands undergoan e)pansion mostly although they undergo a little it of compression initially' loose sandsundergo compression. here is a significant volume change the amount of compression or theamount of e)pansion is quiet significant under shear.

    Suppose we prevent this volume change then what will happen is the shear stress that isresponsile for causing this volume change will not orne y the soil and the soil will tend toehave like a liquid and thats known as liquefaction. We also saw last time' having talked aout

    liquefaction what is critical void ratio and how that can e used to decide whether soil willliquefy under a given condition or not. he concept of critical void ratio can e very easilyunderstood from the statement which & made a few minutes ack that dense sand undergoe)pansion' loose sand undergo compression.

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    (Refer Slide ime! 13!$3 min%

    (Refer Slide ime! 17!#3 min%

    Suppose you have a sand which is neither loose nor dense relatively speaking then if the void

    ratio remains unchanged throughout shear then thats the soil which will undergo liquefactionecause it doesnt undergo volume change during shear and since it is unale to undergo volumechange during shear it will liquefy. So whats that void ratio at which the soil will ehave likethis that can e otained y plotting void ratio change with a)ial strain as shown here and findingout which is the void ratio where the volume change will e nullified and that is known as ,9R.*epending upon the type of tria)ial test we do' there are different values of ,9R. he mostpopularly used one is the ,asagrande ,9R which is nothing ut the void ratio efore application

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    of sigma# and constant sigma#,9R is also possile' constant volume ,9R is also possile todetermine.

    (Refer Slide ime! "1!"1 min%

    &n the second instant where the constant volume ,9R test is conducted e 1after the application ofsigma#is used as the critical void ratio. :ow in todays lecture we will see an assortment oftopics which & am calling as special topics' all of which are very important. Some of them areslightly advanced and & will try to give an e)posure to these topics rather than go into greatdetails. +or e)ample let me start with a list of these so called special topics.

    (Refer Slide ime! "1!#7 min%

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    ;eak and residual shear strengths is something which we need to understand very well which isvery important from the point of view of particularly slope staility or staility of any structurealso. hen we need to understand something known as pore pressure coefficients then what arestress paths< & had mentioned that shear strength of a clayey soil particularly depends upon thestress history. So one needs to know what is the path through which this stresses have changed.

    hen we need to know something aout failure' we have only een seeing so far the =ohrcoulom theory ut there are other theories as well ecause sometime the soils do not ehavestrictly according to the =ohr coulom theory although it is universally valid and universallyused also. /ut there are theories which are particularly applicale to clays especially in theplastic deformation stage and those are known etter as yield theories. We will have to see thosealso. And then finally we will see a very important test called cyclic tria)ial test. hese are dayswere awareness aout earthquakes and their damage has increased like anything ecause of pastrecent e)periences.

    here is one test known as the cyclic tria)ial test which is useful in determining the liquefactionstrength of a soil. &t helps us to predict under what conditions a soil will liquefy and whether the

    present condition is indicative of a potential to liquefy or not. So site characteri8ation' siteassessment from point of view of liquefaction can e carried out using this cyclic tria)ial test andit indirectly ovious the need for using the concept of ,9R. he concept of ,9R is nowgradually eing replaced y another concept ased on the cyclic tria)ial test. 4et us start withpeak and the residual shear strength and their utility.

    (Refer Slide ime! "-!>3 min%

    &f you take a normally consolidated soil' you will have a shear strength verses displacementcurve which will e something this which shows a peak. hen it comes down and shows aresidual value' with large displacements the value goes eyond and then comes down andremains almost constant. So is the case with over consolidated clays also. here is a peak andthen a residual value ut what is ovious from these two figures is that when the soil is over

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    consolidated' the fall to the residual is much more. After reaching the peak the drop in shearstrength is much more with displacement over consolidated clays.(Refer Slide ime! "#!$# min%

    :ow if were to plot the =ohr envelops for the normally consolidated clay and the overconsolidated clay for peak shear strength as well as for residual shear strength. hen for thenormally consolidated clay under drained conditions' the peak shear strength envelop will edepicted y let us say this yellow line. hen the residual friction angle will e at least slightlylesser and that will fall somewhere elow like this. Whereas if we have an over consolidatedclay' the peak will e much more than in the normally consolidated soil and let leave and show a

    small amount of cohesion intercept.

    As & mentioned' the drop from peak to the residual in the case of overconsolidated soil is muchmore than in the case of normally consolidated soil and therefore the overconsolidated residualstrength envelop will invarialy e elow the normally consolidated residual strength envelop.he sequence in which the envelop keeps coming down is practically constant and valid for allclays. he over consolidated peak strength is very high' overconsolidated residual strength isvery low' the normally consolidated strength lye elow etween this two line.

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    (Refer Slide ime! ">!"3 min%

    So some of the points that arise from this discussion are? there is a more pronounced reduction inoverconsolidated clays from peak to residual and another point that is worth mentioning is higherthe clay content' greater is the reduction from peak to residual. he dropping strength withdisplacement is more if the clay content is more in the soil. he residual strength indirectincidentally is independent of stress history' it is only proportional to the effective stress thatsapplied and you can see that in fact that the residual strength remains almost practically constantwith displacement and irrespective of what stress history the soil has e)perienced and it willeventually have a certain residual strength eyond a certain level of displacement.

    @enerally the residual strength envelop is lower than the peak strength envelop foroverconsolidated and therefore also the normally consolidated clays. :ow we will pass on to thene)t special topic' the pore pressure coefficient. ne of the important things in a tria)ial test ismeasurement of pore pressure during the application of the confining pressure as well as thedeviatoric stress. &n fact we have seen in such great length in our previous lectures' how the porepressure varies during the conduct of the test for consolidated drained' consolidated undrainedand unconsolidated undrained tests. We had mapped very systematically as how the confiningpressure is applied and the deviatoric stress is applied' how the pore pressure changes andtherefore how the total stress and the effective stresses change and what is important therefore isto know how the pore pressure changes during the course of a test as the stresses are applied.

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    (Refer Slide ime! "6!-" min%

    ne other way is to say is? is it possile for us to know what the pore pressure change is likely toe not necessarily only y conducting a test ut y using a simple relationship' fundamentalrelationship. Skemptonin "7>$ derived a very fundamental relationship from asic considerationfor evaluating the change in pore pressure during a tria)ial test due to the application of theconfining stress sigma#and then the deviatoric stress. he relationship is e)tremely simple. *elta0 the change in pore pressure is simply a function of delta sigma"and delta sigma#which areapplied. So here it is' there is a tria)ial specimen we first apply sigma #and then we apply thedeviatoric stress. he change in pore pressure' net change that will e produced will oviously eproportional first to the incremental stress in the lateral direction that is delta sigma # and an

    increment component of increment due to the deviatoric stress' this ovious relationship etweenthe change in pore pressure and the applied stresses.

    :ow the constant / and A are what are known as skempton pore pressure parameters and thoseare the parameters which are very important to know and if you know those parameters for agiven clay' then it is possile to make a reasonaly good guess aout the change in pore pressuredue to change in stresses. his is a very important convenience when it comes to solving oraddressing practical prolems were given a stress change' you will e ale to predict what wille the change in pore pressure that is likely to occur. 4et us take a look at what these parametersare or what are the likely values. viously parameter /' since it depends upon on delta sigma#itis equal to one for a fully saturated soil.

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    (Refer Slide ime! "!>6 min%

    &n a fully saturated soil the moment you apply delta sigma#an equal amount of pore pressure wille generated and therefore / has to e equal to one for a fully saturated soil ideally. hen ingeneral a saturation improves' / will go on increasing until under full saturation it ecomes one'under partly saturated conditions / will e less than one. As per the parameter A' the parameterwhich is of interest is really a value at failure ecause on the application of the deviatoric stressthis sigma" sigma# and ringing the sample to failure' whats the ma)imum amount of porepressure that would have got generated and whats the corresponding coefficient A fat failure thatis what of interest.

    (Refer Slide ime! "7!"" min%

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    (Refer Slide ime! -1!$# min%

    +or normally consolidated clays this Afis appro)imately equal to one that is whatever deviatoricstress you apply also gets converted into the pore pressure. /ut if it is an over consolidated soilAfis a function of ,R and for heavily overconsolidated clays it could even e negative. ypicalvalue is 1.- or # and so on. 4et us take a simple e)ample of the use of the pore pressureparameter' let me read out the e)ample. A tria)ial test specimen from a saturated clay sample issu5ected to a cell pressure of 61 kilo newton per meter square. hat means a clay sample istaken and a specimen is e)tracted from that and it is su5ected to a cell pressure of 61 kilo:ewton per meter square in a tria)ial e)periment.

    (Refer Slide ime! -"!$6 min%

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    &f & apply this aove equation & know c dash is 8ero' phi dash is -1' sigma #dash we have 5ustseen is 1 minus final pore pressure. So if & sustitute that here' & will have c eing 8ero' c cot phiecomes 8ero and sigma#is nothing ut in terms of effective stress 1 0f and the denominatorwill ecome C0fplus the deviatoric stress *fand this must e equal to " sine phi y "sine phiand phi is -1 degrees which gives us that the right hand side is equal to 1.$7. :ow from this wefind that *fis $>. kilo newton per meter square and the shear strength undrained shear strengthwhich we were interested in is --.7 kilo newton per meter square.

    (Refer Slide ime! -$!#$ min%

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    So this illustrated the use of the pore pressure parameter in determining the final pore pressureand hence the strength of the soil. :ow we pass on to another important topic known as stresspulse.

    (Refer Slide ime! -$!>1 min%

    What is a stress path< 4et us take a look at the asic definition. &t is the locus of all points that isits a line 5oining a certain points which depict a specific stress state as loading graduallyprogresses through various stages and causes finally failure. Suppose we are testing a sample insay tria)ial test' it is su5ected to loading at different stages' different amounts of loading atdifferent stages. he stage through which the loading goes and leads to failure is what is known

    as the stress path. 4et us take an e)ample.

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    (Refer Slide ime! ->!$# min%

    2ere is an e)ample of a stress path which is showing how the ma)imum shear stress is varyingas loading changes in typical tria)ial test. 2ere is the sigma tow plot and here is thecorresponding p q plot. +rom the sigma plot we know very well that if # tests are conducted #tria)ial test' # successive =ohrs circles can e otained and if & take the stress statecorresponding to the ma)imum shear stress that will oviously e the point at the top of thesemicircle here.

    hen as the loading has changed during the test with every =ohrs circle' the position ofma)imum shears stress point has changed from here to here to here and in a p q graph we can see

    the same path through which ma)imum shear stress has changed with loading. &t is this what isknown as the stress path' this stress path is for ma)imum shear stress and how it varies withloading. :ow here no failure is involved at the moment.

    (Refer Slide ime! -6!$3 min%

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    Suppose & continue the test and a few more tests are conducted' few more specimens are testedand ultimately we reach a set of sigma#' sigma"comination which makes the soil fail that meansthe last =ohr circle shown here is touching the failure envelop. Same thing is true here in the p qgraph. :ow if & take the points corresponding to ma)imum shear stress from each one these=ohrs circles and plot' you will find the progression of the ma)imum shear during the test from

    a low value here' gradually to a point which sits on the envelop indicating that is the point offailure.

    So this is an e)ample of stress path for ma)imum shear stress upto failure. &f one knows thestress path' one knows what kind of stress history the soil has gone through and what has led tofailure and thats very important ecause that throws insight into the ehavior of a soil su5ectedto loading and how it gradually tends to move towards a state of failure.

    (Refer Slide ime! -3!"" min%

    2ere is another e)ample of stress paths. We do a lot of tests continuously one after the other on asame specimen ut with constant sigma#and only sigma"goes on changing and again let us takethe ma)imum shear stress point those will e at constant sigma#A / , * D.

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    (Refer Slide ime! -3!#6 min%

    &f you now plot them on the p q graph' you get a straight line at $> degrees and this is nothingut a ma)imum shear stress path corresponding to sigma#constant state.

    (Refer Slide ime! -3!>$ min%

    :ow here is another diagram' a p q plot which shows different stress path other different formsof stress paths under different other conditions. We 5ust saw that at sigma #equal to constant wehave a stress path which is at $> degrees' if on the other hand sigma "is kept constant' the stresspath would have een as shown here on the left of this diagram. &f on the other hand we had keptsigma"Bsigma#y - as constant and conducted the test and plotted the points of ma)imum shear

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    stress then we would have found that those would have led to a stress path which is goingstraight upward.

    (Refer Slide ime! -7!#- min%

    4et me now give you a very interesting illustration of the application of this stress path and howit has led to a etter understanding of the effect of rainfall on slope and susequent failure andlandslides. So an interesting illustration of significant stress path can e quoted as the effect ofrainfall. When rain occurs on a slope' infiltration is caused. he rain water gradually flows intothe earth. Soil is initially partly saturated and therefore as infiltration takes place' gradually thesuction pressure which is there in the partly saturated soils is neutrali8ed. ;ositive pore pressure

    gradually start developing and ecause of this the initial tension which was inding the particlesin the partly saturated conditions' capillary tension which was holding the particles togetherduring the partly saturated state now gradually gets nullified and the effective stress thereforegets gradually reduced.

    Eou know what will happen if effective stress goes getting reduced' the shear resistance goesdecreasing and therefore ultimately a stage is reached where the effective stress is reduced tosuch an e)tent that the slope fails finally. /ut during all this process we must rememer that thetotal stress in the system remain unchanged' neither an e)ternal stress is applied nor an internalstress is distured' 5ust the pore pressure goes on changing with total pressure remaining thesame. his is a very interesting concept that has een developed y /rand to e)plain the

    phenomenon of rainfall induced landslides. So this implies that total stress remaining same' theslope is gradually coming to failure ecause of increase in pore pressure that means the stresspath has to e hori8ontal that we will see in the ne)t slide.

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    (Refer Slide ime! #"!$- min%

    he conventional laoratory test unfortunately does not help you to simulate this hori8ontalstress path and therefore we will have to device a special testing technique' cyclic tria)ial testcould proaly e an alternative. &n the ne)t slide we will see how the stress really varies' whatkind of stress path we get in our conventional test and what should e the real stress path thatshould e simulated.

    (Refer Slide ime! #-!13 min%

    2ere we have the stress path which actually e)ists during rainfall. his arrow going from right toleft shows how the initial values of p dash and q dash gradually change to the final failure state pdash q dash with total stress remaining same. So this is the true stress path in the field. he soil

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    undergoes change in stress according to this path. :ow suppose this is the failure envelop' thesoil reaches failure or the slope in question reaches failure through this hori8ontal stress pathwith total stress remaining same and it touches failure envelop finally.

    &f on the other hand you do a consolidated undrained test or a consolidated drained test ecause

    the slope has een an e)istence for sometime consolidation has taken place and now suddenlyrainfalls occur and the slope fails. So consolidated drained or consolidated undrained are theappropriate tests depending upon the rate of infiltration of water or the rate of rainfall. 0suallythe rain storms are so heavy' consolidated undrained conditions are more preferred as therepresentative ones rather than the consolidated drains.

    So if you see here the stress paths for consolidated drained test is given y this green line andthat goes and meet the failure envelop like this. his is the path that will e followed whereas ifit is consolidated undrained this is the path that will e followed to failure and none of thesepaths unfortunately depict what is really taking place in the field. herefore we need to know andhave a method y which we can simulate this' if you really want to determine the correct shear

    strength parameters for a slope which has failed due to rainfall.

    (Refer Slide ime! #$!1# min%

    :ow we pass on to an important topic from point of view of failure of soils since we are nowtaking aout failure' octahedral stress failure theories. What are octahedral stresses and what are

    failure theories associated with that

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    (Refer Slide ime! #$!-"min%

    (Refer Slide ime! #$!-- min%

    2ere is a very simple illustration of three dimensional stress state sigma"' sigma-' sigma#. his is

    a stress cue which shows what are the stresses on different planes. We are in interested in oneparticular plane depicted here as A/,. What is special aout this plane is that this is plane onwhich sigma"' sigma-and sigma#are equal. &f you take perpendicular to this plane thats a lineshowing hydrostatic stress state' the perpendicular to this plane will have sigma"equal to sigma-equal to sigma#along it. Such a plane is a very special plane known as the octahedral plane andthe octahedral normal shear stress on this particular plane are given y these two equations. Whatis special aout this again is this is total stress in terms of octahedral stresses and if you consider

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    pore pressure then pore pressure will get sutracted from each one of these stresses and thereforeeffective sigma octahedral would simply e this minus 0.

    Whereas effective shear stress octahedral will not change' it will e the same as the totaloctahedral shear stress. he octahedral stress state is a very important stress state to know

    ecause this plane A/, is often very close to the actual failure oserved in the soil underdifferent stress state and many of the failure theories make use of the octahedral stress which isknown and which can e determined and from that they derive their yield theories and predictwhether or not a soil will yield.

    We take a look now at one important failure theory other than the =ohr coulom theory. his isknown as the *rucker ;rager theory and it is quiet applicale to soils. We know that in the=ohrs strength theory' the failure envelop is a straight line in two dimensional stress state. &nthree dimensional stress state its not a straight line. &f you look at this figure for e)ample this isa three dimensional stress state figure sigma"sigma-sigma#are shown a)is.

    (Refer Slide ime! #6!> min%

    Suppose you plot all those points which corresponds to failure as predicted y the =ohr coulomtheory then what you will get e in three dimensions' a he)agon like this. A he)agonal hollowsurface thats what you will get' in two dimensions it looks as a straight line. What *rucker;rager theory says is that =ohr coulom theory does not predict failure appropriately in certain

    conditions and they have modified the =ohr coulom theory particularly to account forehaviors such as stain hardening with displacement and they have given a slightly modifiedequation for predicting the strength or yield.

    2ere is one equation and here is another? they are actually similar they are merely two differentforms of writing the same failure condition as per the *rucker ;rager theory. /oth =ohrcoulom surface and *rucker ;rager surface are shown in this figure. he he)agonal surfacecorresponds to =ohr coulom yield theory whereas the conical surface refers to *rucker ;rager

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    yield theory. &n cross section it will look like this' the he)agon and the *rucker ;rager circle.:ow do these theories match at all' how deviant from each other. he *rucker ;rager theory cane fitted to the he)agon of the =ohr coulom theory either y fitting a circle whichcircumscries the he)agon or y inscriing a circle within the he)agon.

    Accordingly we will have some equivalence etween =ohr coulom theory and the *rucker;rager theory' the equivalence is of course will vary depending upon whether we are using outercircle or the inner circle. &n this two equations which are given here' k is a material parameter'alpha is a material parameter so is eta and F -' = s' these are matrices involving the mean stressor the stress invariance.

    (Refer Slide ime! #7!$1 min%

    We shall not go into very great detail aout what is stress invariant' what are mean stresses'suffice it to know that there is a theory as the *rucker ;rager theory and it is quiet widelyapplied in order to predict yield in soils particularly clayey soils. :ow if & use the circumscriedcircle over the outer ascises (Refer Slide ime! $1!-# min% of the =ohr coulom he)agon.

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    (Refer Slide ime! $1!13 min%

    hen the values of alpha and k that will need to e determined and used in the *rucker ;ragertheory will e given y this formula. viously if you know the friction angle for the soil' youcan determine the parameters that will go in to the *rucker ;rager theory' alpha and k. heparameters that will go into the *rucker ;rager theory depend upon the conditions under whichthe soil is su5ected to various different types of stresses. Suppose if the soil is su5ected to threedimensional stress state' alpha will e given y - sin phi y root of # into # minus sin phi and ky another similar equation.

    &f a plane strain condition prevails then this tan phi y 7 B"- tan square phi to the power of half

    is the corresponding applicale value of alpha and if on the other hand we use the inscriedcircle which is within the he)agon then alpha and k will ecome values shown in the last twocolumns of the tale here. *epending upon whether we are analy8ing a plane strain prolem orwhether we want an upper ound or a lower ound' we can choose an appropriate values of alphaand k in a typical analysis of say a tunnel or a slope using a finite element technique for e)amplethere are other model as well although it is eyond the scope of this lecture to go into greatdetails aout these other models.

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    (Refer Slide ime! $"!$7 min%

    ne model is of special interest and it is worth making a mention aout it although not much indetail. his is known as the critical state model. here are two different models proposed underthis' one is known the cam clay model another is known as modified cam clay model. hesewere proposed yAndrew Schofieldin ,amridge.

    he asic considerations in development of these models were what is the strength of the soil'the change in volume that takes place during shear then what are those critical statescorresponding to large deformation without stress change or volume change. &f you think ofliquefaction it is very easy to comprehend the meaning of the statement' liquefaction is nothing

    ut a case of large deformation at constant stress or constant volume. herefore the cam claymodel attempts to e)plain these large deformations which may takes under constant stress orconstant volume conditions. he state of the soil is characteri8ed y three parameters.

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    (Refer Slide ime! $#!1" min%

    As in the *rucker ;rager theory we had - parameters' here we have # parameters. he =ohrcoulom theory has again only - parameters c and phi. he cam clay model on other hand uses pdash and of course q dash ut it also uses the specific volume v.

    (Refer Slide ime! $#!-> min%

    2ere are the yield curves which will result if we use the cam clay model. 2ere is the yieldsurface on the p q plot for cam clay' this is for the modified cam clay and this is the line whichtells us which are those critical stress state at which a soil will either according to cam claymodel or according to the modified clay model corresponds to a situation where there is yield orthere is a large deformation.

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    So this point or this point will represent' the point either this or this will represent a situation oflarge unlimited deformation as in the case of liquefaction. his can e predicted reasonaly wellusing the cam clay or the modified cam clay model.

    (Refer Slide ime! $$!"3 min%

    :ow lastly we come to the test called cyclic tria)ial test aout which & mentioned sometime ackwhich is a very useful tria)ial test and is a very innovative relatively recent development andcompletely digital usually and is of interest to know.

    (Refer Slide ime! $$!# min%

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    2ere is a simple illustration of a cyclic tria)ial equipment. his has the facility to apply cyclicloading and thats why it is called cyclic tria)ial test. &t is a conventional tria)ial test e)cept thatthe load that can e applied can e cycled. Suppose you take earthquake as an e)ample then theload due an earthquake can e considered to e somewhat cyclic in nature. &t is repetitive andtherefore if we want to really understand the effects of the earthquake on shear strength of soil

    we must conduct the tria)ial test under the conditions of cycled loading' cyclic loading must epossile. So this equipment permits precisely that' you can apply load in terms of forces andhave control over the force applied' you can apply the load indirectly y inducing a strain. Adirect shear e)ample is a test where we induce strain and produce the shear forces appropriately.So this could e strain controlled' it could e displacement control also. A different type ofloadings is permitted ecause this is usually digitally computer control.

    he different types of loading that can e applied? it can also e such as to simulate a givenearthquake. A given earthquake loading if you simulate and conduct the test under thoseconditions' the parameters which you get from this tria)ial test will e the est that one can get inorder to truly analy8e the ehavior of the soil under earthquake conditions. &t permits' the

    equipment has the facility to saturate the sample to consolidate the sample and to apply cyclicloading and then finally to shear the sample. hats what the cyclic tria)ial testing system iscapale of doing. :ow this can e used to understand and predict liquefaction in a veryingenious way. his is a typical scenario that was found during the /hu5 earthquake on -6Fanuary -11".

    (Refer Slide ime! $6!>" min%

    his shows large scale flow of soil and this is e)actly what is known as liquefaction. his is theearthquake spectrum which produced this kind of liquefaction. Suppose you know the spectrum'suppose you want to identify the potential for liquefaction in a given soil then this cyclic tria)ialtest is a wonderful test for this purpose.

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    (Refer Slide ime! $!1" min%

    2ere is a typical plot that you might otain from typical cyclic tria)ial test. Eou apply a numerof cycles and find out when the liquefactions starts then plot the numer of cycles which causesinitially liquefaction against a ratio known as the cyclic stress ratio sigma dy - sigma1dash.4ets see in the ne)t slide what this represents.

    (Refer Slide ime! $!-1 min%

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    (Refer Slide ime! $!$ min%

    &nitial liquefaction is the state in which pore pressure uilds up to a value equal to the initialconfining pressure that means it completely neutrali8es the initial confining pressure andtherefore the soil starts liquefying or where strains reach very high values for e)ample >G. :owif you do a cyclic tria)ial test and identify the numer of cycles at which one of these conditionsis satisfied and liquefaction takes place and then plot the cyclic stress ratio against it' you get arelationship from which you will know whether or not liquefaction can take place. he cyclicstress ratio for simple shear stress is given y this parameter.

    (Refer Slide ime! $3!"3 min%

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    (Refer Slide ime! $3!$# min%

    +or tria)ial shear stress it is given y sigma dy - sigma1dash where each one of these symolrepresent different shear and vertical stresses under different amplitude conditions as listed here.Suppose you want to relate the simple shear tests results with tria)ial test results thats possiley relating the cyclic stress ratio through a parameter alphacand incidentally this alphacdependsupon octahedral dynamic shear stress divided y effective static octahedral normal stress. So thisshows how in yield theories' the octahedral shear stress plays an important role.(Refer Slide ime! $7!"1 min%

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    :ow we come to the end of the chapter of shear strength of soils and here are a few questionswhich you are well advised to try to answer. Suppose you go ack through this lectures then youwill find an answer to a question' is undrained test valid for sand and if so when. What isliquefaction and when does it occur and why< A slope has een an e)istence let us say for quietsometime. :ow what kind of test will you do in order to get the strength parameters which are

    required to analy8e its present state of health< ry to answer these questions. he answers haveeen given y me during the course of my lecture. hey are availale in the slides which hadeen displayed some time ack. ry to also solve simple prolem for e)ample here.

    (Refer Slide ime! >1!1 min%

    he unconfined compressive strength of a saturated clay is given and when it is su5ected to atria)ial shear test' the pore pressure coefficient Af is found to e 1.-' c dash and phi dashdetermined from the test are > and -1. :ow try to find out the undrained shear strength. :owincidentally' in this particular case although the values of c dash phi dash are given' A fis given.&ts very ovious right from the statement of the question that its the compressive strength "-1divided y - which give you the shear strength. herefore the undrained shear stress is readilyavailale here as half of the compressive strength 61 kilo newton per meter square. Eou canunder this etter if you go ack to my illustration and e)planation of the unconsolidatedundrained test.

    :ow the other data that is given in this particular prolem is useful for determining the initial

    pore pressure and you can determine this in a manner similar to the recent prolem which wesaw in this lecture that is prolem numer "' using the similar e)pression you can find out thatinitial pore pressure is .>> kilo newton per meter square in this particular case.

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    (Refer Slide ime! >"!-7 min%

    :ow suppose if you have - different vanes and you test the same soil for strength with these -different vanes' how do we get the shear strength of the soil in the hori8ontal and the verticaldirection planes< his can e done y going ack to fundamental equations of torque generatedin a vane shear test.

    (Refer Slide ime! >"!>- min%

    /y using the individual equations for torque for each vane and the ratio of the torques and yknowing the geometry of the vanes' one can easily determine Suhand Suvwhich are the requiredshear strengths in the hori8ontal and the vertical directions.

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    (Refer Slide ime! >-!17 min%

    We have used a numer of terminology' you are advised to go through complete list ofterminology and understand their real meaning and significance in practice. he terminology arestrain and stress controlled test' constant sigma#test' ,*' ,0 and 00 test' critical void ratio andliquefaction' peak and residual strengths' stress paths' cyclic stress ratio' octahedral stresses andyield theories.(Refer Slide ime! >-!# min%

    2ere is a list of ooks which will e useful to you for going through and understanding in greaterdetail whatever & have attempted to cover in these lectures.

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    (Refer Slide ime! >-!$6 min%

    2ere are a few more references' with this we come to the end.hank you.