Students we go forward now to the sixth lecture on the topic of shear strength of soils. As

per my habit I will first try to recollect what we have discussed in the previous lectures,

especially the last fifth lecture before we go on to the topic of todays lecture. So

to start with, in the last lecture I had mentioned that we have divided this topic into 3 aspects.

The aspect being the type of equipment used and the kind of test that are done to determine

the shear strength of soils. Out of those we had discussed in detail the direct shear

test, triaxial shear test, unconfined compressive strength test and in the just completed last

lecture we saw in detail the vane shear test and then an example on how to determine the

vane shear strength from the results of a typical vane shear test.

Then we went on to the second topic or second subtopic within this main topic of shear strength

that is drained and undrained test. That is the importance of drainage and how the tests

are conducted either with or without drainage and what is their significances. And lets

recapitulate as usual very briefly what we did in this 2 sub topics.

First the vane shear test, as I mentioned last time a vane is a device which has 2 vanes

like this, two blades or vanes like this which are attach to a steel rod which can be rotated.

There is a handle that is a torque can be applied. It has got dimensions of H and d

as shown. The values of d and H usually vary in this range in the field device whereas

in the miniature lab device the values are much smaller, d is only 12.5 mm and H the

height is 25 mm.

This device is immersed or pushed into the soil gently and a torque is applied and the

device is rotated. When the device is rotated it produces a cylindrical failure surface

in the soil. Based on the torque required to produce this failure surface we are in

a position to calculate from the geometry of the failure surface, the unit shear resistance

offered by the soil which is nothing but the shear strength of the soil.

So we have a formula which computes the shear resistance along the cylindrical surface of

the soil. The surface which has been cut by the vane and the resistance experienced by

the vane at the top and the bottom surfaces. These two together contribute to the total

shear strength which must be overcome by the torque so that failure can be produced inside

the soil. So what is the moment of these forces and the sum of these moments and calculate

that and put it equal to the torque, that will give you a relationship from which the

unit shear resistance towf at failure, shear strength at failure can be calculated and

thats nothing but the shear strength. We had seen that there are some corrections that

need to applied depending upon what kind of distribution of shear resistance we assume

at the top and the bottom and also depending on whether the vane is fully immersed in the

soil or not. We saw a simple example where the undrained cohesion of the soil was computed

from the torque and the geometrical dimensions which were available to us in this problem.

Now let us go to drained and undrained test. Let us take the basic scheme which we used

last time to understand the importance of drainage. Here is a typical triaxial test

specimen, it is subjected to two stages of loading as we know in the conventional triaxial

test, I must mention here at this point of time, there are several different types of

triaxial test.

The one which we normally use is the one which is called the CTC or the conventional triaxial

compression test. The CTC test for example involves applying a uniform lateral pressure

sigma3 in the water that surrounds the specimen in the test cylinder. And then this pressure

is in the second stage of loading increased to a value sigma1 by adding what is known

as the deviatoric stress sigma1 - sigma3. Thus the second stage of loading leads to

the specimen experiencing a lateral pressure sigma3 in the horizontal direction and a net

axial pressure or a total axial pressure sigma1 in the vertical direction and while applying

these stresses we can either allow drainage or prevent. If drainage is allowed there will

be a volume change and consolidation will take place. That is we are assuming that the

soil is saturated.

If on the other hand consolidation is not permitted that is drainage is not allowed

then the soil remains as unconsolidated in the first stage of loading. And then when

we go on to the second stage if we allow volume change through drainage then thats known

as drained test whereas if you do not allow any volume change, if you prevent the drainage

then it is called the undrained test.

We have three combinations, well known combinations of test. The consolidated drained, consolidated

undrained, unconsolidated undrained. The first is commonly called the CD test, the second

one the CU test, the third the UU test or sometimes these are also known as the S slow

test and the Q the quick time test.

Let us take a look at the mechanics of the consolidated drained test quickly. Here is

this table which we discussed last time which shows the pressure variation with the loading.

Here for example when the confining pressure is applied the total stress is equal to the

confining pressure in both direction because this specimen is surrounded by water and water

transmits pressure equally in all directions. Water does not take any pressure because drainage

is permitted, water is allowed to be expelled, pore pressure or neutral stress is zero.

Effective stress is simply equal to the applied stress sigma3 in both directions. So let us

take a look at the mechanics of the first type test that is the consolidated drained

or CD test. Of course we had seen it last time also but let us quickly recapitulate.

There are two stages of loading as we had seen last time and the stresses vary through

out the test as shown in this table. First stage of loading involves application of the

confining pressure sigma3, when the confining pressure is applied the total stress becomes

equal to the confining pressure both in the horizontal and the vertical direction because

water transmits pressure equally in both directions.

On the other hand since drainage is permitted water itself does not take any pressure and

the neutral stress u remains zero both in the horizontal and the vertical direction

and the effective stress which is the difference of these two becomes equal to simply the applied

sigma3. On the other hand when the axial load is applied in the next stage, the applied

pressure gets added to the total stress in the vertical direction because no pressure

is applied in the horizontal direction. The neutral stress still remains zero because

during shear drainage is not prevented. The effective stress therefore becomes the difference

of these two equal to sigma1 dash equal to sigma3 plus delta sigma1, sigma3 dash equal

to sigma3.

Thus whatever pressure we apply initially in the first of stage loading in the lateral

direction that remains effective even at the end of the test whereas the vertical effective

stress becomes equal to the applied lateral pressure plus additional axial load delta

sigma1 which we apply. In other words all the load that is applied becomes effective

in this test because its a test in which consolidation is permitted and drainage is

permitted or volume change is permitted during shear.

The Mohr diagram for this looks somewhat like this. This is the typical Mohr envelop which

you see here being a drained test for reasons which I explained last time, the envelop will

pass through the origin if the soil is normally consolidated. Because the very concept of

normally consolidated soil implies that in its initial formation stage when there is

no pressure sigma, it will have no strength, it will be a slurry or a fluid and it will

have no shear strength but then as the stress sigma increases the slurry gets more and more

consolidated leading to ultimately over a long period of time, a soil which is consolidated

fully under the pressure that it has experienced.

If on the other hand the soil is overconsolidated that is it had experienced a pressure in the

past which is larger than the pressure that is now applied then the envelop will not be

passing through the origin because there will be a cohesion in such a soil at pressures

which are less than the past overburden pressure but beyond that this soil is as good as normally

consolidated soil for pressures above the old preconsolidation pressure and therefore

it joins the envelop. And a drained test gives you c and phiD, D standing for drain and these

are effective stresses and therefore we can call them as c dash and phi dash.

The next type of test is the cu test, here now if we want to study the mechanism first

when confining pressure delta sigma3 is applied, the total stress becomes sigma1 equal to sigma3,

sigma3 equal to sigma3 just as in the previous test. Because here again consolidation is

permitted in the first stage of loading. Pore pressure remains zero, effective stress becomes

the applied stress, but the difference comes now in the second stage compared to the previous

test. When the axial load is applied the total stress is of course becomes sigma3 plus delta

sigma1 and sigma3 in the lateral direction but since no drainage is permitted neutral

stresses are non-zero. They are equal to the extra axial stress that we are applying that

is delta sigma1 in both directions because this pressure is taken up by water.

And now the effective pressure is less than the effective pressure that existed before

the application of this undrained load because during undrained load pore pressures are developed

and therefore whatever preexisting effective stress was there at the end of the first stage

loading, gets reduced due to this additional pore pressure. So sigma1 dash become sigma3,

sigma3 dash becomes sigma3 minus delta sigma1.

The corresponding Mohrs diagram if we take a look, the continuous line thick circle represents

the total stress circle that is the total stress sigma3 applied and the corresponding

total stress sigma1. If we know the pore pressure which incidentally is equal to delta sigma1

that is the additional axial load applied thats which the pore pressure is known

in this particular test.

If we subtract this pore pressure then we get the effective stress dotted line diagram,

the Mohr circle represented in dotted lines which says that the effective stress at point

A which is the minor principle stress is sigma3 minus delta sigma1 that at this point E, E

is equal to sigma3. Now if I draw an envelope to this drained Mohrs circle I will get

a line an envelop passing through the origin with an angle phiD drainage, D standing for

drains and cohesion equal to zero. But if I plot a tangent to a total stress Mohrs

circle of course it is implied that we will be doing atleast three such test with a total

stress measurements and plotting a common tangent. Here for convenience I have only

shown one such Mohrs circle. If this were the common tangent to three such tests, this

would pass through the y axis at a point giving an intercept equal to ccu where cu stands

of consolidated undrained, the corresponding angle of friction would be phi cu.

Next if we take the unconsolidated undrained test UU, here since even during the application

of confining pressure, consolidation is not permitted. We are not allowing any effective

stress to develop at all. Then whats the stress thats going to be there in the soil?

It will obviously be the stress under which the soil has been formed that is the overburden

stress, preexisting stress on the soil before the test, before the specimen is extracted

and this test is conducted and therefore in this case in order to know the stress state

at the end of the test. We have got to take the initial overburden effective stress that

was present in the soil before the test was conducted. In fact the testing procedure is

such that initially we subject the soil to a pressure equal to the overburden pressure

and then we continue the test. So overburden pressure is let us say sigma0 dash then that

is the effective pressure sigma1 and sigma3, no neutral pressure initially and the effective

stress is sigma0 dash as shown here.

Now the moment we apply the confining pressure, total stresses become equal to this original

sigma0 dash plus the applied confining pressure. Pore pressure is the applied confining pressure

obviously undrained conditions are prevailing. The effective stress therefore does not change;

it is the same that was existing in the field. When you now apply the axial load delta sigma1,

it just gets added to the total stress in the vertical direction, no change in the total

stress in the horizontal direction, pore pressure also increase because no volume change or

undrained conditions are maintained. So u becomes delta sigma3 plus delta sigma1 in

both directions. The effective stress in the vertical direction remains as sigma0 dash

therefore whereas in the horizontal direction it reduces by an amount equal to delta sigma1

the pore pressure that is generated by the axial load.

If we were to plot a Mohrs diagram based on this then the Mohr diagrams will be shifted

from each other depending upon the sigma0 dash value and delta sigma3 that we use but

the diameter of the Mohr circle will remain all the time same. That is equal to the axial

load delta sigma1 that is applied. Therefore here we have two circles corresponding to

2 different initial confining pressures applied, their common tangent will make an angle phi

undrained equal to zero and giving a cohesion c undrained equal to the intercept on the

y axis and this is what an unconsolidated undrained test and its Mohr diagram look like.

Now let us go on to the subject matter of todays lecture. In todays lecture we

shall be proceeding further and seeing some more details about the drained and undrained

test and also take up a few examples. And we will also see the third aspect that I was

planning to cover in this series of lectures. That is test on cohesion less and the cohesive

soils. Now the next slide talks about the practical significance of the drained and

undrained condition that are used in the test.

The practical significance of drainage can be very nicely understood from the example

of the lifetime performance of an earth dam or a fill. So let us take the example of an

earth dam. What are the major stages in its lifetime? First is the construction, as you

know an earth dam is constructed by compaction layer by layer. And during this process of

construction water is added to the soil and compaction is carried out and in this process

some pore pressures are bound to be generated and therefore during the short term of construction

period, if the pore pressures which are generated are high then even due to the self weight

and its shear stress component in some specific particular surface of weakness could cause

a slide or a failure of the earth dam. What I mean is suppose this is the earth dam which

is being constructed layer by layer, if pore pressures are allowed to generate due to rapid

construction and non-permission of water to expel then it quiet possible that the pore

pressure so generated cause shear stresses along some potential failure plane which exceed

the shear resistance of the soil and failure may take place.

Why does the shear resistance come down is because the pore pressure produce a net effective

stress which is very small due to which the resistance that is created is also small.

This kind of failure is a short term type of failure. On the other hand if you take

a slightly longer duration of the earth dam, during its lifetime the reservoir gets filled

or emptied depending upon whether its a monsoon season or its a reason for irrigation

for cropping. So there is a seasonal variation of the water level behind the earth dam in

the reservoir and if this water level goes down rapidly for any reason then again it

is a converse of what happened in the during the construction. Again due to rapid drawdown

there could be high pore pressure generated which might take time to dissipate and if

they do not have the time to dissipate because of very low permeability of the soil then

the effective stresses could considerably reduce.

And the shear resistance of the soil may come down and there could be a plane, a potential

failure plane along which a slide might occur. On the other hand the long term behavior stability

and safety of the earth dam is perhaps the most important aspect and we have to ensure

that the dam remains stable for the estimated or the design life of its span of performance.

Lets take a look at what kind of shear strength test needs to be done, what shear

strength parameters need to be used under these constructions, drawdown and long term

conditions? Its interesting to note that each of these corresponds to a different set

of drainage conditions and therefore a different test result has to be used. If construction

is slow and pore pressures are allowed to dissipate then there is consolidation during

construction and drainage also.

Hence it is a consolidated drained situation and CD test results can be used that is shear

strength of the soil will be c dash plus sigma dash tan phi dash along any particular plane.

If now consider the case of construction being done rapidly. Then pore pressures are generated

and are not allowed to dissipate. There is therefore no consolidation and therefore no

shear drainage during shear.

Ideal test for simulating this condition would be the unconsolidated undrained test. The

results of this test are the most appropriate for use during evaluation of the safety of

the dam during rapid construction. So s is equal to cu plus sigma tan phiu where sigma

stands for the total stresses. Under the drawdown condition again pore pressure are not allowed

to be dissipated.

There is consolidation because sometime has passed since the dam has been constructed

but there is no drainage during shear. Hence the consolidated undrained test would fit

the requirement and shear strength will be ccu plus sigma tan phicu.

In this test as we had pointed out or seen a little earlier, the pore pressures are known.

They are simply equal to the additional axial load applied during the undrained stage, the

second stage of loading. Since the pore pressures are known although we get the total stresses

and shear strength parameters corresponding to the total stresses, we can always easily

the effective stresses and the effective stress parameters and use them. And in fact often

this test is preferred to the simple consolidated undrained test. On the other hand the long-term

behavior as I have mentioned already is a situation where the construction is complete.

The pore pressures have been totally dissipated, the dam is now in a condition which can be

equated to the totally drained condition that is consolidated drained and the CD test results

again can be used as we did for construction period where slow construction was permitted,

s is equal to c dash plus sigma dash tan phi dash.

Therefore to summarize during the life of an earth dam, slow construction and long term

stability would both involve determining drained shear strength parameters through the CD test.

For sudden drawdown CU test and for rapid construction and the corresponding short term

stability UU test would be appropriate. Lets see an additional comment on this aspects

involving drainage. We have mentioned sometime back that undrained strength are usually determined

only for clay soils which are low in permeability but actually in principle undrained test can

be determined for all soils but they may not apply to sands they may not apply to all the

soils though it is possible to determine that. What would apply perhaps to highly pervious

soils like sands and gravels would be only the drained strength. So effective strength

parameters can be used to check the stability at any time if the pore pressures are known.

It is not necessary that we need to use only the total stress parameters for a situation

where the UU test results are required.

If we know the pore pressures at any point of time, the pore pressures are measured in

the test then always the effective strength parameter can be calculated and they can be

applied for computing the stability of the dam at any stage not necessarily at any particular

stage. Now for sands and gravel because they are freely draining in nature, the effective

strength parameters can be used even to check the short term stability because short term

stability under rapid construction demands the UU test.

On the other hand if it is a pervious sand or gravel then effective strength parameters

can still be used because pore pressures would not be allowed to generate. A total stresses

analysis would not be therefore recommended for such soils.

For clayey soils a total stress analysis is the only correct way to access the stability.

Now there are a number of factors other than this drainage which affect the shear strength

of a soil. Primarily we can identify 6 such factors.

These six factors are listed in front of you here. The type of the particles are responsible

for the kind of shear resistance that can be mobilized between the particles when they

experience the interlocking pressure or the intergranular stresses. The density of packing

up the soil also is a parameter to reckon with because the density of packing indicates

how close the particles are and how much in contact they are. Confinement again is an

indicator of the degree of contact between the particles.

The first 3 factors therefore are best applied to coarse grain soils whereas the next three

aspects dealing with drainage, the degree of saturation or relative rate of loading

compared to the relative rate of drainage all these things are specifically applicable

to fine grain soils. Lets take a look at what the effects of these parameters are.

Effects of particle shape and gradation- significant only for granular soils. If the particles

are angular if they have sharp edges then they exhibit lot of interlocking. Therefore

the angle of friction would be high. If on the other hand they are rounded as in the

case river bound soils or flaky as in the case of clayey soils, the friction is likely

to be very low. And coming to gradation a soil which is well graded obviously will have

the highest possible frictional resistance.

Here is the table which gives the approximate value of friction for different degrees of

density for different particle shapes, rounded and uniform have angle of friction varying

from 30 to 37 as the density increases. Well graded soils can have very high friction angle

even as high as 45 when they are densely packed.

Now effect of relative density is more or less rewording of what we saw just in the

previous slide. When the relative density is zero, the soil is in its loosest stage,

e the void ratio is minimum and friction is lowest possible. When the relative density

is 100% which of course is difficult, the soil is in its densest state, e is maximum

and the friction is high.

There is approximate relationship between the angle of friction and the relative density,

as the relative density percentage increases from 0 to 100, the angle of friction also

increases as shown in this figure.

Now confinement is also again significant for granular soils, the frictional component

depends upon the normal stress sigma and the normal stress sigma depends upon the confining

pressure where sigma is the normal pressure on any potential plane of sliding. Therefore

confinement increases phi.

The effect of drainage has been seen by us rather in detail during the course of last

lecture and todays lecture and it is significant for fine grain soils depending upon whether

we have drainage and volume change permitted or not permitted, we will have effective stress

sigma dash or total stress sigma and corresponding angles of friction and cohesion.

Water content and degree of saturation will also play an important, when the water content

increases cohesions decreases. When the degree of saturation increases also, the cohesion

increases to an optimum value and then decreases.

The rate of loading is extremely important because the so called drainage, so called

drained or undrained condition is a relative term as I have stressed in the last lecture.

The loading if it is rapid compared to the rate of dissipation of pore pressure then

it is an undrained condition. So fast rate of loading, less time to failure indicates

undrained conditions and therefore it gives a relatively higher shear strength corresponding

to the total stress parameters especially for a value of cohesion.

Now let us take a few examples on different types of triaxial tests. We have already seen

a number of examples in lecture number 4. It will be recommended that you please go

through the examples that I have discussed in lecture number 4. However I will quickly

run through those examples for the sake of continuity.

There was an example in which we had the cell pressure, the deviatoric stress at failure

and the pore water pressure known and the shear strength parameters were to be determined.We

could either determine the total stresses or the effective stresses and using either

the Mohr circle or a formula we could get the cohesions and friction.

Here is the Mohr circle, by drawing the Mohr circle we can get the cohesion and friction

corresponding to this envelop and if it is drained test, c will be zero.In terms of total

stresses for this particular problem we get phi equal to 14.5 and in terms of effective

stresses phi is equal to 26.4.There was another example which was very similar to this.

And another example 8 in which we had sigma3 and the deviatoric stress sigma1-sigma3. We

computed c and phi either using the Mohrs circle plot or both by using the Mohrs

circle plot and by using the p q plot. Here is the Mohrs circle plot and from the Mohrs

circle plot and using the analytical relationship and also by using the p q plot, we got the

values of cohesion and friction.

Here the p q plots are shown and the cu and phiu values corresponding to total stress

p q plots are also displayed. And the cu and phiu values from the Mohrs circle plot

and the p q plot are shown here. Example 9 again involved knowing sigma3, sigma1-sigma3and

the pore pressure and the computation of the shear strength. Here we have the pore pressure

calculation and the effective stress calculation. Then we plot a Mohrs circle with the above

values of effective stresses.

We get the values of cu and phi u or c dash and phi dash depending upon whether we plot

total stress diagram or effective stress diagram. These are the data for the corresponding p

q plot and the p q plot also gives you for total stress a value of cU and a value of

phiU and for effective stresses also a value of c dash and phi dash.

This is the summary of the results. This is again another example which we saw last time

where sigma3 dash and the deviatoric stress effective were both known from which we calculated

the sigma1 dash and computed c and phi by an effective stress p q plot.

Now we go on to another example in which a clay specimen is tested under fully drained

condition in a triaxial test. The cell pressure used was 80 kilo newton per meters, the shear

strength parameters were known c dash and phi dash. What is the compressive strength?

The compressive strength can be calculated as equal to the distance AB the deviatoric

stress.

And so from the Mohr diagram and by using this relationship which can be obtained from

geometry, we can get the deviatoric stress Df which is nothing but the compressive strength.

This is similar to a problem which we saw in the lecture number 3 and therefore I would

recommend that you take a look at the formula and its application as it has already been

discussed in lecture number 3.

Here we have now new examples introduced in this particular lecture. Here is an example

in which sigma3 and sigma1 are given for 3 samples, c and phi are required. This is very

similar to the example number 10 which we have just gone through and therefore using

the p q plot for example we can get c and phi. Now what are these values of c and phi?

Being a drained test the values which we obtained are nothing but the effective stress parameters

but he result shows that there is a value of cohesion equal to 23.5 kilo newton per

meter square. We had seen that in a CD test there is no cohesion at sigma equal to zero,

for a normally consolidated soil. If there is cohesion then the soil is over consolidated

soil. Therefore we conclude that this particular soil in this example is an overconsolidated

soil because c dash is not equal to zero.

The next example 15 gives you the shear strength of a saturated clay sample from an unconfined

compressive strength. Since the unconfined compressive strength is very similar to the

UU test except that the sigma3 value is zero, we can use this result to calculate the axial

pressure required to cause failure in an unconfined compressive strength if we know the sigma3.

So here it is from the unconfined compressive strength test we get sigma1- sigma3 by 2 where

sigma3 is zero. Since this remains constant in the UU test, from here we can calculate

sigma1- sigma3 + sigma3 and get sigma1 the axial stress which comes to 330 kilo newton

per meter square in this particular case. This is a total pressure because its an

unconsolidated undrained test.

The next test is again an unconsolidated undrained test. Here we have the total strength measured

as 17.5 kilo pascals and the cell pressure used, if the effective strength parameters

are c dash equal to zero, phi dash equal to 26 degrees and the pore pressure equal to

43 kilo Pascals, what would be the cell pressure? We know the undrained strength that

must be equal to sigma1- sigma3 by 2 or sigma1 dash minus sigma3 dash by 2, both are equal.

We use this failure criterion which gives us a relationship between sigma1 dash and

sigma3 dash. Since we know the pore pressure value and also the undrained strength sigma1

- sigma3 by 2. We can use this relationship and calculate sigma1 dash and sigma3 dash

and once we know sigma3 dash, using the value of pore pressure 43 kilo Pascals, we can

calculate the total cell pressure which is the unknown.

Now lets take a quick look at cohesion less and cohesive soils and their behavior. We

have seen this slide before in lecture 3. Here is a typical direct shear specimen when

we conduct a test on sand, if the sand is dense we get shear stresses versus shear displacement

diagram like this and if the soil is loose sand then the shear stress verses shear displacement

diagram becomes like this.

The volume slightly decreases, there is a slight compression initially and then the

soil starts expanding if the soil is dense sand. On the other hand it continues to show

a compression if it is loose sand and which means that we will have peak value of tan

phi equal to tow upon sigma peak or a tow upon sigma ultimate corresponding to the portion

beyond the peak. And in the case of loose sand only the second aspect will apply.

Now if you take the triaxial test again dense sand and loose sand will behave in a similar

manner. If we plot sigma1 upon sigma3 and the shear displacement or the axial displacement

in this case, we will get a diagram very similar to what we got for the direct shear test and

this again shows that there is a peak value of phi and an ultimate value of phi both of

which can be computed from this simple relationships. Here again there is a compression in the early

stages, in the case of dense sand this gets converted into an expansion as the shear displacement

increases, whereas in the case of loose sand continuous compression goes on taking place.

We can summarize our observations like this. There is a peak and an ultimate value when

sand is subjected to either direct shear or triaxial shear. Dense sands undergo an expansion,

loose sand undergo compression. There is a significant volume change in both cases, prevention

of this volume change is not going to permit development of the effective stresses and

the friction angle and therefore it will lead to a kind of failure which arises because

this behaves like a liquid, shear resistance is not allowed to develop due to volume change

prevention and this phenomenon is known as liquefaction.

Now let us take a look at the concept of critical void ratio and the meaning of liquefaction,

significance of liquefaction and how to determine it. When volume change is prevented there

is a gain in strength in dense sand because as we have seen in the compression behavior

of the dense sand there is a compression initially and if volume change is prevented this compression

does not takes place and therefore there is a gain in strength.

Whereas on the front of loose sands, if we do not permit volume change we are preventing

the sand from becoming denser and gaining more strength. Therefore it actually loses

strength but this shows that there must be a void ratio at which sand neither is dense

nor is loose and therefore neither gain strength nor loses strength. Take a look at this diagram

on the above slide. This shows void ratio verses the axial strain in a typical triaxial

test. As the axial strain goes on taking place and compression goes on taking place, if we

do not permit volume change, a loose sand will go on showing decreasing void ratio whereas

dense sand will go on showing slight increase in void ratio. Ultimately if we compare these

void ratios with a void ratio corresponding to the zero line, we find that there is a

void ratio here at which the axial strain is such that no volume change takes place.

That means void ratio is a critical value at which neither gain in strength is there

nor loss in strength is there and the soil is neither dense nor loose and it is this

void ratio at which volume change does not take place and therefore soil undergoes liquefaction

during shear. This void ratio actually is defined differently by different persons;

Casagrande has proposed a critical void ratio of his own which is nothing but the void ratio

which is determined before the application of the confining pressure sigma3.

On the other hand if we do what is known as the constant sigma3 tests that is the conventional

triaxial test where sigma3 is kept constant in the first stage of loading then the e0

after the application of sigma3 if that is used the corresponding CVR is known as constant

sigma3 CVR. If we use what is known as a constant volume test that is through out the test we

adjust sigma1 in such a way that there is no volume in the triaxial sample then thats

known as the constant volume CVR.

Here is a method to determine the critical void ratio, Casagrande and constant sigma3

CVR can both be determined by conducting a typical triaxial test plotting the volume

change either contraction or expansion as against the void ratio along the y axis. Here

is the vertical line which signifies that on the left of this there is contraction,

on the right of this there is expansion depending upon void ratio. Then this line also shows

that there is no volume change either contraction or expansion, if the volume change if the

void ratio is such that it falls on this line and on the curve shown here. This curve is

the e0 versus volume change curve and this vertical line is a no volume change line and

therefore this point of intersection is nothing but the critical void ratio. This critical

void ratio is the void ratio as per Casagrande or constant sigma3test because Casagrande

test is also a constant sigma3 test.

If we do a constant volume test on the other hand then we do not end up with any contraction

or expansion here, volume remains the same and therefore what we need to do is to plot

a parameter A that is nothing but sigma3 at the end of the test against the initial void

ratio and get this curve for different values of initial void ratio. Then the critical void

ratio is that which corresponds to a value of A which is equal to (sigma3)I where (sigma3)I

is the initial confining pressure that is applied to the soil. This is a constant volume

test remember, therefore whatever (sigma3)I we apply gets adjusted by us and goes on changing

during the test. So that value of (sigma3)I which does not change which remains the same

at a particular void ratio gives us the critical void ratio. So this is how we determine the

critical void ratio.

With this we come to the end of todays lecture, we have studied shear strength during

the course of todays lecture with specific reference to the drained and undrained test

and with respect to tests on cohesionless soils and cohesive soils. But actually so

far we have covered in explicit detail only test on cohesion less soils, the reason is

whatever we have studied with respect to drained and undrained test themselves are nothing

but the explanations for test on cohesive soils. In the next lecture, we will continue

a little bit with tests on cohesionless and cohesive soils and also take up a few special

topics. Thank you.